1.1 Fields · Matrix Theory

In $ℝ$ we can add, subtract, multiply, and divide by any nonzero number. In $ℝ^{2}$ we can add two vectors and scale a vector by a real number, but there is no useful way to multiply or divide two vectors. Linear algebra keeps these roles apart. Vectors only need to be added and scaled, while the numbers that do the scaling, the scalars, need the full arithmetic of $ℝ$ . Before we define vector spaces in Section 1.2, this section makes “full arithmetic” precise. The resulting notion, a field, is all we will ever ask of our scalars, so the theory applies verbatim over $ℚ$ , $ℝ$ , $ℂ$ , and number systems as unfamiliar as $𝔽_{2}$ , where $1 + 1 = 0$ .

1.1.1The field axioms

Intuition (What we need from scalars). The basic computational problem of linear algebra is solving systems of linear equations. Take the smallest case: one equation $𝑎 𝑥 + 𝑐 = 𝑑$ in one unknown, with $𝑎 \neq 0$ . Over $ℝ$ we subtract $𝑐$ and divide by $𝑎$ without a second thought. Spelled out, subtracting $𝑐$ means adding $- 𝑐$ to both sides and simplifying the left side with

(𝑎 𝑥 + 𝑐) + (- 𝑐) = 𝑎 𝑥 + (𝑐 + (- 𝑐)) = 𝑎 𝑥 + 0 = 𝑎 𝑥,

and dividing by $𝑎$ means multiplying by $𝑎^{- 1}$ and simplifying with

𝑎^{- 1} (𝑎 𝑥) = (𝑎^{- 1} 𝑎) 𝑥 = 1 𝑥 = 𝑥 .

Gauss–Jordan elimination (Section 3.3) solves systems of any size with the same moves: it adds a multiple of one equation to another and divides by nonzero pivots. So we want two operations that can be undone, that can be regrouped and reordered freely, that have neutral elements $0$ and $1$ , and that are linked by distributivity, so that multiplying a sum means multiplying each term. The field axioms are exactly this list.

Definition 1.1.1 (Field). A field is a set $𝐹$ together with two operations, addition $+ : 𝐹 \times 𝐹 \to 𝐹$ and multiplication $\cdot : 𝐹 \times 𝐹 \to 𝐹$ , that satisfy the following field axioms.

Commutativity. For all $𝑎, 𝑏 \in 𝐹$ , $𝑎 + 𝑏 = 𝑏 + 𝑎$ and $𝑎 𝑏 = 𝑏 𝑎$ .
Associativity. For all $𝑎, 𝑏, 𝑐 \in 𝐹$ , $(𝑎 + 𝑏) +$ $𝑐 =$ $𝑎 +$ $(𝑏 + 𝑐)$ and $(𝑎 𝑏) 𝑐 = 𝑎 (𝑏 𝑐)$ .
Distributivity. For all $𝑎, 𝑏, 𝑐 \in 𝐹$ , $𝑎 (𝑏 + 𝑐) =$ $𝑎 𝑏 +$ $𝑎 𝑐$ .
Additive identity. There is an element $0 \in 𝐹$ such that $𝑎 + 0 = 𝑎$ for all $𝑎 \in 𝐹$ .
Multiplicative identity. There is an element $1 \in 𝐹$ with $1 \neq 0$ such that $𝑎 \cdot 1 = 𝑎$ for all $𝑎 \in 𝐹$ .
Additive inverses. For every $𝑎 \in 𝐹$ there is some $𝑏 \in 𝐹$ with $𝑎 + 𝑏 = 0$ .
Multiplicative inverses. For every $𝑎 \in 𝐹$ with $𝑎 \neq 0$ there is some $𝑏 \in 𝐹$ with $𝑎 𝑏 = 1$ .

The elements of $𝐹$ are called scalars. We write $𝑎 𝑏$ for $𝑎 \cdot 𝑏$ .

Remark 1.1.2. Since $+$ and $\cdot$ are functions $𝐹 \times 𝐹 \to 𝐹$ , sums and products of elements of $𝐹$ lie in $𝐹$ . When we check that a candidate set is a field, this closure has to be checked along with the axioms.

We stated the identity, inverse, and distributive axioms on one side only. Commutativity supplies the other side: $0 + 𝑎 = 𝑎$ , $1 𝑎 = 𝑎$ , $𝑏 + 𝑎 = 0$ and $𝑏 𝑎 = 1$ for the inverses $𝑏$ above, and

(𝑎 + 𝑏) 𝑐 = 𝑐 (𝑎 + 𝑏) = 𝑐 𝑎 + 𝑐 𝑏 = 𝑎 𝑐 + 𝑏 𝑐 .

We use these mirrored forms freely and cite them by the axiom they come from.

Finally, read the quantifiers carefully. The identity axioms say that some element $0$ satisfies $𝑎 + 0 = 𝑎$ for every $𝑎$ , and similarly for $1$ . The axioms after them then refer to “ $0$ ” and “ $1$ ” as if these were determined. If a field could contain two different additive identities, the condition $𝑎 \neq 0$ would be ambiguous. Theorem 1.1.9 shows that this cannot happen, and the proof of its part (a) uses only commutativity and the identity axioms, so the argument is not circular.

Example 1.1.3 (

ℚ

and

ℝ

ℝ

, with its usual addition and multiplication, is a field; we take the arithmetic of real numbers as known. So is

ℚ

: sums, products, and negatives of rational numbers are rational,

1 / 𝑞

is rational for every nonzero rational

𝑞

, and identities such as

𝑎 (𝑏 + 𝑐) =

𝑎 𝑏 +

𝑎 𝑐

hold in

ℚ

because they hold in

ℝ

. (Proposition 1.1.33 makes this last argument general.)

Example 1.1.4 (The field with two elements). Let $𝔽_{2} = {0, 1}$ , with

0 + 0 = 1 + 1 = 0, 0 + 1 = 1 + 0 = 1, 1 \cdot 1 = 1, 0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0 .

These are the parity rules for integers if we read $0$ as “even” and $1$ as “odd”: odd plus odd is even, odd times odd is odd, and so on. Both operations are commutative by inspection, $0$ and $1$ are identities with $0 \neq 1$ , the additive inverse of $0$ is $0$ and that of $1$ is $1$ (as $1 + 1 = 0$ ), and $1$ is its own multiplicative inverse. Associativity and distributivity can be checked case by case, or deduced from Lemma 1.1.21 below, since $𝔽_{2}$ is $ℤ / 2 ℤ$ .

So a field’s arithmetic can look quite unlike that of $ℝ$ . An argument about general fields that silently assumes $1 + 1 \neq 0$ is not valid in $𝔽_{2}$ .

Non-example 1.1.5 (

ℤ

). The integers satisfy every field axiom except the existence of multiplicative inverses. That one fails for

𝑎 = 2

: for every

𝑏 \in ℤ

the product

2 𝑏

is even, so

2 𝑏 \neq 1

. Correspondingly, the equation

2 𝑥 = 1

has no solution in

ℤ

Non-example 1.1.6 (Componentwise multiplication on $ℝ^{2}$ ). Give $ℝ^{2}$ the componentwise operations

(𝑎, 𝑏) + (𝑐, 𝑑) = (𝑎 + 𝑐, 𝑏 + 𝑑), (𝑎, 𝑏) (𝑐, 𝑑) = (𝑎 𝑐, 𝑏 𝑑) .

Every axiom except multiplicative inverses holds, because each holds in each coordinate separately; the identities are $(0, 0)$ and $(1, 1)$ . But $(1, 0) \neq (0, 0)$ has no inverse, since $(1, 0) (𝑐, 𝑑) =$ $(𝑐, 0) \neq$ $(1, 1)$ for every $(𝑐, 𝑑)$ . So the obvious way to multiply vectors in $ℝ^{2}$ does not give a field. Section 1.1.4 gives $ℝ^{2}$ a less obvious multiplication that does.

Intuition (What each axiom is for). Here is what goes wrong when one axiom is dropped.

Without multiplicative inverses we cannot divide by pivots. In $ℤ$ , the equation $2 𝑥 = 1$ has no solution.
Without additive inverses we cannot subtract one equation from another. In ${0, 1, 2, \dots}$ , the equation $𝑥 + 1 = 0$ has no solution.
Without associativity, multiplying $𝑎 𝑥 = 𝑏$ by $𝑎^{- 1}$ gives $𝑎^{- 1} (𝑎 𝑥) = 𝑎^{- 1} 𝑏$ , and we could not regroup the left side as $(𝑎^{- 1} 𝑎) 𝑥 = 𝑥$ .
Without commutativity of multiplication, $𝑎 𝑥 = 𝑏$ and $𝑥 𝑎 = 𝑏$ are different equations. Square matrices multiply like this (Section 2.4).
Without distributivity, nothing links $+$ to $\cdot$ , since it is the only axiom that mentions both. Even $0 𝑎 = 0$ (Proposition 1.1.15) can fail: Exercise 1.1.55 builds a structure satisfying every other axiom in which $0 \cdot 0 \neq 0$ .
Without $1 \neq 0$ , the one-element set ${0}$ with $0 + 0 = 0 \cdot 0 = 0$ would be a field. It would also be the only new example, since $1 = 0$ forces $𝐹 = {0}$ (Exercise 1.1.39).

1.1.2Consequences of the axioms

From now on, $𝐹$ denotes a field. All we know about $𝐹$ is the axioms, so facts that are second nature in $ℝ$ , such as $0 𝑎 = 0$ and $(- 𝑎) (- 𝑏) = 𝑎 𝑏$ , need proofs. In exchange, each such proof establishes the fact in every field at once.

Technique 1.1.7 (Proving from axioms). When objects are defined by axioms, as fields are here and vector spaces will be in Section 1.2, the only facts available are the axioms and what has already been proved from them. A fact that is obvious for real numbers is not available until it is proved, because a general field need not look like $ℝ$ .

The move is to write the argument as a chain of equalities that changes one thing at a time, and to justify each link by the axiom or earlier result it uses. Watch for steps that move a term past another (commutativity) or regroup (associativity): in $ℝ$ we make these without noticing.

Reach for it whenever the hypotheses are an axiom system and the claim is an identity or an elementary property. If you are stuck, ask which axioms can connect the objects in the claim. For example, distributivity is the only axiom that involves both $+$ and $\cdot$ , so any fact relating $0$ (an additive object) to multiplication must go through it. The canonical example is $0 𝑎 = 0$ (Proposition 1.1.15).

Used in 29 places and 1 practice exercise

Theorem 1.1.9, Proposition 1.1.13, Proposition 1.1.15, Proposition 1.1.16, Proposition 1.1.17, Proposition 1.1.30, Proposition 1.1.33, Exercise 1.1.38, Exercise 1.1.39, Exercise 1.1.58, Proposition 1.2.5, Proposition 1.2.10, Proposition 1.2.22, Proposition 1.2.27, Proposition 1.2.28, Proposition 1.2.30, Proposition 1.2.32, Proposition 1.2.33, Proposition 1.2.34, Proposition 1.2.35, Exercise 1.2.54, Exercise 1.2.55, Exercise 1.2.56, Exercise 1.7.14, Problem 1.2, Lemma 2.1.22, Proposition 4.6.2, Theorem 4.6.22, Proposition 6.1.4; 1 practice exercise.

Uniqueness

The identity and inverse axioms assert that certain elements exist, not that they are unique. Uniqueness is what entitles us to speak of the zero of $𝐹$ and the inverse of $𝑎$ .

Technique 1.1.8 (Uniqueness via two candidates). To show that at most one object has a property $𝑃$ , suppose $𝑥$ and $𝑥^{'}$ both have $𝑃$ and prove $𝑥 = 𝑥^{'}$ . Usually one evaluates a single well-chosen expression in two ways, once using that $𝑥$ has $𝑃$ and once using that $𝑥^{'}$ has $𝑃$ .

Reach for it whenever a claim says “the”, “unique”, or “exactly one”. Existence and uniqueness are separate claims: prove existence by producing an object, and uniqueness by this technique. The canonical example is the uniqueness of $0$ in Theorem 1.1.9, where the expression is $0 + 0^{'}$ .

Used in 67 places and 5 practice exercises

Theorem 1.1.9, Exercise 1.1.11, Corollary 1.1.14, Proposition 1.1.16, Exercise 1.1.37, Exercise 1.1.46, Exercise 1.1.47, Proposition 1.2.27, Proposition 1.2.28, Proposition 1.2.30, Proposition 1.2.34, Exercise 1.2.43, Exercise 1.2.52, Proposition 1.6.16, Exercise 1.7.6, Exercise 1.7.7, Exercise 1.7.11, Problem 1.3, Problem 1.17, Proposition 2.1.4, Theorem 2.2.1, Proposition 2.3.10, Proposition 2.3.17, Proposition 2.4.3, Theorem 2.5.10, Exercise 2.5.37, Exercise 2.5.49, Exercise 2.6.42, Exercise 2.6.43, Exercise 2.6.49, Theorem 3.3.17, Corollary 3.3.18, Proposition 3.4.27, Problem 3.2, Problem 3.17, Exercise 4.3.44, Proposition 4.4.31, Proposition 4.5.16, Proposition 4.6.7, Theorem 4.6.30, Exercise 4.6.39, Exercise 4.6.54, Theorem 4.7.22, Proposition 4.7.32, Proposition 5.2.5, Proposition 5.3.17, Theorem 6.4.6, Theorem 6.4.24, Theorem 6.4.38, Proposition 6.5.27, Theorem 6.6.3, Proposition 6.6.9, Exercise 6.6.37, Theorem 7.1.14, Corollary 7.1.17, Proposition 7.1.29, Theorem 7.1.45, Exercise 7.1.51, Exercise 7.1.58, Theorem 7.6.2, Notation 8.3.10, Theorem 8.3.23, Theorem A.1.14, Proposition A.2.4, Exercise A.2.9, Theorem A.3.10, Proposition B.1.2; 5 practice exercises.

Theorem 1.1.9 (Uniqueness of identities and inverses). Let $𝐹$ be a field.

(a) $𝐹$ has exactly one additive identity and exactly one multiplicative identity.
(b)Every $𝑎 \in 𝐹$ has exactly one additive inverse, and every nonzero $𝑎 \in 𝐹$ has exactly one multiplicative inverse.

Proof.routine

Uses. Definition 1.1.1 (Field)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. Suppose

0

and

0^{'}

are both additive identities. Find one expression that involves both.

Hint 2. Evaluate

0 + 0^{'}

twice: once using that

0^{'}

is an identity, once using that

0

is.

Hint 3. For inverses, suppose

𝑎 + 𝑏 = 0

and

𝑎 + 𝑏^{'} = 0

. Start from

𝑏 = 𝑏 + 0

, replace

0

𝑎 + 𝑏^{'}

, and regroup.

Sketch. Each identity absorbs the other:

0 = 0 + 0^{'} = 0^{'} + 0 = 0^{'}

. If

𝑏

and

𝑏^{'}

are additive inverses of

𝑎

, insert

0 = 𝑎 + 𝑏^{'}

into

𝑏 = 𝑏 + 0

and regroup to get

(𝑏 + 𝑎) +

𝑏^{'} =

0 +

𝑏^{'} =

𝑏^{'}

. The multiplicative statements are the same arguments with

\cdot

and

1

in place of

+

and

0

(a) Suppose $0$ and $0^{'}$ both satisfy the additive identity axiom. Then

\begin{aligned} 0 & = 0 + 0^{'} & (as 0^{'} is an additive identity) \\ = 0^{'} + 0 & (commutativity) \\ = 0^{'} & (as 0 is an additive identity) . \end{aligned}

The same argument, with $1$ and $1^{'}$ in place of $0$ and $0^{'}$ and multiplication in place of addition, shows that the multiplicative identity is unique. Realize that we used only commutativity and the identity axioms. So from here on “ $0$ ” and “ $1$ ” are unambiguous, and the remaining axioms say what they were meant to say.

(b) Let $𝑎 \in 𝐹$ , and suppose $𝑎 + 𝑏 = 0$ and $𝑎 + 𝑏^{'} = 0$ . Then

\begin{aligned} 𝑏 & = 𝑏 + 0 & (additive identity) \\ = 𝑏 + (𝑎 + 𝑏^{'}) & (as 𝑎 + 𝑏^{'} = 0) \\ = (𝑏 + 𝑎) + 𝑏^{'} & (associativity) \\ = 0 + 𝑏^{'} & (as 𝑏 + 𝑎 = 𝑎 + 𝑏 = 0) \\ = 𝑏^{'} & (additive identity), \end{aligned}

as desired. If $𝑎 \neq 0$ and $𝑎 𝑏 = 𝑎 𝑏^{'} = 1$ , the same chain with multiplication in place of addition and $1$ in place of $0$ gives $𝑏 = 𝑏^{'}$ . □

Notation 1.1.10. By Theorem 1.1.9 we may write $- 𝑎$ for the additive inverse of $𝑎$ , and $𝑎^{- 1}$ for the multiplicative inverse of $𝑎 \neq 0$ . Subtraction and division are defined by

𝑎 - 𝑏 = 𝑎 + (- 𝑏), \frac{𝑎}{𝑏} = 𝑎 𝑏^{- 1} (𝑏 \neq 0) .

For a positive integer $𝑛$ , $𝑎^{𝑛}$ is the product of $𝑛$ copies of $𝑎$ , and $𝑎^{0} = 1$ .

Exercise 1.1.11.routine Show that

- 0 = 0

, and that

- (𝑎 + 𝑏) =

(- 𝑎) +

(- 𝑏)

for all

𝑎, 𝑏 \in 𝐹

Solution.

Uses. Theorem 1.1.9 (Uniqueness of identities and inverses), Definition 1.1.1 (Field)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. By Theorem 1.1.9, to show that

𝑦 = - 𝑥

it suffices to show that

𝑥 + 𝑦 = 0

Hint 2. For the second claim, add

(𝑎 + 𝑏) +

((- 𝑎) + (- 𝑏))

and regroup so that

𝑎

meets

- 𝑎

and

𝑏

meets

- 𝑏

Sketch.

0 + 0 = 0

, so

0

is an additive inverse of

0

. Regrouping gives

(𝑎 + 𝑏) +

((- 𝑎) + (- 𝑏)) =

(𝑎 + (- 𝑎)) +

(𝑏 + (- 𝑏)) =

0

. Both claims follow from uniqueness of additive inverses.

Since $0 + 0 = 0$ by the additive identity axiom, $0$ is an additive inverse of $0$ , so $- 0 = 0$ by Theorem 1.1.9.

For the second claim,

\begin{aligned} (𝑎 + 𝑏) + ((- 𝑎) + (- 𝑏)) & = 𝑎 + (𝑏 + ((- 𝑎) + (- 𝑏))) & (associativity) \\ = 𝑎 + ((𝑏 + (- 𝑎)) + (- 𝑏)) & (associativity) \\ = 𝑎 + (((- 𝑎) + 𝑏) + (- 𝑏)) & (commutativity) \\ = 𝑎 + ((- 𝑎) + (𝑏 + (- 𝑏))) & (associativity) \\ = (𝑎 + (- 𝑎)) + (𝑏 + (- 𝑏)) & (associativity) \\ = 0 + 0 = 0 & (additive inverses, additive identity) . \end{aligned}

So $(- 𝑎) + (- 𝑏)$ is an additive inverse of $𝑎 + 𝑏$ , and by Theorem 1.1.9 it equals $- (𝑎 + 𝑏)$ .

Remark 1.1.12. Regrouping arguments like the solution of Exercise 1.1.11 get long, so we record a convenience. Associativity implies that all ways of parenthesizing a sum

𝑎_{1} + 𝑎_{2} + \dots + 𝑎_{𝑛}

give the same element: by strong induction on

𝑛

, each of them equals

(\dots ((𝑎_{1} + 𝑎_{2}) + 𝑎_{3}) + \dots) +

𝑎_{𝑛}

. Commutativity then implies that the order of the terms does not matter either, since any reordering is a sequence of swaps of adjacent terms. The same holds for products. From now on we drop such parentheses and reorder sums and products freely.

Arithmetic in a field

We now return to the problem that motivated the definition.

Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ ). Let $𝑎, 𝑏 \in 𝐹$ .

(a)The equation $𝑎 + 𝑥 = 𝑏$ has exactly one solution $𝑥 \in 𝐹$ , namely $𝑥 = 𝑏 - 𝑎$ .
(b)If $𝑎 \neq 0$ , the equation $𝑎 𝑥 = 𝑏$ has exactly one solution $𝑥 \in 𝐹$ , namely $𝑥 = 𝑏 / 𝑎$ .

Proof.routine

Uses. Definition 1.1.1 (Field), Notation 1.1.10

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Existence and uniqueness are separate claims. For existence, substitute the proposed solution.

Hint 2. For uniqueness, take any solution

𝑥

𝑎 + 𝑥 = 𝑏

, add

- 𝑎

to both sides on the left, and regroup to isolate

𝑥

Sketch. Substituting

𝑥 = 𝑏 - 𝑎

and regrouping gives

𝑎 +

(𝑏 - 𝑎) =

(𝑎 + (- 𝑎)) +

𝑏 =

𝑏

. Conversely, if

𝑎 + 𝑥 = 𝑏

, then

𝑥 =

((- 𝑎) + 𝑎) +

𝑥 =

(- 𝑎) +

(𝑎 + 𝑥) =

(- 𝑎) +

𝑏 =

𝑏 -

𝑎

. Part (b) is the same argument with

𝑎^{- 1}

and

1

in place of

- 𝑎

and

0

(a) First, $𝑥 = 𝑏 - 𝑎$ is a solution:

\begin{aligned} 𝑎 + (𝑏 - 𝑎) & = 𝑎 + ((- 𝑎) + 𝑏) & (definition of subtraction, commutativity) \\ = (𝑎 + (- 𝑎)) + 𝑏 & (associativity) \\ = 0 + 𝑏 = 𝑏 & (additive inverse, additive identity) . \end{aligned}

Conversely, suppose $𝑎 + 𝑥 = 𝑏$ . Then

\begin{aligned} 𝑥 & = 0 + 𝑥 & (additive identity) \\ = ((- 𝑎) + 𝑎) + 𝑥 & (additive inverse) \\ = (- 𝑎) + (𝑎 + 𝑥) & (associativity) \\ = (- 𝑎) + 𝑏 = 𝑏 - 𝑎 & (as 𝑎 + 𝑥 = 𝑏, commutativity), \end{aligned}

so $𝑏 - 𝑎$ is the only solution.

(b) Suppose $𝑎 \neq 0$ , so that $𝑎^{- 1}$ exists. Then $𝑥 = 𝑏 / 𝑎 = 𝑏 𝑎^{- 1}$ is a solution, since by commutativity, associativity, and the inverse and identity axioms,

𝑎 (𝑏 𝑎^{- 1}) = 𝑎 (𝑎^{- 1} 𝑏) = (𝑎 𝑎^{- 1}) 𝑏 = 1 𝑏 = 𝑏 .

Conversely, if $𝑎 𝑥 = 𝑏$ , the same axioms give

𝑥 = 1 𝑥 = (𝑎^{- 1} 𝑎) 𝑥 = 𝑎^{- 1} (𝑎 𝑥) = 𝑎^{- 1} 𝑏 = 𝑏 / 𝑎,

so $𝑏 / 𝑎$ is the only solution. □

Part (b) is what the section set out to guarantee: in any field, a linear equation in one unknown with a nonzero coefficient has exactly one solution. The uniqueness half is often used in the following form.

Corollary 1.1.14 (Cancellation). Let

𝑎, 𝑏, 𝑐 \in 𝐹

. If

𝑎 + 𝑏 = 𝑎 + 𝑐

, then

𝑏 = 𝑐

. If

𝑎 𝑏 = 𝑎 𝑐

and

𝑎 \neq 0

, then

𝑏 = 𝑐

Proof.routine

Uses. Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ )

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Both

𝑏

and

𝑐

solve one equation of the kind in Proposition 1.1.13. Which one?

Hint 2. They both solve

𝑎 + 𝑥 = 𝑑

, where

𝑑 = 𝑎 + 𝑏

. Apply the uniqueness in part (a) of Proposition 1.1.13.

Sketch.

𝑏

and

𝑐

are two solutions of

𝑎 + 𝑥 = 𝑎 + 𝑏

, which has only one solution. Likewise

𝑏

and

𝑐

both solve

𝑎 𝑥 = 𝑎 𝑏

when

𝑎 \neq 0

Suppose

𝑎 + 𝑏 = 𝑎 + 𝑐

. Then

𝑏

and

𝑐

are both solutions of the equation

𝑎 + 𝑥 = 𝑎 + 𝑏

, which has exactly one solution by part (a) of Proposition 1.1.13. Hence

𝑏 = 𝑐

. If

𝑎 𝑏 = 𝑎 𝑐

with

𝑎 \neq 0

, then

𝑏

and

𝑐

both solve

𝑎 𝑥 = 𝑎 𝑏

, and part (b) of Proposition 1.1.13 gives

𝑏 = 𝑐

. □

Intuition. No axiom says anything about a product with

0

. The only fact we have about

0

is additive, namely

0 + 0 = 0

, and distributivity is the only axiom that converts additive facts into multiplicative ones. So multiply

0 + 0 = 0

𝑎

and see what distributivity says.

Proposition 1.1.15 (Multiplying by zero). For every

𝑎 \in 𝐹

0 𝑎 = 𝑎 0 = 0

Proof.routine

Uses. Definition 1.1.1 (Field), Corollary 1.1.14 (Cancellation)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Which axiom could say anything about a product involving

0

? Only one axiom mentions both

+

and

\cdot

Hint 2. Multiply both sides of

0 + 0 = 0

𝑎

and expand with distributivity.

Hint 3. You get

𝑎 0 + 𝑎 0 = 𝑎 0

. Compare it with

𝑎 0 + 0 = 𝑎 0

and cancel.

Sketch. Distributivity turns

0 + 0 = 0

into

𝑎 0 + 𝑎 0 = 𝑎 0 = 𝑎 0 + 0

. Cancelling

𝑎 0

(Corollary 1.1.14) gives

𝑎 0 = 0

, and commutativity gives

0 𝑎 = 0

Let $𝑎 \in 𝐹$ . Then

\begin{aligned} 𝑎 0 + 𝑎 0 & = 𝑎 (0 + 0) & (distributivity) \\ = 𝑎 0 & (additive identity) \\ = 𝑎 0 + 0 & (additive identity) . \end{aligned}

Cancelling $𝑎 0$ from both sides (Corollary 1.1.14) gives $𝑎 0 = 0$ . By commutativity, $0 𝑎 = 𝑎 0 = 0$ . □

The proof converted the additive fact $0 + 0 = 0$ into the multiplicative fact $𝑎 0 + 𝑎 0 = 𝑎 0$ . The same trick shows that $0 𝐯 = 𝟎$ for vectors in Section 1.2.

Intuition. Every sign rule below comes from one observation: to show that an element equals

- 𝑥

, it is enough to show that it is an additive inverse of

𝑥

, since there is only one. In particular

(- 1) 𝑎

is an additive inverse of

𝑎

, and once

- 𝑎 = (- 1) 𝑎

is known, signs are just factors of

- 1

that can be moved around.

Proposition 1.1.16 (Sign rules). For all $𝑎, 𝑏 \in 𝐹$ :

(a) $- (- 𝑎) = 𝑎$ ;
(b) $(- 1) 𝑎 = - 𝑎$ ;
(c) $(- 𝑎) 𝑏 =$ $- (𝑎 𝑏) =$ $𝑎 (- 𝑏)$ ;
(d) $(- 𝑎) (- 𝑏) = 𝑎 𝑏$ .

Proof.routine

Uses. Theorem 1.1.9 (Uniqueness of identities and inverses), Proposition 1.1.15 (Multiplying by zero), Definition 1.1.1 (Field)

Techniques. Uniqueness via two candidates (Technique 1.1.8), Proving from axioms (Technique 1.1.7)

Hint 1. To show that an element equals

- 𝑥

, show that it is an additive inverse of

𝑥

and quote Theorem 1.1.9.

Hint 2. For (b), compute

𝑎 + (- 1) 𝑎

by writing

𝑎 = 1 𝑎

and using distributivity and Proposition 1.1.15.

Hint 3. For (c) and (d), rewrite each negative as

(- 1)

times something using (b), and move the factors

- 1

around with associativity and commutativity.

Sketch.

𝑎

is an additive inverse of

- 𝑎

, which gives (a). Distributivity and

0 𝑎 = 0

give

𝑎 +

(- 1) 𝑎 =

(1 + (- 1)) 𝑎 =

0

, which gives (b). Then (c) is (b) plus associativity and commutativity, and (d) is (c) twice followed by (a).

(a) Since $(- 𝑎) +$ $𝑎 =$ $𝑎 +$ $(- 𝑎) =$ $0$ , the element $𝑎$ is an additive inverse of $- 𝑎$ . By Theorem 1.1.9, $𝑎 = - (- 𝑎)$ .

(b) We compute

\begin{aligned} 𝑎 + (- 1) 𝑎 & = 1 𝑎 + (- 1) 𝑎 & (multiplicative identity) \\ = (1 + (- 1)) 𝑎 & (distributivity) \\ = 0 𝑎 & (additive inverse) \\ = 0, \end{aligned}

where the last step is Proposition 1.1.15. So $(- 1) 𝑎$ is an additive inverse of $𝑎$ , and $(- 1) 𝑎 = - 𝑎$ by Theorem 1.1.9.

(c) By (b), associativity, and (b) again, $(- 𝑎) 𝑏 =$ $((- 1) 𝑎) 𝑏 =$ $(- 1) (𝑎 𝑏) =$ $- (𝑎 𝑏)$ . Using commutativity and what we just proved, $𝑎 (- 𝑏) =$ $(- 𝑏) 𝑎 =$ $- (𝑏 𝑎) =$ $- (𝑎 𝑏)$ .

(d) By (c) twice and then (a), $(- 𝑎) (- 𝑏) =$ $- (𝑎 (- 𝑏)) =$ $- (- (𝑎 𝑏)) =$ $𝑎 𝑏$ . □

In particular $(- 1) (- 1) = 1$ , and $(𝑎 - 𝑏) 𝑐 =$ $𝑎 𝑐 -$ $𝑏 𝑐$ by distributivity and (c). One more consequence is used constantly, for instance whenever we factor an expression and set each factor equal to zero.

Proposition 1.1.17 (No zero divisors). If

𝑎, 𝑏 \in 𝐹

and

𝑎 𝑏 = 0

, then

𝑎 = 0

𝑏 = 0

. Equivalently, a product of nonzero elements of

𝐹

is nonzero.

Proof.routine

Uses. Definition 1.1.1 (Field), Proposition 1.1.15 (Multiplying by zero)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. To prove “

𝑎 = 0

𝑏 = 0

”, assume one of them is nonzero and prove that the other is zero.

Hint 2. If

𝑎 \neq 0

, then

𝑎^{- 1}

exists. Multiply

𝑎 𝑏 = 0

by it.

Sketch. If

𝑎 \neq 0

, then

𝑏 =

1 𝑏 =

(𝑎^{- 1} 𝑎) 𝑏 =

𝑎^{- 1} (𝑎 𝑏) =

𝑎^{- 1} 0 =

0

Suppose $𝑎 𝑏 = 0$ . If $𝑎 = 0$ we are done, so suppose $𝑎 \neq 0$ . Then $𝑎^{- 1}$ exists, and

\begin{aligned} 𝑏 & = 1 𝑏 & (multiplicative identity) \\ = (𝑎^{- 1} 𝑎) 𝑏 & (multiplicative inverse) \\ = 𝑎^{- 1} (𝑎 𝑏) & (associativity) \\ = 𝑎^{- 1} 0 & (as 𝑎 𝑏 = 0) \\ = 0, \end{aligned}

where the last step is Proposition 1.1.15. So $𝑎 = 0$ or $𝑏 = 0$ . □

The integers also have no zero divisors, so this property alone does not make a field. In Section 1.1.3 we meet number systems that have zero divisors, and there the uniqueness in part (b) of Proposition 1.1.13 fails.

Exercise 1.1.18.routine Find all

𝑥 \in 𝐹

with

𝑥^{2} = 𝑥

Solution.

Uses. Proposition 1.1.17 (No zero divisors), Proposition 1.1.16 (Sign rules), Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ )

Hint 1. Move everything to one side and factor.

Hint 2.

𝑥^{2} - 𝑥 = 𝑥 (𝑥 - 1)

. Now use Proposition 1.1.17.

Sketch.

𝑥^{2} = 𝑥

if and only if

𝑥 (𝑥 - 1) = 0

, which by Proposition 1.1.17 happens exactly when

𝑥 = 0

𝑥 = 1

By distributivity and Proposition 1.1.16,

𝑥 (𝑥 - 1) =

𝑥^{2} +

𝑥 (- 1) =

𝑥^{2} -

𝑥

. By Proposition 1.1.13,

𝑥^{2} = 𝑥

holds if and only if

𝑥^{2} - 𝑥 = 0

, that is, if and only if

𝑥 (𝑥 - 1) = 0

. By Proposition 1.1.17 this happens if and only if

𝑥 = 0

𝑥 - 1 = 0

, and

𝑥 - 1 = 0

means

𝑥 = 1

(again by Proposition 1.1.13). So the solutions are exactly

𝑥 = 0

and

𝑥 = 1

, which are distinct since

0 \neq 1

1.1.3The fields $𝔽_{𝑝}$

$𝔽_{2}$ is the first of an infinite family of finite fields, built from arithmetic with remainders.

Definition 1.1.19 (Integers modulo $𝑛$ ). Let $𝑛 \geq 2$ be an integer. For $𝑥 \in ℤ$ , write $𝑥 mod 𝑛$ for the remainder of $𝑥$ on division by $𝑛$ , that is, the unique $𝑟 \in {0, 1, \dots, 𝑛 - 1}$ such that $𝑛 ∣ 𝑥 - 𝑟$ . The integers modulo $𝑛$ are the set $ℤ / 𝑛 ℤ =$ ${0, 1, \dots, 𝑛 - 1}$ with the operations

𝑎 +_{𝑛} 𝑏 = (𝑎 + 𝑏) mod 𝑛, 𝑎 \cdot_{𝑛} 𝑏 = (𝑎 𝑏) mod 𝑛 .

This is clock arithmetic: in $ℤ / 12 ℤ$ , $9 +_{12} 5 = 2$ and $5 \cdot_{12} 5 = 1$ . Outside of proofs we drop the subscripts and write $𝑎 + 𝑏$ and $𝑎 𝑏$ , once it is clear which $𝑛$ we work with.

All the field axioms except multiplicative inverses hold in $ℤ / 𝑛 ℤ$ for every $𝑛$ . Structures like this are common enough to have a name.

Definition 1.1.20 (Commutative ring). A commutative ring is a set with an addition and a multiplication that satisfy all the field axioms, except possibly the existence of multiplicative inverses and the condition

1 \neq 0

So every field is a commutative ring. The integers and componentwise $ℝ^{2}$ (Non-example 1.1.6) are commutative rings that are not fields, and so are polynomials with coefficients in a field (Section 7.1).

Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ ). Let $𝑛 \geq 2$ .

(a)For all $𝑥, 𝑦 \in ℤ$ , $(𝑥 mod 𝑛) +_{𝑛} (𝑦 mod 𝑛) =$ $(𝑥 + 𝑦) mod 𝑛$ and $(𝑥 mod 𝑛) \cdot_{𝑛} (𝑦 mod 𝑛) =$ $(𝑥 𝑦) mod 𝑛$ .
(b) $ℤ / 𝑛 ℤ$ is a commutative ring with identities $0$ and $1$ . The additive inverse of $𝑎 \neq 0$ is $𝑛 - 𝑎$ .

Proof.standard

Uses. Definition 1.1.19 (Integers modulo $𝑛$ ), Definition 1.1.20 (Commutative ring), the division algorithm: $𝑥 mod 𝑛 = 𝑦 mod 𝑛$ if and only if $𝑛 ∣ 𝑥 - 𝑦$

Hint 1. Every axiom except inverses is either an identity, like associativity, or asks for an identity element. The identities hold in

ℤ

; you need a way to carry them over to

ℤ / 𝑛 ℤ

Hint 2. For (a),

𝑥

and

𝑥 mod 𝑛

differ by a multiple of

𝑛

. What does that say about

(𝑥 + 𝑦)

versus

(𝑥 mod 𝑛) +

(𝑦 mod 𝑛)

Hint 3. For (b), use (a) to write both sides of each identity in

ℤ / 𝑛 ℤ

as the remainder of the corresponding side of the same identity in

ℤ

Sketch. Since

𝑥

and

𝑥 mod 𝑛

differ by a multiple of

𝑛

, so do

𝑥 + 𝑦

and

(𝑥 mod 𝑛) +

(𝑦 mod 𝑛)

, and likewise for products; this is (a). By (a), each side of an identity such as associativity in

ℤ / 𝑛 ℤ

is the remainder of the corresponding side in

ℤ

, where the identity holds. The identities are

0

and

1

, and

𝑎 +_{𝑛} (𝑛 - 𝑎) = 0

Throughout we use the fact, from the division algorithm, that $𝑥 mod 𝑛 = 𝑦 mod 𝑛$ if and only if $𝑛 ∣ 𝑥 - 𝑦$ . In particular $𝑛 ∣ 𝑥 - (𝑥 mod 𝑛)$ for every $𝑥 \in ℤ$ .

(a) Let $𝑥, 𝑦 \in ℤ$ , and write $𝑥^{'} = 𝑥 mod 𝑛$ and $𝑦^{'} = 𝑦 mod 𝑛$ . Then

(𝑥 + 𝑦) - (𝑥^{'} + 𝑦^{'}) = (𝑥 - 𝑥^{'}) + (𝑦 - 𝑦^{'})

is a sum of multiples of $𝑛$ , so $(𝑥^{'} + 𝑦^{'}) mod 𝑛 =$ $(𝑥 + 𝑦) mod 𝑛$ . That is, $𝑥^{'} +_{𝑛} 𝑦^{'} =$ $(𝑥 + 𝑦) mod 𝑛$ . Similarly,

𝑥 𝑦 - 𝑥^{'} 𝑦^{'} = (𝑥 - 𝑥^{'}) 𝑦 + 𝑥^{'} (𝑦 - 𝑦^{'})

is a multiple of $𝑛$ , so $𝑥^{'} \cdot_{𝑛} 𝑦^{'} =$ $(𝑥 𝑦) mod 𝑛$ .

(b) Every element of $ℤ / 𝑛 ℤ$ is its own remainder, $𝑎 mod 𝑛 = 𝑎$ , so (a) applies to elements of $ℤ / 𝑛 ℤ$ directly. Let $𝑎, 𝑏, 𝑐 \in ℤ / 𝑛 ℤ$ . For associativity of addition, (a) gives

(𝑎 +_{𝑛} 𝑏) +_{𝑛} 𝑐 = ((𝑎 + 𝑏) mod 𝑛) +_{𝑛} (𝑐 mod 𝑛) = ((𝑎 + 𝑏) + 𝑐) mod 𝑛,

and likewise $𝑎 +_{𝑛} (𝑏 +_{𝑛} 𝑐) =$ $(𝑎 + (𝑏 + 𝑐)) mod 𝑛$ . These agree because $(𝑎 + 𝑏) +$ $𝑐 =$ $𝑎 +$ $(𝑏 + 𝑐)$ in $ℤ$ . Associativity of $\cdot_{𝑛}$ , and commutativity of both operations, follow in the same way from the corresponding laws in $ℤ$ . For distributivity, by (a) twice,

𝑎 \cdot_{𝑛} (𝑏 +_{𝑛} 𝑐) = (𝑎 (𝑏 + 𝑐)) mod 𝑛 = (𝑎 𝑏 + 𝑎 𝑐) mod 𝑛 = (𝑎 \cdot_{𝑛} 𝑏) +_{𝑛} (𝑎 \cdot_{𝑛} 𝑐) .

Next, $𝑎 +_{𝑛} 0 = 𝑎 mod 𝑛 = 𝑎$ and $𝑎 \cdot_{𝑛} 1 = 𝑎 mod 𝑛 = 𝑎$ , and $0 \neq 1$ because $𝑛 \geq 2$ . Finally, $0 +_{𝑛} 0 = 0$ , and for $1 \leq 𝑎 \leq 𝑛 - 1$ the integer $𝑛 - 𝑎$ lies in $ℤ / 𝑛 ℤ$ and $𝑎 +_{𝑛} (𝑛 - 𝑎) =$ $𝑛 mod 𝑛 =$ $0$ . So every axiom of a commutative ring holds. □

Remark 1.1.22. Look back at which results of Section 1.1.2 used multiplicative inverses. Uniqueness of the identities and of additive inverses (Theorem 1.1.9), part (a) of Proposition 1.1.13, additive cancellation, Proposition 1.1.15, and Proposition 1.1.16 did not, so they hold in every commutative ring with the same proofs. We will use this in Section 7.3, where determinants have polynomial entries. Proposition 1.1.17 and multiplicative cancellation did use inverses, and Non-example 1.1.23 shows that they can fail.

Compare the multiplication tables of $ℤ / 5 ℤ$ and $ℤ / 6 ℤ$ .

$\cdot$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$

Table 1.1.1. Multiplication in

ℤ / 5 ℤ

. Every nonzero row contains each element exactly once, in particular a

1

(highlighted), so every nonzero element has an inverse.

$\cdot$	$0$	$1$	$2$	$3$	$4$	$5$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$	$5$
$2$	$0$	$2$	$4$	$0$	$2$	$4$
$3$	$0$	$3$	$0$	$3$	$0$	$3$
$4$	$0$	$4$	$2$	$0$	$4$	$2$
$5$	$0$	$5$	$4$	$3$	$2$	$1$

Table 1.1.2. Multiplication in

ℤ / 6 ℤ

. The rows of

2

3

, and

4

contain

0

in a nonzero column (highlighted) and never contain

1

Non-example 1.1.23 (

ℤ / 6 ℤ

). In

ℤ / 6 ℤ

we have

2 \cdot 3 = 6 mod 6 = 0

with

2 \neq 0

and

3 \neq 0

. Fields have no zero divisors (Proposition 1.1.17), so

ℤ / 6 ℤ

is not a field. The failure is visible in Table 1.1.2: the row of

2

0, 2, 4, 0, 2, 4

, which never contains

1

, so

2

has no inverse. Linear equations misbehave accordingly:

2 𝑥 = 1

has no solution, and

2 𝑥 = 4

has two,

𝑥 = 2

and

𝑥 = 5

Intuition. By Lemma 1.1.21, the only question is which nonzero

𝑎

have inverses, that is, which rows of the multiplication table contain a

1

. If

𝑛 = 𝑟 𝑠

with

1 < 𝑟, 𝑠 < 𝑛

, then

𝑟 𝑠 = 0

ℤ / 𝑛 ℤ

, a zero divisor, which no field has. If

𝑛

is prime, we give two proofs that every nonzero

𝑎

is invertible. The first constructs

𝑎^{- 1}

from Bézout’s identity. The second never finds

𝑎^{- 1}

: it shows that the row of

𝑎

has no repeated entries, and then counting forces every element,

1

included, to appear in it.

Proposition 1.1.24 (

ℤ / 𝑛 ℤ

is a field if and only if

𝑛

is prime). Let

𝑛 \geq 2

be an integer. Then

ℤ / 𝑛 ℤ

is a field if and only if

𝑛

is prime.

Proof (via Bézout’s identity).standard

Uses. Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ ), Proposition 1.1.17 (No zero divisors), Bézout’s identity: if $gcd (𝑎, 𝑛) = 1$ , then $𝑎 𝑥 + 𝑛 𝑦 = 1$ for some $𝑥, 𝑦 \in ℤ$

Hint 1. By Lemma 1.1.21, only multiplicative inverses are in question. If

𝑛 = 𝑟 𝑠

with

1 < 𝑟, 𝑠 < 𝑛

, look at

𝑟 \cdot_{𝑛} 𝑠

Hint 2. For

𝑛 = 𝑝

prime, an inverse of

𝑎

is an integer

𝑏

with

𝑝 ∣ 𝑎 𝑏 - 1

, that is,

𝑎 𝑏 + 𝑝 𝑦 = 1

for some integer

𝑦

. Which theorem produces such integers?

Hint 3. Bézout’s identity applies because

gcd (𝑎, 𝑝) = 1

. Reduce the resulting

𝑥

modulo

𝑝

Sketch. If

𝑛 = 𝑟 𝑠

with

1 < 𝑟, 𝑠 < 𝑛

, then

𝑟 \cdot_{𝑛} 𝑠 = 0

is a product of nonzero elements, impossible in a field. If

𝑛 = 𝑝

is prime and

0 < 𝑎 < 𝑝

, then

gcd (𝑎, 𝑝) = 1

, so

𝑎 𝑥 + 𝑝 𝑦 = 1

for some integers

𝑥, 𝑦

, and

𝑥 mod 𝑝

is an inverse of

𝑎

by part (a) of Lemma 1.1.21.

By Lemma 1.1.21, $ℤ / 𝑛 ℤ$ satisfies every field axiom except possibly the existence of multiplicative inverses. So it is a field if and only if every nonzero element has a multiplicative inverse.

( $\Rightarrow$ ) We prove the contrapositive. Suppose $𝑛$ is not prime. Since $𝑛 \geq 2$ , we can write $𝑛 = 𝑟 𝑠$ with integers $1 < 𝑟, 𝑠 < 𝑛$ . Then $𝑟$ and $𝑠$ are nonzero elements of $ℤ / 𝑛 ℤ$ with $𝑟 \cdot_{𝑛} 𝑠 = 𝑛 mod 𝑛 = 0$ . A field has no zero divisors (Proposition 1.1.17), so $ℤ / 𝑛 ℤ$ is not a field.

( $\Leftarrow$ ) Suppose $𝑛 = 𝑝$ is prime, and let $𝑎 \in {1, \dots, 𝑝 - 1}$ . The only positive divisors of $𝑝$ are $1$ and $𝑝$ , and $𝑝 ∤ 𝑎$ because $0 < 𝑎 < 𝑝$ , so $gcd (𝑎, 𝑝) = 1$ . By Bézout’s identity there are integers $𝑥, 𝑦$ with $𝑎 𝑥 + 𝑝 𝑦 = 1$ . Let $𝑏 = 𝑥 mod 𝑝$ . By part (a) of Lemma 1.1.21,

𝑎 \cdot_{𝑝} 𝑏 = (𝑎 mod 𝑝) \cdot_{𝑝} (𝑥 mod 𝑝) = (𝑎 𝑥) mod 𝑝 = (1 - 𝑝 𝑦) mod 𝑝 = 1,

where the last step holds because $1 - 𝑝 𝑦$ and $1$ differ by a multiple of $𝑝$ . So $𝑏$ is a multiplicative inverse of $𝑎$ , as desired. □

Proof (by counting).standard

Uses. Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ ), Proposition 1.1.17 (No zero divisors), Euclid’s lemma: if $𝑝$ is prime and $𝑝 ∣ 𝑎 𝑐$ , then $𝑝 ∣ 𝑎$ or $𝑝 ∣ 𝑐$ , an injective function from a finite set to itself is surjective

Hint 1. Fix

𝑎 \neq 0

and look at its row in the multiplication table, that is, the function

𝑥 \mapsto 𝑎 \cdot_{𝑝} 𝑥

. You want

1

to appear in it.

Hint 2. Show that the row has no repeated entries: if

𝑎 \cdot_{𝑝} 𝑥 = 𝑎 \cdot_{𝑝} 𝑦

, then

𝑝 ∣ 𝑎 (𝑥 - 𝑦)

. Use Euclid’s lemma.

Hint 3. A row with

𝑝

entries taken from a set of

𝑝

elements, with no repeats, contains every element.

Sketch. The direction (

\Rightarrow

) is as in the first proof. For

𝑝

prime and

𝑎 \neq 0

, the map

𝑥 \mapsto 𝑎 \cdot_{𝑝} 𝑥

ℤ / 𝑝 ℤ

is injective, since

𝑝 ∣ 𝑎 (𝑥 - 𝑦)

and

𝑝 ∤ 𝑎

force

𝑝 ∣ 𝑥 - 𝑦

, hence

𝑥 = 𝑦

. An injective map from a finite set to itself is surjective, so

𝑎 \cdot_{𝑝} 𝑏 = 1

for some

𝑏

The direction ( $\Rightarrow$ ) is proved as in the first proof. For ( $\Leftarrow$ ), let $𝑝$ be prime and $𝑎 \in {1, \dots, 𝑝 - 1}$ , and define $𝑚 : ℤ / 𝑝 ℤ \to ℤ / 𝑝 ℤ$ by $𝑚 (𝑥) = 𝑎 \cdot_{𝑝} 𝑥$ . We will show that $𝑚$ is injective. Suppose $𝑚 (𝑥) = 𝑚 (𝑦)$ , that is, $(𝑎 𝑥) mod 𝑝 =$ $(𝑎 𝑦) mod 𝑝$ . Then $𝑝 ∣$ $𝑎 𝑥 -$ $𝑎 𝑦 =$ $𝑎 (𝑥 - 𝑦)$ . Since $𝑝$ is prime and $𝑝 ∤ 𝑎$ , Euclid’s lemma gives $𝑝 ∣ 𝑥 - 𝑦$ . But $𝑥, 𝑦 \in {0, \dots, 𝑝 - 1}$ , so $| 𝑥 - 𝑦 | < 𝑝$ , and the only multiple of $𝑝$ strictly between $- 𝑝$ and $𝑝$ is $0$ . Hence $𝑥 = 𝑦$ .

So $𝑚$ is an injective function from the finite set $ℤ / 𝑝 ℤ$ to itself. Its image then has $𝑝$ elements, so it is all of $ℤ / 𝑝 ℤ$ , and in particular $1 = 𝑚 (𝑏) = 𝑎 \cdot_{𝑝} 𝑏$ for some $𝑏$ . That is, $𝑏$ is a multiplicative inverse of $𝑎$ . □

The two proofs differ in what they produce. The first computes $𝑎^{- 1}$ , and running the Euclidean algorithm to find Bézout’s $𝑥$ is how one inverts elements of $𝔽_{𝑝}$ in practice. The second only proves that $𝑎^{- 1}$ exists, but it uses just two facts about $ℤ / 𝑝 ℤ$ : it is finite, and $𝑎 𝑥 = 𝑎 𝑦$ with $𝑎 \neq 0$ forces $𝑥 = 𝑦$ . On a finite set, a function to itself is injective if and only if it is surjective, so a multiplication that can be cancelled can also be undone. The same principle returns in Section 4.2 with dimension in place of the number of elements: a linear map from a finite-dimensional space to itself is injective if and only if it is surjective.

Notation 1.1.25 (

𝔽_{𝑝}

). For a prime

𝑝

, the field

ℤ / 𝑝 ℤ

is denoted

𝔽_{𝑝}

. Later chapters use a bare

𝔽

to mean

ℝ

ℂ

;

𝔽_{𝑝}

always carries its subscript, so the two uses never clash.

Example 1.1.26 (Arithmetic in

𝔽_{5}

). From Table 1.1.1, the inverses in

𝔽_{5}

are

1^{- 1} = 1

2^{- 1} = 3

3^{- 1} = 2

, and

4^{- 1} = 4

. To solve

3 𝑥 + 4 = 1

𝔽_{5}

, follow Proposition 1.1.13: first

3 𝑥 = 1 - 4 = - 3 = 2

, and then

𝑥 = 3^{- 1} \cdot 2 = 2 \cdot 2 = 4

. Check:

3 \cdot 4 + 4 = 16 mod 5 = 1

Exercise 1.1.27.routine Find the multiplicative inverse of every nonzero element of

𝔽_{7}

. Then solve

3 𝑥 + 5 = 2

𝔽_{7}

Solution.

Uses. Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime), Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ )

Hint 1. Multiply pairs of nonzero elements and reduce modulo

7

; you want products that leave remainder

1

Hint 2. For the equation, follow Proposition 1.1.13: subtract

5

, then multiply by

3^{- 1}

Sketch.

2 \cdot 4 = 8

3 \cdot 5 = 15

, and

6 \cdot 6 = 36

all leave remainder

1

. Then

3 𝑥 = 2 - 5 = 4

, so

𝑥 = 5 \cdot 4 = 6

Since $2 \cdot 4 = 8 = 7 + 1$ , $3 \cdot 5 = 15 = 14 + 1$ , and $6 \cdot 6 = 36 = 35 + 1$ , the inverses are

1^{- 1} = 1, 2^{- 1} = 4, 3^{- 1} = 5, 4^{- 1} = 2, 5^{- 1} = 3, 6^{- 1} = 6 .

For the equation, $3 𝑥 + 5 = 2$ holds if and only if $3 𝑥 = 2 - 5 = - 3 = 4$ , which holds if and only if $𝑥 =$ $3^{- 1} \cdot$ $4 =$ $5 \cdot$ $4 =$ $20 mod 7 =$ $6$ (Proposition 1.1.13). Check: $3 \cdot 6 + 5 = 23 = 21 + 2$ , so $3 \cdot 6 + 5 = 2$ in $𝔽_{7}$ .

Aside (Characteristic). In $𝔽_{𝑝}$ , adding $1$ to itself $𝑝$ times gives $0$ , while in $ℚ$ , $ℝ$ , and $ℂ$ no number of $1$ ′s adds up to $0$ . For a positive integer $𝑚$ and $𝑎 \in 𝐹$ , write $𝑚 \cdot 𝑎$ for $𝑎 + 𝑎 + \dots + 𝑎$ ( $𝑚$ terms). This is repeated addition, not multiplication by an element of $𝐹$ .

Definition 1.1.28 (Characteristic). The characteristic of a field

𝐹

, written

char 𝐹

, is the least positive integer

𝑚

such that

𝑚 \cdot 1 = 0

, or

0

if there is no such

𝑚

So $char 𝔽_{𝑝} = 𝑝$ and $char ℚ =$ $char ℝ =$ $char ℂ =$ $0$ . A nonzero characteristic is always prime (Exercise 1.1.48). If $char 𝐹 = 𝑝 > 0$ , then $𝑝 \cdot 𝑎 = (𝑝 \cdot 1) 𝑎 = 0$ for every $𝑎 \in 𝐹$ , by distributivity.

The case that matters most later is characteristic $2$ , where $1 + 1 = 0$ . There $- 𝑎 = 𝑎$ for every $𝑎$ , and we cannot divide by $1 + 1$ . Some later arguments do divide by $1 + 1$ : the converse of “alternating implies antisymmetric” for determinants (Section 5.1), and the passage between quadratic forms and symmetric matrices (Section 12.2). Those results assume $1 + 1 \neq 0$ , and $𝔽_{2}$ is where to look for counterexamples when the assumption is dropped.

1.1.4The complex numbers

Componentwise multiplication does not make $ℝ^{2}$ a field (Non-example 1.1.6), but a different multiplication does. We give only what the next few chapters need: the definition and the field axioms. Conjugation, the modulus, polar form, and roots of unity are developed in Section B.1.

Definition 1.1.29 (Complex numbers). The complex numbers are the set $ℂ = ℝ^{2}$ , whose element $(𝑎, 𝑏)$ we write as $𝑎 + 𝑏 𝑖$ , with the operations

\begin{aligned} (𝑎 + 𝑏 𝑖) + (𝑐 + 𝑑 𝑖) & = (𝑎 + 𝑐) + (𝑏 + 𝑑) 𝑖, \\ (𝑎 + 𝑏 𝑖) (𝑐 + 𝑑 𝑖) & = (𝑎 𝑐 - 𝑏 𝑑) + (𝑎 𝑑 + 𝑏 𝑐) 𝑖 . \end{aligned}

The real part of $𝑎 + 𝑏 𝑖$ is $𝑎$ and its imaginary part is $𝑏$ . We identify the real number $𝑎$ with $𝑎 + 0 𝑖$ , and write $𝑏 𝑖$ for $0 + 𝑏 𝑖$ and $𝑖$ for $0 + 1 𝑖$ .

Intuition. The multiplication rule is forced. Suppose we want a field that contains $ℝ$ and an element $𝑖$ with $𝑖^{2} = - 1$ . Then the field axioms leave no choice:

(𝑎 + 𝑏 𝑖) (𝑐 + 𝑑 𝑖) = 𝑎 𝑐 + 𝑎 𝑑 𝑖 + 𝑏 𝑐 𝑖 + 𝑏 𝑑 𝑖^{2} = (𝑎 𝑐 - 𝑏 𝑑) + (𝑎 𝑑 + 𝑏 𝑐) 𝑖 .

So the definition is the only candidate, and what remains is to check that it satisfies the axioms. In particular $𝑖^{2} =$ $(0 \cdot 0 - 1 \cdot 1) +$ $(0 \cdot 1 + 1 \cdot 0) 𝑖 =$ $- 1$ .

The identification of $𝑎 \in ℝ$ with $𝑎 + 0 𝑖$ respects the operations: $(𝑎 + 0 𝑖) +$ $(𝑐 + 0 𝑖) =$ $(𝑎 + 𝑐) +$ $0 𝑖$ and $(𝑎 + 0 𝑖) (𝑐 + 0 𝑖) =$ $𝑎 𝑐 +$ $0 𝑖$ . Also $𝑎 (𝑐 + 𝑑 𝑖) =$ $𝑎 𝑐 +$ $𝑎 𝑑 𝑖$ , so multiplying by a real number scales both coordinates.

Proposition 1.1.30 ( $ℂ$ is a field). $ℂ$ is a field, with identities $0 = 0 + 0 𝑖$ and $1 = 1 + 0 𝑖$ . The additive inverse of $𝑎 + 𝑏 𝑖$ is $- 𝑎 - 𝑏 𝑖$ , and if $𝑎 + 𝑏 𝑖 \neq 0$ , then

{(𝑎 + 𝑏 𝑖)}^{- 1} = \frac{𝑎}{𝑎^{2} + 𝑏^{2}} - \frac{𝑏}{𝑎^{2} + 𝑏^{2}} 𝑖 .

Proof, part 1 of 2: the algebraic laws.routine

Uses. Definition 1.1.29 (Complex numbers), the field axioms of $ℝ$

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Write

𝑧 = 𝑎 + 𝑏 𝑖

𝑤 = 𝑐 + 𝑑 𝑖

, and

𝑢 = 𝑒 + 𝑓 𝑖

with real

𝑎, \dots, 𝑓

, and compute both sides of each law from Definition 1.1.29.

Hint 2. Addition is coordinatewise, so its laws hold coordinate by coordinate. For multiplication, expand both sides and compare real parts and imaginary parts separately.

Sketch. Addition is coordinatewise addition in

ℝ^{2}

, so its laws come from those of

ℝ

. For each multiplicative law, expand both sides with the definition; the real parts agree, and so do the imaginary parts, by the field axioms of

ℝ

Let $𝑧 = 𝑎 + 𝑏 𝑖$ , $𝑤 = 𝑐 + 𝑑 𝑖$ , and $𝑢 = 𝑒 + 𝑓 𝑖$ , where $𝑎,$ $𝑏,$ $𝑐,$ $𝑑,$ $𝑒,$ $𝑓 \in$ $ℝ$ . Addition in $ℂ$ is coordinatewise addition in $ℝ^{2}$ , so it is commutative and associative because addition in $ℝ$ is.

For commutativity of multiplication, $𝑤 𝑧 =$ $(𝑐 𝑎 - 𝑑 𝑏) +$ $(𝑐 𝑏 + 𝑑 𝑎) 𝑖$ , which equals $𝑧 𝑤 =$ $(𝑎 𝑐 - 𝑏 𝑑) +$ $(𝑎 𝑑 + 𝑏 𝑐) 𝑖$ by commutativity in $ℝ$ . For associativity, expanding with the definition gives

\begin{aligned} (𝑧 𝑤) 𝑢 & = ((𝑎 𝑐 - 𝑏 𝑑) 𝑒 - (𝑎 𝑑 + 𝑏 𝑐) 𝑓) + ((𝑎 𝑐 - 𝑏 𝑑) 𝑓 + (𝑎 𝑑 + 𝑏 𝑐) 𝑒) 𝑖 \\ = (𝑎 𝑐 𝑒 - 𝑎 𝑑 𝑓 - 𝑏 𝑐 𝑓 - 𝑏 𝑑 𝑒) + (𝑎 𝑐 𝑓 + 𝑎 𝑑 𝑒 + 𝑏 𝑐 𝑒 - 𝑏 𝑑 𝑓) 𝑖, \end{aligned}

and

\begin{aligned} 𝑧 (𝑤 𝑢) & = (𝑎 (𝑐 𝑒 - 𝑑 𝑓) - 𝑏 (𝑐 𝑓 + 𝑑 𝑒)) + (𝑎 (𝑐 𝑓 + 𝑑 𝑒) + 𝑏 (𝑐 𝑒 - 𝑑 𝑓)) 𝑖 \\ = (𝑎 𝑐 𝑒 - 𝑎 𝑑 𝑓 - 𝑏 𝑐 𝑓 - 𝑏 𝑑 𝑒) + (𝑎 𝑐 𝑓 + 𝑎 𝑑 𝑒 + 𝑏 𝑐 𝑒 - 𝑏 𝑑 𝑓) 𝑖, \end{aligned}

so $(𝑧 𝑤) 𝑢 = 𝑧 (𝑤 𝑢)$ . For distributivity,

\begin{aligned} 𝑧 (𝑤 + 𝑢) & = (𝑎 (𝑐 + 𝑒) - 𝑏 (𝑑 + 𝑓)) + (𝑎 (𝑑 + 𝑓) + 𝑏 (𝑐 + 𝑒)) 𝑖 \\ = ((𝑎 𝑐 - 𝑏 𝑑) + (𝑎 𝑒 - 𝑏 𝑓)) + ((𝑎 𝑑 + 𝑏 𝑐) + (𝑎 𝑓 + 𝑏 𝑒)) 𝑖 \\ = 𝑧 𝑤 + 𝑧 𝑢 . \end{aligned}

Intuition. The remaining axioms ask for identities and inverses, and only the multiplicative inverse needs an idea. For $𝑧 = 𝑎 + 𝑏 𝑖$ , multiplying by $𝑎 - 𝑏 𝑖$ produces a real number,

(𝑎 + 𝑏 𝑖) (𝑎 - 𝑏 𝑖) = (𝑎^{2} + 𝑏^{2}) + (- 𝑎 𝑏 + 𝑏 𝑎) 𝑖 = 𝑎^{2} + 𝑏^{2},

and a nonzero real number can be divided by. Dividing $𝑎 - 𝑏 𝑖$ by $𝑎^{2} + 𝑏^{2}$ gives the inverse in the statement.

Proof, part 2 of 2: identities and inverses.routine

Uses. Definition 1.1.29 (Complex numbers), $𝑥^{2} + 𝑦^{2} > 0$ for real $(𝑥, 𝑦) \neq (0, 0)$

Hint 1. Check directly that

0 + 0 𝑖

and

1 + 0 𝑖

are identities and that

- 𝑎 - 𝑏 𝑖

is an additive inverse of

𝑎 + 𝑏 𝑖

Hint 2. For the multiplicative inverse, first explain why

𝑎^{2} + 𝑏^{2} \neq 0

when

𝑎 + 𝑏 𝑖 \neq 0

. Then multiply

𝑎 + 𝑏 𝑖

by the proposed inverse.

Sketch. The identities and additive inverses are immediate from the definition. If

𝑎 + 𝑏 𝑖 \neq 0

, then

𝑎^{2} + 𝑏^{2} > 0

, and multiplying out shows that

(𝑎 - 𝑏 𝑖) / (𝑎^{2} + 𝑏^{2})

is an inverse.

For $𝑧 = 𝑎 + 𝑏 𝑖$ we have $𝑧 + (0 + 0 𝑖) = 𝑧$ and $𝑧 (1 + 0 𝑖) =$ $(𝑎 \cdot 1 - 𝑏 \cdot 0) +$ $(𝑎 \cdot 0 + 𝑏 \cdot 1) 𝑖 =$ $𝑧$ , and $0 + 0 𝑖 \neq 1 + 0 𝑖$ . Also $(𝑎 + 𝑏 𝑖) +$ $(- 𝑎 - 𝑏 𝑖) =$ $0 +$ $0 𝑖$ .

Now suppose $𝑧 \neq 0$ , that is, $(𝑎, 𝑏) \neq (0, 0)$ . Then $𝐷 = 𝑎^{2} + 𝑏^{2} > 0$ , since squares of real numbers are nonnegative and at least one of $𝑎^{2}$ and $𝑏^{2}$ is positive. Let $𝑤 =$ $𝑎 / 𝐷 -$ $(𝑏 / 𝐷) 𝑖$ . Then

𝑧 𝑤 = (𝑎 \cdot \frac{𝑎}{𝐷} - 𝑏 \cdot \frac{- 𝑏}{𝐷}) + (𝑎 \cdot \frac{- 𝑏}{𝐷} + 𝑏 \cdot \frac{𝑎}{𝐷}) 𝑖 = \frac{𝑎^{2} + 𝑏^{2}}{𝐷} + 0 𝑖 = 1,

so $𝑤$ is a multiplicative inverse of $𝑧$ . □

From now on we compute in $ℂ$ with the field axioms and $𝑖^{2} = - 1$ , as in the intuition above; the product formula never needs to be memorized.

Example 1.1.31 (Complex arithmetic). $(2 + 3 𝑖) (1 - 𝑖) =$ $2 -$ $2 𝑖 +$ $3 𝑖 -$ $3 𝑖^{2} =$ $5 +$ $𝑖$ . By Proposition 1.1.30, ${(1 + 2 𝑖)}^{- 1} =$ $(1 - 2 𝑖) / 5$ , so the unique solution of $(1 + 2 𝑖) 𝑧 = 3 - 𝑖$ (Proposition 1.1.13) is

𝑧 = \frac{(3 - 𝑖) (1 - 2 𝑖)}{5} = \frac{3 - 6 𝑖 - 𝑖 + 2 𝑖^{2}}{5} = \frac{1 - 7 𝑖}{5} .

The number $𝑎 - 𝑏 𝑖$ in the proof is the complex conjugate of $𝑎 + 𝑏 𝑖$ , and $\sqrt{𝑎^{2} + 𝑏^{2}}$ is the modulus of $𝑎 + 𝑏 𝑖$ . Both reappear in Chapter 6, where inner products over $ℂ$ need them.

1.1.5Subfields

The operations of $ℂ$ , restricted to the real numbers $𝑎 + 0 𝑖$ , are the operations of $ℝ$ , and the operations of $ℝ$ restricted to $ℚ$ are those of $ℚ$ . Fields sitting inside other fields like this come up whenever we compare the same vectors over two fields, for instance over $ℝ$ and over $ℂ$ (Section 1.5).

Definition 1.1.32 (Subfield). Let

𝐹

be a field. A subset

𝐾 \subseteq 𝐹

is a subfield of

𝐹

0, 1 \in 𝐾

and

𝐾

is closed under the operations of

𝐹

: for all

𝑎, 𝑏 \in 𝐾

, we have

𝑎 + 𝑏 \in 𝐾

𝑎 𝑏 \in 𝐾

- 𝑎 \in 𝐾

, and

𝑎^{- 1} \in 𝐾

𝑎 \neq 0

Proposition 1.1.33 (A subfield is a field). A subfield

𝐾

of a field

𝐹

is a field under the operations of

𝐹

restricted to

𝐾

Proof.routine

Uses. Definition 1.1.32 (Subfield), Definition 1.1.1 (Field)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Sort the axioms into those of the form “for all elements, an identity holds” and those of the form “there exists an element”.

Hint 2. The first kind hold in

𝐾

because they hold in

𝐹

. For the second kind, the element whose existence is claimed must lie in

𝐾

, and that is what the closure conditions provide.

Sketch. Closure under

+

and

\cdot

makes them operations on

𝐾

. Identities that hold for all elements of

𝐹

hold for all elements of

𝐾

. The identities

0, 1

and the inverses required by the existence axioms lie in

𝐾

by assumption.

Since

𝐾

is closed under

+

and

\cdot

, these restrict to operations

𝐾 \times 𝐾 \to 𝐾

. Commutativity, associativity, and distributivity hold for all elements of

𝐹

, so in particular for all elements of

𝐾

. We have

0, 1 \in 𝐾

, the equations

𝑎 + 0 = 𝑎

and

𝑎 \cdot 1 = 𝑎

hold for

𝑎 \in 𝐾

because they hold in

𝐹

, and

0 \neq 1

. Finally, for

𝑎 \in 𝐾

we have

- 𝑎 \in 𝐾

with

𝑎 + (- 𝑎) = 0

, and if

𝑎 \neq 0

then

𝑎^{- 1} \in 𝐾

with

𝑎 𝑎^{- 1} = 1

. So every field axiom holds in

𝐾

. □

Example 1.1.34 (Subfields and non-subfields).

ℚ

is a subfield of

ℝ

, and

ℝ

, viewed as the set of

𝑎 + 0 𝑖

, is a subfield of

ℂ

; the identification above shows the closure conditions, with

{(𝑎 + 0 𝑖)}^{- 1} =

𝑎^{- 1} +

0 𝑖

. On the other hand,

{0, 1}

is not a subfield of

ℝ

, since

1 + 1 = 2 \notin {0, 1}

, even though

{0, 1}

with the operations of

𝔽_{2}

is a field: a subfield must use the operations of the bigger field. Similarly

ℤ

is not a subfield of

ℚ

, since

2^{- 1} = 1 / 2 \notin ℤ

Exercises

Exercise 1.1.35.routine Solve the system

𝑥 + 2 𝑦 = 3, 3 𝑥 + 4 𝑦 = 1

(a) over $𝔽_{7}$ , and (b) over $𝔽_{2}$ , reading each integer as its remainder modulo $2$ .

Solution.

Uses. Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime), Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ )

Hint 1. Eliminate

𝑥

as you would over

ℝ

, by subtracting

3

times the first equation from the second. Every step is field arithmetic, so it is valid over

𝔽_{7}

Hint 2. In

𝔽_{7}

2^{- 1} = 4

. Over

𝔽_{2}

, first rewrite the coefficients:

2 = 4 = 0

and

3 = 1

Sketch. Over

𝔽_{7}

, elimination gives

- 2 𝑦 = - 1

, so

𝑦 = 4

and

𝑥 = 2

; this pair checks. Over

𝔽_{2}

the system collapses to

𝑥 = 1

twice, so

𝑦

is free and there are two solutions.

(a) We work in $𝔽_{7}$ , where $2^{- 1} = 4$ since $2 \cdot 4 = 8 = 7 + 1$ . Suppose $(𝑥, 𝑦)$ is a solution. Subtracting $3$ times the first equation from the second gives

(3 𝑥 + 4 𝑦) - 3 (𝑥 + 2 𝑦) = 1 - 3 \cdot 3, that is, - 2 𝑦 = - 8 .

In $𝔽_{7}$ , $- 8 = - 1$ , so $2 𝑦 = 1$ by Proposition 1.1.16, and $𝑦 = 2^{- 1} = 4$ . The first equation then gives $𝑥 = 3 - 2 \cdot 4 = - 5 = 2$ . So $(2, 4)$ is the only possible solution, and it is one: in $𝔽_{7}$ , $2 + 2 \cdot 4 = 10 = 3$ and $3 \cdot 2 + 4 \cdot 4 = 22 = 1$ .

(b) In $𝔽_{2}$ we have $2 = 4 = 0$ and $3 = 1$ , so the system reads $𝑥 = 1$ , $𝑥 = 1$ . Its solutions are $(𝑥, 𝑦) = (1, 0)$ and $(1, 1)$ . The elimination in (a) is unavailable here, because it divided by $2$ , which is $0$ in $𝔽_{2}$ .

Exercise 1.1.36.routine Write each of the following in the form

𝑎 + 𝑏 𝑖

with

𝑎, 𝑏 \in ℝ

: (a)

(3 - 2 𝑖) (1 + 4 𝑖)

; (b)

{(2 + 𝑖)}^{- 1}

; (c)

(1 + 𝑖) / (1 - 𝑖)

; (d)

{(1 + 𝑖)}^{8}

Solution.

Uses. Proposition 1.1.30 ( $ℂ$ is a field)

Hint 1. Expand products with distributivity and use

𝑖^{2} = - 1

Hint 2. For (b) and (c), use the inverse formula of Proposition 1.1.30, or multiply by

𝑎 - 𝑏 𝑖

where

𝑎 + 𝑏 𝑖

is the number being inverted. For (d), square repeatedly.

Sketch. (a)

3 +

12 𝑖 -

2 𝑖 +

8 =

11 +

10 𝑖

. (b)

(2 - 𝑖) / 5

. (c)

{(1 + 𝑖)}^{2} / 2 = 𝑖

. (d)

{(1 + 𝑖)}^{2} = 2 𝑖

, so

{(1 + 𝑖)}^{8} = {(2 𝑖)}^{4} = 16

(a) $(3 - 2 𝑖) (1 + 4 𝑖) =$ $3 +$ $12 𝑖 -$ $2 𝑖 -$ $8 𝑖^{2} =$ $11 +$ $10 𝑖$ .

(b) By Proposition 1.1.30 with $𝑎 = 2$ , $𝑏 = 1$ , and $𝑎^{2} + 𝑏^{2} = 5$ , ${(2 + 𝑖)}^{- 1} = \frac{2}{5} - \frac{1}{5} 𝑖$ . Check: $(2 + 𝑖) (2 - 𝑖) =$ $4 +$ $1 =$ $5$ .

(c) ${(1 - 𝑖)}^{- 1} = (1 + 𝑖) / 2$ , so $(1 + 𝑖) / (1 - 𝑖) =$ ${(1 + 𝑖)}^{2} / 2 =$ $(1 + 2 𝑖 + 𝑖^{2}) / 2 =$ $𝑖$ .

(d) ${(1 + 𝑖)}^{2} = 2 𝑖$ , so ${(1 + 𝑖)}^{4} = {(2 𝑖)}^{2} = - 4$ and ${(1 + 𝑖)}^{8} = {(- 4)}^{2} = 16$ .

Exercise 1.1.37.routine Let

𝑎, 𝑏 \in 𝐹

be nonzero. Show that

𝑎^{- 1} \neq 0

and

{(𝑎^{- 1})}^{- 1} = 𝑎

, and that

𝑎 𝑏 \neq 0

and

{(𝑎 𝑏)}^{- 1} = 𝑎^{- 1} 𝑏^{- 1}

Solution.

Uses. Theorem 1.1.9 (Uniqueness of identities and inverses), Proposition 1.1.15 (Multiplying by zero), Proposition 1.1.17 (No zero divisors)

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Before writing

{(𝑎^{- 1})}^{- 1}

you must know that

𝑎^{- 1} \neq 0

, because the inverse axiom only applies to nonzero elements. What would

𝑎^{- 1} = 0

give?

Hint 2. To identify an inverse, exhibit a candidate and invoke uniqueness: check that

𝑎

is an inverse of

𝑎^{- 1}

, and that

𝑎^{- 1} 𝑏^{- 1}

is an inverse of

𝑎 𝑏

Sketch. If

𝑎^{- 1} = 0

, then

1 = 𝑎 𝑎^{- 1} = 0

. Since

𝑎^{- 1} 𝑎 = 1

𝑎

is the inverse of

𝑎^{- 1}

. By Proposition 1.1.17,

𝑎 𝑏 \neq 0

, and

(𝑎 𝑏) (𝑎^{- 1} 𝑏^{- 1}) =

(𝑎 𝑎^{- 1}) (𝑏 𝑏^{- 1}) =

1

If $𝑎^{- 1} = 0$ , then $1 = 𝑎 𝑎^{- 1} = 𝑎 0 = 0$ by Proposition 1.1.15, contradicting $1 \neq 0$ . So $𝑎^{- 1} \neq 0$ , and $𝑎^{- 1}$ has a multiplicative inverse. Since $𝑎^{- 1} 𝑎 = 𝑎 𝑎^{- 1} = 1$ , the element $𝑎$ is a multiplicative inverse of $𝑎^{- 1}$ , so ${(𝑎^{- 1})}^{- 1} = 𝑎$ by Theorem 1.1.9.

By Proposition 1.1.17, $𝑎 𝑏 \neq 0$ . Reordering the product,

(𝑎 𝑏) (𝑎^{- 1} 𝑏^{- 1}) = (𝑎 𝑎^{- 1}) (𝑏 𝑏^{- 1}) = 1 \cdot 1 = 1,

so $𝑎^{- 1} 𝑏^{- 1}$ is a multiplicative inverse of $𝑎 𝑏$ , and ${(𝑎 𝑏)}^{- 1} = 𝑎^{- 1} 𝑏^{- 1}$ by Theorem 1.1.9.

Exercise 1.1.38.routine Let $𝑎, 𝑏, 𝑐, 𝑑 \in 𝐹$ with $𝑏 \neq 0$ and $𝑑 \neq 0$ . Show that

\frac{𝑎}{𝑏} + \frac{𝑐}{𝑑} = \frac{𝑎 𝑑 + 𝑏 𝑐}{𝑏 𝑑}, \frac{𝑎}{𝑏} \cdot \frac{𝑐}{𝑑} = \frac{𝑎 𝑐}{𝑏 𝑑},

and that $𝑎 / 𝑏 = 𝑐 / 𝑑$ if and only if $𝑎 𝑑 = 𝑏 𝑐$ .

Solution.

Uses. Notation 1.1.10, Exercise 1.1.37, Definition 1.1.1 (Field)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Rewrite every fraction

𝑥 / 𝑦

𝑥 𝑦^{- 1}

, and note that

𝑏 𝑑 \neq 0

with

{(𝑏 𝑑)}^{- 1} = 𝑏^{- 1} 𝑑^{- 1}

(Exercise 1.1.37).

Hint 2. Then expand with distributivity and reorder products freely. The factors

𝑏 𝑏^{- 1}

and

𝑑 𝑑^{- 1}

collapse to

1

Sketch. With

{(𝑏 𝑑)}^{- 1} = 𝑏^{- 1} 𝑑^{- 1}

, the right-hand sides expand to

𝑎 𝑏^{- 1} (𝑑 𝑑^{- 1}) +

𝑐 𝑑^{- 1} (𝑏 𝑏^{- 1})

and

(𝑎 𝑏^{- 1}) (𝑐 𝑑^{- 1})

. For the last claim, multiply

𝑎 / 𝑏 = 𝑐 / 𝑑

𝑏 𝑑

, and

𝑎 𝑑 = 𝑏 𝑐

𝑏^{- 1} 𝑑^{- 1}

By Exercise 1.1.37, $𝑏 𝑑 \neq 0$ and ${(𝑏 𝑑)}^{- 1} = 𝑏^{- 1} 𝑑^{- 1}$ . Using distributivity and reordering products,

\frac{𝑎 𝑑 + 𝑏 𝑐}{𝑏 𝑑} = (𝑎 𝑑 + 𝑏 𝑐) 𝑏^{- 1} 𝑑^{- 1} = 𝑎 𝑏^{- 1} (𝑑 𝑑^{- 1}) + 𝑐 𝑑^{- 1} (𝑏 𝑏^{- 1}) = 𝑎 𝑏^{- 1} + 𝑐 𝑑^{- 1} = \frac{𝑎}{𝑏} + \frac{𝑐}{𝑑} .

Similarly,

\frac{𝑎}{𝑏} \cdot \frac{𝑐}{𝑑} = 𝑎 𝑏^{- 1} 𝑐 𝑑^{- 1} = (𝑎 𝑐) (𝑏^{- 1} 𝑑^{- 1}) = (𝑎 𝑐) {(𝑏 𝑑)}^{- 1} = \frac{𝑎 𝑐}{𝑏 𝑑} .

For the last claim, suppose $𝑎 𝑏^{- 1} = 𝑐 𝑑^{- 1}$ . Multiplying both sides by $𝑏 𝑑$ gives $𝑎 𝑑 (𝑏 𝑏^{- 1}) =$ $𝑏 𝑐 (𝑑 𝑑^{- 1})$ , that is, $𝑎 𝑑 = 𝑏 𝑐$ . Conversely, if $𝑎 𝑑 = 𝑏 𝑐$ , multiplying both sides by $𝑏^{- 1} 𝑑^{- 1}$ gives $𝑎 𝑏^{- 1} (𝑑 𝑑^{- 1}) =$ $𝑐 𝑑^{- 1} (𝑏 𝑏^{- 1})$ , that is, $𝑎 / 𝑏 = 𝑐 / 𝑑$ .

Exercise 1.1.39.routine Let

𝐹

be a set with operations

+

and

\cdot

satisfying all the field axioms except that

1 \neq 0

is not required, and suppose that in fact

1 = 0

. Show that

𝐹

has exactly one element.

Solution.

Uses. Proposition 1.1.15 (Multiplying by zero)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Compute

𝑎 \cdot 1

for an arbitrary

𝑎 \in 𝐹

, in two ways.

Hint 2. You will want

𝑎 0 = 0

. Check that the proof of Proposition 1.1.15 never used

1 \neq 0

Sketch.

𝑎 = 𝑎 \cdot 1 = 𝑎 \cdot 0 = 0

for every

𝑎

, where

𝑎 0 = 0

holds because the proof of Proposition 1.1.15 used only distributivity, the additive axioms, and cancellation.

The set $𝐹$ is nonempty because it contains $0$ . The proof of Proposition 1.1.15 used distributivity, the additive identity axiom, and additive cancellation, which in turn rests only on the additive axioms; none of these involve $1 \neq 0$ . So $𝑎 0 = 0$ for every $𝑎 \in 𝐹$ . Now for every $𝑎 \in 𝐹$ ,

𝑎 = 𝑎 \cdot 1 = 𝑎 \cdot 0 = 0,

using the multiplicative identity axiom and $1 = 0$ . Hence $𝐹 = {0}$ .

Exercise 1.1.40.routine Prove or disprove: in every field, if

𝑎^{2} = 𝑏^{2}

, then

𝑎 = 𝑏

𝑎 = - 𝑏

Solution.

Uses. Proposition 1.1.16 (Sign rules), Proposition 1.1.17 (No zero divisors)

Hint 1. Move everything to one side and try to factor

𝑎^{2} - 𝑏^{2}

Hint 2.

(𝑎 - 𝑏) (𝑎 + 𝑏) =

𝑎^{2} -

𝑏^{2}

in every field. Now use Proposition 1.1.17.

Sketch. True:

(𝑎 - 𝑏) (𝑎 + 𝑏) =

𝑎^{2} -

𝑏^{2} =

0

, so

𝑎 - 𝑏 = 0

𝑎 + 𝑏 = 0

The claim is true. Expanding with distributivity and Proposition 1.1.16,

(𝑎 - 𝑏) (𝑎 + 𝑏) = 𝑎^{2} + 𝑎 𝑏 - 𝑏 𝑎 - 𝑏^{2} = 𝑎^{2} - 𝑏^{2},

since $𝑎 𝑏 -$ $𝑏 𝑎 =$ $𝑎 𝑏 -$ $𝑎 𝑏 =$ $0$ . If $𝑎^{2} = 𝑏^{2}$ , then $𝑎^{2} - 𝑏^{2} = 0$ , so $(𝑎 - 𝑏) (𝑎 + 𝑏) = 0$ . By Proposition 1.1.17, $𝑎 - 𝑏 = 0$ or $𝑎 + 𝑏 = 0$ , that is, $𝑎 = 𝑏$ or $𝑎 = - 𝑏$ .

In $𝔽_{2}$ we have $- 𝑏 = 𝑏$ , so the conclusion just says $𝑎 = 𝑏$ . That is consistent: $0^{2} = 0$ and $1^{2} = 1$ are different.

Exercise 1.1.41.routine Prove or disprove: for every prime

𝑝

, the equation

𝑥^{2} + 1 = 0

has a solution in

𝔽_{𝑝}

Solution.

Uses. Definition 1.1.19 (Integers modulo $𝑛$ )

Hint 1. Try small primes one at a time. Each

𝔽_{𝑝}

has only

𝑝

elements to test.

Hint 2.

𝑝 = 2

and

𝑝 = 5

work. Try

𝑝 = 3

Sketch. False: in

𝔽_{3}

𝑥^{2} + 1

takes the values

1, 2, 2

𝑥 = 0, 1, 2

The claim is false. In $𝔽_{3}$ ,

0^{2} + 1 = 1, 1^{2} + 1 = 2, 2^{2} + 1 = 5 mod 3 = 2,

so $𝑥^{2} + 1 = 0$ has no solution. (It does for some primes: $𝑥 = 1$ in $𝔽_{2}$ , and $𝑥 = 2$ or $𝑥 = 3$ in $𝔽_{5}$ .)

Exercise 1.1.42.routine Find the error in the following argument.

Claim. In every field, the equation $𝑥^{2} = 1$ has exactly two solutions.

Proof. We have $𝑥^{2} -$ $1 =$ $(𝑥 - 1) (𝑥 + 1)$ , so by Proposition 1.1.17, $𝑥^{2} = 1$ holds if and only if $𝑥 = 1$ or $𝑥 = - 1$ . These are the two solutions.

Solution.

Uses. Proposition 1.1.17 (No zero divisors), Example 1.1.4 (The field with two elements)

Hint 1. The factorization and the use of Proposition 1.1.17 are correct. Test the conclusion in the smallest field.

Hint 2. In

𝔽_{2}

, what is

- 1

Sketch. The argument shows that the solutions are exactly

1

and

- 1

, but these coincide when

1 + 1 = 0

. In

𝔽_{2}

there is only one solution.

The argument correctly shows that the solutions of

𝑥^{2} = 1

are exactly

𝑥 = 1

and

𝑥 = - 1

. The error is the last sentence:

1

and

- 1

need not be different. In

𝔽_{2}

1 + 1 = 0

, so

- 1 = 1

, and

𝑥^{2} = 1

has exactly one solution. In general

1 = - 1

if and only if

1 + 1 = 0

, so the correct statement is that

𝑥^{2} = 1

has exactly two solutions when

1 + 1 \neq 0

, and exactly one when

1 + 1 = 0

Exercise 1.1.43.routine Let

𝑎, 𝑏 \in 𝐹

, and define

𝑓 : 𝐹 \to 𝐹

𝑓 (𝑥) = 𝑎 𝑥 + 𝑏

. Show that

𝑓

is a bijection if and only if

𝑎 \neq 0

, and find

𝑓^{- 1}

in that case.

Solution.

Uses. Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ ), Proposition 1.1.15 (Multiplying by zero)

Hint 1.

𝑓

is a bijection if and only if, for every

𝑦 \in 𝐹

, the equation

𝑎 𝑥 + 𝑏 = 𝑦

has exactly one solution

𝑥

Hint 2. For

𝑎 \neq 0

, use Proposition 1.1.13 twice. For

𝑎 = 0

𝑓

is constant. Why is a constant function on

𝐹

not injective?

Sketch. If

𝑎 \neq 0

, then

𝑎 𝑥 + 𝑏 = 𝑦

if and only if

𝑥 = 𝑎^{- 1} (𝑦 - 𝑏)

, so

𝑓

is bijective with

𝑓^{- 1} (𝑦) =

𝑎^{- 1} (𝑦 - 𝑏)

. If

𝑎 = 0

, then

𝑓 (0) = 𝑓 (1) = 𝑏

with

0 \neq 1

Suppose $𝑎 \neq 0$ , and let $𝑦 \in 𝐹$ . By part (a) of Proposition 1.1.13, applied with unknown $𝑎 𝑥$ , the equation $𝑎 𝑥 + 𝑏 = 𝑦$ holds if and only if $𝑎 𝑥 = 𝑦 - 𝑏$ , and by part (b) this holds if and only if $𝑥 = 𝑎^{- 1} (𝑦 - 𝑏)$ . So every $𝑦$ has exactly one preimage, that is, $𝑓$ is a bijection, and $𝑓^{- 1} (𝑦) =$ $𝑎^{- 1} (𝑦 - 𝑏)$ .

Suppose $𝑎 = 0$ . Then $𝑓 (𝑥) = 0 𝑥 + 𝑏 = 𝑏$ for every $𝑥$ , by Proposition 1.1.15. In particular $𝑓 (0) = 𝑓 (1)$ , and $0 \neq 1$ , so $𝑓$ is not injective.

Exercise 1.1.44.standard

(a)Show that the only complex numbers $𝑧$ with $𝑧^{2} = - 1$ are $𝑖$ and $- 𝑖$ .
(b)Find all $𝑧 \in ℂ$ with $𝑧^{2} = 𝑖$ .

Solution.

Uses. Proposition 1.1.17 (No zero divisors), Proposition 1.1.30 ( $ℂ$ is a field), Exercise 1.1.40

Hint 1. For (a), factor

𝑧^{2} + 1

using

𝑖^{2} = - 1

, then use that a field has no zero divisors.

Hint 2. For (b), write

𝑧 = 𝑎 + 𝑏 𝑖

with

𝑎, 𝑏 \in ℝ

and compare the real and imaginary parts of

𝑧^{2} = 𝑖

Hint 3. You get

𝑎^{2} = 𝑏^{2}

and

2 𝑎 𝑏 = 1

. So

𝑏 = \pm 𝑎

, and one sign is impossible.

Sketch. (a)

𝑧^{2} +

1 =

(𝑧 - 𝑖) (𝑧 + 𝑖)

. (b)

𝑧^{2} =

(𝑎^{2} - 𝑏^{2}) +

2 𝑎 𝑏 𝑖

, so

𝑎^{2} = 𝑏^{2}

and

2 𝑎 𝑏 = 1

; this forces

𝑏 = 𝑎

and

𝑎 = \pm 1 / \sqrt{2}

, so

𝑧 = \pm (1 + 𝑖) / \sqrt{2}

(a) Since $𝑖^{2} = - 1$ , we have $(𝑧 - 𝑖) (𝑧 + 𝑖) =$ $𝑧^{2} -$ $𝑖^{2} =$ $𝑧^{2} +$ $1$ . So $𝑧^{2} = - 1$ if and only if $(𝑧 - 𝑖) (𝑧 + 𝑖) = 0$ , which by Proposition 1.1.17 holds if and only if $𝑧 = 𝑖$ or $𝑧 = - 𝑖$ . These are different, since $𝑖 - (- 𝑖) = 2 𝑖 \neq 0$ .

(b) Write $𝑧 = 𝑎 + 𝑏 𝑖$ with $𝑎, 𝑏 \in ℝ$ . Then $𝑧^{2} =$ $(𝑎^{2} - 𝑏^{2}) +$ $2 𝑎 𝑏 𝑖$ , so $𝑧^{2} = 𝑖$ if and only if $𝑎^{2} - 𝑏^{2} = 0$ and $2 𝑎 𝑏 = 1$ . The first equation gives $𝑏 = 𝑎$ or $𝑏 = - 𝑎$ (Exercise 1.1.40, in $ℝ$ ). If $𝑏 = - 𝑎$ , then $2 𝑎 𝑏 = - 2 𝑎^{2} \leq 0$ , which is not $1$ . So $𝑏 = 𝑎$ and $2 𝑎^{2} = 1$ , giving $𝑎 = \pm 1 / \sqrt{2}$ . Hence

𝑧 = \frac{1 + 𝑖}{\sqrt{2}} or 𝑧 = - \frac{1 + 𝑖}{\sqrt{2}},

and both work: ${((1 + 𝑖) / \sqrt{2})}^{2} =$ $(1 + 2 𝑖 + 𝑖^{2}) / 2 =$ $𝑖$ .

Once one square root $𝑤$ of $𝑖$ is known, (a)‘s method finds all of them: $𝑧^{2} = 𝑤^{2}$ means $(𝑧 - 𝑤) (𝑧 + 𝑤) = 0$ , so $𝑧 = \pm 𝑤$ .

Exercise 1.1.45.standard Find the error in the following argument.

Claim. $ℤ / 𝑛 ℤ$ is a field for every $𝑛 \geq 2$ .

Proof. By Lemma 1.1.21, only multiplicative inverses need checking. Let $𝑎 \neq 0$ in $ℤ / 𝑛 ℤ$ , and define $𝑚 (𝑥) = 𝑎 𝑥$ . If $𝑚 (𝑥) = 𝑚 (𝑦)$ , then $𝑎 𝑥 = 𝑎 𝑦$ , so $𝑥 = 𝑦$ by cancellation (Corollary 1.1.14). So $𝑚$ is an injective function from a finite set to itself, hence surjective, and $𝑎 𝑥 = 1$ for some $𝑥$ .

Solution.

Uses. Corollary 1.1.14 (Cancellation), Non-example 1.1.23 ( $ℤ / 6 ℤ$ ), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime)

Hint 1. Every step but one is valid for all

𝑛

. Check what each cited result assumes.

Hint 2. Where did Corollary 1.1.14 come from, and does

ℤ / 6 ℤ

satisfy it? Try

𝑎 = 2

Sketch. Multiplicative cancellation was proved for fields, using

𝑎^{- 1}

. Applying it to

ℤ / 𝑛 ℤ

assumes the inverse we are trying to produce. It fails in

ℤ / 6 ℤ

2 \cdot 0 = 2 \cdot 3

The error is the appeal to Corollary 1.1.14. That result was proved for fields, and its multiplicative part multiplies by

𝑎^{- 1}

, which is exactly what the argument is trying to show exists. So the argument is circular. Cancellation genuinely fails in

ℤ / 6 ℤ

2 \cdot 0 = 0 = 2 \cdot 3

but

0 \neq 3

, so

𝑚

is not injective there (Non-example 1.1.23). The rest of the argument is fine, and the counting proof of Proposition 1.1.24 repairs it by proving injectivity from Euclid’s lemma, which is where primality of

𝑛

enters.

Exercise 1.1.46.standard

(a)Let $𝐹$ be a field with exactly three elements, $𝐹 = {0, 1, 𝑐}$ . Show that
$1 + 1 = 𝑐, 1 + 𝑐 = 0, 𝑐 + 𝑐 = 1, 𝑐 \cdot 𝑐 = 1 .$
Conclude that the addition and multiplication tables of $𝐹$ are those of $𝔽_{3}$ with $𝑐$ in place of $2$ .
(b)Find all fields $𝐹$ in which $𝑥^{3} = 𝑥$ for every $𝑥 \in 𝐹$ .

Solution.

Uses. Corollary 1.1.14 (Cancellation), Theorem 1.1.9 (Uniqueness of identities and inverses), Proposition 1.1.15 (Multiplying by zero), Proposition 1.1.17 (No zero divisors), Exercise 1.1.40

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. In (a), determine

1 + 1

first. Rule out

1 + 1 = 1

with cancellation, and

1 + 1 = 0

by asking what

1 + 𝑐

could then be.

Hint 2. The other entries are forced the same way. Products involving

0

1

are already known from the axioms and Proposition 1.1.15.

Hint 3. In (b), factor

𝑥^{3} -

𝑥 =

𝑥 (𝑥 - 1) (𝑥 + 1)

and use Proposition 1.1.17 to show

𝐹 \subseteq {0, 1, - 1}

Sketch. (a) Each sum is pinned down by cancellation or by uniqueness of inverses;

𝑐^{2} = 1

because

𝑐^{2} = 0

and

𝑐^{2} = 𝑐

are impossible. (b)

𝑥^{3} = 𝑥

forces

𝑥 \in {0, 1, - 1}

, so

𝐹

has two or three elements, and it is

𝔽_{2}

𝔽_{3}

up to renaming; both satisfy

𝑥^{3} = 𝑥

(a) Sums involving $0$ , and products involving $0$ or $1$ , are determined by the identity axioms and Proposition 1.1.15, so we need $1 + 1$ , $1 + 𝑐$ , $𝑐 + 𝑐$ , and $𝑐 \cdot 𝑐$ . Throughout, $0$ , $1$ , $𝑐$ are distinct.

$1 + 1$ . If $1 + 1 = 1 = 1 + 0$ , then $1 = 0$ by cancellation, which is false. Suppose $1 + 1 = 0$ . If $1 + 𝑐 = 1 = 1 + 0$ , then $𝑐 = 0$ ; if $1 + 𝑐 = 𝑐 = 0 + 𝑐$ , then $1 = 0$ ; both are false, so $1 + 𝑐 = 0$ . But then $1$ and $𝑐$ are both additive inverses of $1$ , so $𝑐 = 1$ by Theorem 1.1.9, which is false. Hence $1 + 1 = 𝑐$ .

$1 + 𝑐$ . As above, $1 + 𝑐 = 1$ forces $𝑐 = 0$ and $1 + 𝑐 = 𝑐$ forces $1 = 0$ . So $1 + 𝑐 = 0$ , and in particular $- 𝑐 = 1$ .

$𝑐 + 𝑐$ . If $𝑐 + 𝑐 = 𝑐 = 𝑐 + 0$ , then $𝑐 = 0$ . If $𝑐 + 𝑐 = 0$ , then $- 𝑐 = 𝑐$ , but $- 𝑐 = 1$ , so $𝑐 = 1$ . Both are false, so $𝑐 + 𝑐 = 1$ .

$𝑐 \cdot 𝑐$ . If $𝑐^{2} = 0$ , then $𝑐 = 0$ by Proposition 1.1.17. If $𝑐^{2} = 𝑐 = 𝑐 \cdot 1$ , then $𝑐 = 1$ by cancellation (as $𝑐 \neq 0$ ). So $𝑐^{2} = 1$ .

In $𝔽_{3}$ , $1 + 1 = 2$ , $1 + 2 = 3 mod 3 = 0$ , $2 + 2 = 4 mod 3 = 1$ , and $2 \cdot 2 = 4 mod 3 = 1$ . So the tables of $𝐹$ are those of $𝔽_{3}$ with $𝑐$ in place of $2$ .

(b) Suppose $𝑥^{3} = 𝑥$ for all $𝑥 \in 𝐹$ . As in Exercise 1.1.40, $(𝑥 - 1) (𝑥 + 1) =$ $𝑥^{2} -$ $1$ , so $𝑥 (𝑥 - 1) (𝑥 + 1) =$ $𝑥^{3} -$ $𝑥 =$ $0$ for every $𝑥$ . By Proposition 1.1.17 (applied twice), every $𝑥$ equals $0$ , $1$ , or $- 1$ . So $𝐹 \subseteq {0, 1, - 1}$ .

If $1 + 1 = 0$ , then $- 1 = 1$ and $𝐹 = {0, 1}$ . In a field with two elements every sum and product is forced as above, with $1 + 1 = 0$ because $1 + 1 = 1$ would give $1 = 0$ ; so $𝐹$ has the tables of $𝔽_{2}$ . If $1 + 1 \neq 0$ , then $- 1 \neq 1$ , and $- 1 \neq 0$ , so $𝐹 = {0, 1, - 1}$ has exactly three elements, and by (a) $𝐹$ has the tables of $𝔽_{3}$ (with $- 1$ in place of $2$ ).

Conversely, $𝑥^{3} = 𝑥$ holds in $𝔽_{2}$ ( $0^{3} = 0$ , $1^{3} = 1$ ) and in $𝔽_{3}$ ( $2^{3} = 8 mod 3 = 2$ ). So the fields with $𝑥^{3} = 𝑥$ for all $𝑥$ are exactly $𝔽_{2}$ and $𝔽_{3}$ , up to renaming the elements.

Exercise 1.1.47.standard Show that the following are equivalent for a field $𝐹$ :

(i) $1 + 1 = 0$ ;
(ii) $𝑎 = - 𝑎$ for every $𝑎 \in 𝐹$ ;
(iii)the map $𝐹 \to 𝐹$ , $𝑥 \mapsto 𝑥^{2}$ , is injective.

Solution.

Uses. Proposition 1.1.15 (Multiplying by zero), Proposition 1.1.16 (Sign rules), Theorem 1.1.9 (Uniqueness of identities and inverses), Exercise 1.1.40

Techniques. Uniqueness via two candidates (Technique 1.1.8)

Hint 1. Prove the cycle (i)

\Rightarrow

(ii)

\Rightarrow

(iii)

\Rightarrow

(i).

Hint 2. (i)

\Rightarrow

(ii):

𝑎 + 𝑎 = (1 + 1) 𝑎

. (ii)

\Rightarrow

(iii): use Exercise 1.1.40. (iii)

\Rightarrow

(i): compare the squares of

1

and

- 1

Sketch. If

1 + 1 = 0

, then

𝑎 + 𝑎 = (1 + 1) 𝑎 = 0

, so

𝑎 = - 𝑎

. If

𝑎 = - 𝑎

always, then

𝑥^{2} = 𝑦^{2}

gives

𝑥 = \pm 𝑦 = 𝑦

. If squaring is injective, then

1^{2} = {(- 1)}^{2}

gives

1 = - 1

, so

1 + 1 = 0

We prove (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i).

(i) $\Rightarrow$ (ii). Let $𝑎 \in 𝐹$ . By distributivity and Proposition 1.1.15, $𝑎 +$ $𝑎 =$ $1 𝑎 +$ $1 𝑎 =$ $(1 + 1) 𝑎 =$ $0 𝑎 =$ $0$ . So $𝑎$ is an additive inverse of $𝑎$ , and $𝑎 = - 𝑎$ by Theorem 1.1.9.

(ii) $\Rightarrow$ (iii). Suppose $𝑥^{2} = 𝑦^{2}$ . By Exercise 1.1.40, $𝑥 = 𝑦$ or $𝑥 = - 𝑦$ , and $- 𝑦 = 𝑦$ by (ii). So $𝑥 = 𝑦$ .

(iii) $\Rightarrow$ (i). By Proposition 1.1.16, ${(- 1)}^{2} = 1 \cdot 1 = 1^{2}$ . Injectivity gives $- 1 = 1$ , so $1 + 1 = 1 + (- 1) = 0$ .

Exercise 1.1.48.standard Show that the characteristic of a field is either

0

or a prime number.

Solution.

Uses. Definition 1.1.28 (Characteristic), Proposition 1.1.17 (No zero divisors)

Hint 1. Suppose

char 𝐹 = 𝑚 > 0

and

𝑚 = 𝑟 𝑠

with

1 < 𝑟, 𝑠 < 𝑚

. What can you say about

(𝑟 \cdot 1) (𝑠 \cdot 1)

Hint 2. Show that

(𝑟 \cdot 1) (𝑠 \cdot 1) =

(𝑟 𝑠) \cdot

1

using distributivity, then use Proposition 1.1.17 and the minimality of

𝑚

Sketch.

(𝑟 \cdot 1) (𝑠 \cdot 1) =

(𝑟 𝑠) \cdot

1

by distributivity. If the characteristic

𝑚

factored as

𝑟 𝑠

with

1 < 𝑟, 𝑠 < 𝑚

, then

(𝑟 \cdot 1) (𝑠 \cdot 1) =

𝑚 \cdot

1 =

0

would force

𝑟 \cdot 1 = 0

𝑠 \cdot 1 = 0

, contradicting the minimality of

𝑚

First, for positive integers $𝑟$ and $𝑠$ , distributivity gives

(𝑟 \cdot 1) (𝑠 \cdot 1) = (𝑟 \cdot 1) (1 + \dots + 1) = (𝑟 \cdot 1) + \dots + (𝑟 \cdot 1) = (𝑟 𝑠) \cdot 1,

where there are $𝑠$ terms in the middle expressions, and the last sum consists of $𝑟 𝑠$ ones.

Suppose $char 𝐹 = 𝑚 > 0$ . Then $𝑚 \neq 1$ , since $1 \cdot 1 = 1 \neq 0$ . Suppose for contradiction that $𝑚$ is not prime, so $𝑚 = 𝑟 𝑠$ with $1 < 𝑟, 𝑠 < 𝑚$ . Then $(𝑟 \cdot 1) (𝑠 \cdot 1) =$ $𝑚 \cdot$ $1 =$ $0$ , so by Proposition 1.1.17, $𝑟 \cdot 1 = 0$ or $𝑠 \cdot 1 = 0$ . Either way some positive integer smaller than $𝑚$ kills $1$ , contradicting the choice of $𝑚$ as the least such integer. Hence $𝑚$ is prime.

Exercise 1.1.49.standard Let

ℚ (\sqrt{2}) =

{𝑎 + 𝑏 \sqrt{2} | 𝑎, 𝑏 \in ℚ}

. Show that

ℚ (\sqrt{2})

is a subfield of

ℝ

Solution.

Uses. Definition 1.1.32 (Subfield), $\sqrt{2}$ is irrational

Hint 1. Check the conditions of Definition 1.1.32 one at a time. All but one are immediate.

Hint 2. For the inverse of

𝑎 + 𝑏 \sqrt{2} \neq 0

, imitate the complex case: multiply by

𝑎 - 𝑏 \sqrt{2}

Hint 3. Why is

𝑎^{2} - 2 𝑏^{2} \neq 0

? Use that

\sqrt{2}

is irrational.

Sketch. Closure under

+

\cdot

, and negation is a direct computation. For

𝑎 + 𝑏 \sqrt{2} \neq 0

(𝑎 + 𝑏 \sqrt{2}) (𝑎 - 𝑏 \sqrt{2}) =

𝑎^{2} -

2 𝑏^{2}

, which is a nonzero rational because

\sqrt{2} \notin ℚ

; dividing

𝑎 - 𝑏 \sqrt{2}

by it gives the inverse.

Write $𝐾 = ℚ (\sqrt{2})$ . Clearly $0 = 0 + 0 \sqrt{2}$ and $1 = 1 + 0 \sqrt{2}$ lie in $𝐾$ . For $𝑎, 𝑏, 𝑐, 𝑑 \in ℚ$ ,

\begin{aligned} (𝑎 + 𝑏 \sqrt{2}) + (𝑐 + 𝑑 \sqrt{2}) & = (𝑎 + 𝑐) + (𝑏 + 𝑑) \sqrt{2}, \\ (𝑎 + 𝑏 \sqrt{2}) (𝑐 + 𝑑 \sqrt{2}) & = (𝑎 𝑐 + 2 𝑏 𝑑) + (𝑎 𝑑 + 𝑏 𝑐) \sqrt{2}, \\ - (𝑎 + 𝑏 \sqrt{2}) & = (- 𝑎) + (- 𝑏) \sqrt{2}, \end{aligned}

and these lie in $𝐾$ because $ℚ$ is closed under its operations.

Now let $𝑧 = 𝑎 + 𝑏 \sqrt{2} \neq 0$ . We claim $𝑎^{2} - 2 𝑏^{2} \neq 0$ . If $𝑏 = 0$ , then $𝑎 = 𝑧 \neq 0$ , so $𝑎^{2} - 2 𝑏^{2} = 𝑎^{2} \neq 0$ . If $𝑏 \neq 0$ and $𝑎^{2} = 2 𝑏^{2}$ , then ${(𝑎 / 𝑏)}^{2} = 2$ , so $\sqrt{2} = | 𝑎 / 𝑏 |$ would be rational, which is false. Hence $𝑞 = 𝑎^{2} - 2 𝑏^{2}$ is a nonzero rational, and

(𝑎 + 𝑏 \sqrt{2}) (\frac{𝑎}{𝑞} - \frac{𝑏}{𝑞} \sqrt{2}) = \frac{𝑎^{2} - 2 𝑏^{2}}{𝑞} = 1 .

So $𝑧^{- 1} =$ $𝑎 / 𝑞 -$ $(𝑏 / 𝑞) \sqrt{2} \in$ $𝐾$ . Thus $𝐾$ is a subfield of $ℝ$ , and by Proposition 1.1.33 it is a field.

Exercise 1.1.50.standard Let $𝐹$ be a field with $1 + 1 \neq 0$ . Write $2 = 1 + 1$ and $4 = 2 \cdot 2$ , and let $𝑎, 𝑏, 𝑐 \in 𝐹$ with $𝑎 \neq 0$ .

(a)Show that $𝑎 𝑥^{2} + 𝑏 𝑥 + 𝑐 = 0$ has a solution $𝑥 \in 𝐹$ if and only if $𝑏^{2} - 4 𝑎 𝑐 = 𝑑^{2}$ for some $𝑑 \in 𝐹$ , and that in that case the solutions are $𝑥 = (- 𝑏 \pm 𝑑) / (2 𝑎)$ .
(b)Show that (a) fails without the hypothesis $1 + 1 \neq 0$ : find $𝑎, 𝑏, 𝑐 \in 𝔽_{2}$ with $𝑎 \neq 0$ such that $𝑏^{2} - 4 𝑎 𝑐$ is a square but $𝑎 𝑥^{2} + 𝑏 𝑥 + 𝑐$ has no root in $𝔽_{2}$ .

Solution.

Uses. Proposition 1.1.17 (No zero divisors), Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ ), Exercise 1.1.40

Hint 1. Complete the square. Multiplying the equation by

4 𝑎

is allowed and reversible, because

4 𝑎 \neq 0

Hint 2. Check that

4 𝑎 (𝑎 𝑥^{2} + 𝑏 𝑥 + 𝑐) =

{(2 𝑎 𝑥 + 𝑏)}^{2} -

(𝑏^{2} - 4 𝑎 𝑐)

Hint 3. For (b), note that

4 = 0

𝔽_{2}

, so

𝑏^{2} - 4 𝑎 𝑐 = 𝑏^{2}

is always a square. Try

𝑥^{2} + 𝑥 + 1

Sketch. Since

4 𝑎 \neq 0

, the equation is equivalent to

{(2 𝑎 𝑥 + 𝑏)}^{2} =

𝑏^{2} -

4 𝑎 𝑐

. A solution

𝑥

gives

𝑑 = 2 𝑎 𝑥 + 𝑏

; conversely

𝑑

gives

𝑥 = (𝑑 - 𝑏) / (2 𝑎)

, and Exercise 1.1.40 shows every solution has this form for

𝑑

- 𝑑

. In

𝔽_{2}

𝑥^{2} + 𝑥 + 1

has discriminant

1 = 1^{2}

but no root.

(a) Since $2 \neq 0$ and $𝑎 \neq 0$ , Proposition 1.1.17 gives $2 𝑎 \neq 0$ and $4 𝑎 = 2 \cdot 2 \cdot 𝑎 \neq 0$ . Expanding,

{(2 𝑎 𝑥 + 𝑏)}^{2} - (𝑏^{2} - 4 𝑎 𝑐) = 4 𝑎^{2} 𝑥^{2} + 4 𝑎 𝑏 𝑥 + 4 𝑎 𝑐 = 4 𝑎 (𝑎 𝑥^{2} + 𝑏 𝑥 + 𝑐) .

Because $4 𝑎 \neq 0$ , a product $4 𝑎 𝑦$ is $0$ exactly when $𝑦 = 0$ . So $𝑥$ is a solution if and only if ${(2 𝑎 𝑥 + 𝑏)}^{2} =$ $𝑏^{2} -$ $4 𝑎 𝑐$ .

If $𝑥$ is a solution, then $𝑑 = 2 𝑎 𝑥 + 𝑏$ satisfies $𝑑^{2} = 𝑏^{2} - 4 𝑎 𝑐$ . Conversely, suppose $𝑑^{2} = 𝑏^{2} - 4 𝑎 𝑐$ . If $𝑥$ is a solution, then ${(2 𝑎 𝑥 + 𝑏)}^{2} = 𝑑^{2}$ , so $2 𝑎 𝑥 + 𝑏 = 𝑑$ or $2 𝑎 𝑥 + 𝑏 = - 𝑑$ (Exercise 1.1.40), and solving for $𝑥$ (Proposition 1.1.13) gives $𝑥 = (- 𝑏 + 𝑑) / (2 𝑎)$ or $𝑥 = (- 𝑏 - 𝑑) / (2 𝑎)$ . Both of these are solutions, since for them ${(2 𝑎 𝑥 + 𝑏)}^{2} =$ ${(\pm 𝑑)}^{2} =$ $𝑑^{2} =$ $𝑏^{2} -$ $4 𝑎 𝑐$ . So the solutions are exactly $𝑥 = (- 𝑏 \pm 𝑑) / (2 𝑎)$ .

(b) Take $𝑎 = 𝑏 = 𝑐 = 1$ . In $𝔽_{2}$ , $4 = 0$ , so $𝑏^{2} - 4 𝑎 𝑐 = 1 = 1^{2}$ is a square. But $0^{2} + 0 + 1 = 1$ and $1^{2} + 1 + 1 = 1$ , so $𝑥^{2} + 𝑥 + 1$ has no root in $𝔽_{2}$ . The proof of (a) breaks at the first step, which multiplies by $4 𝑎 = 0$ .

Exercise 1.1.51.standard Let $𝑝$ be prime and let $𝑎 \in 𝔽_{𝑝}$ be nonzero.

(a)Show that the products $𝑎 \cdot$ $1,$ $𝑎 \cdot$ $2,$ $\dots,$ $𝑎 \cdot$ $(𝑝 - 1)$ are the elements $1, 2, \dots, 𝑝 - 1$ of $𝔽_{𝑝}$ in some order.
(b)Deduce that $𝑎^{𝑝 - 1} = 1$ in $𝔽_{𝑝}$ . (This is Fermat’s little theorem.)

Solution.

Uses. Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime), Proposition 1.1.17 (No zero divisors), Corollary 1.1.14 (Cancellation)

Hint 1. For (a), show that the

𝑝 - 1

products are nonzero and pairwise distinct. You may now use that

𝔽_{𝑝}

is a field.

Hint 2. For (b), multiply all

𝑝 - 1

products together, in two ways.

Hint 3. The product

𝑃 = 1 \cdot 2 \dots (𝑝 - 1)

is nonzero, so it can be cancelled.

Sketch. The products

𝑎 𝑥

for

𝑥 \neq 0

are nonzero and distinct, so they list the

𝑝 - 1

nonzero elements. Multiplying them all gives

𝑎^{𝑝 - 1} 𝑃 = 𝑃

with

𝑃 = 1 \cdot 2 \dots (𝑝 - 1) \neq 0

; cancel

𝑃

By Proposition 1.1.24, $𝔽_{𝑝}$ is a field.

(a) Each product $𝑎 𝑥$ with $𝑥 \in {1, \dots, 𝑝 - 1}$ is nonzero by Proposition 1.1.17. If $𝑎 𝑥 = 𝑎 𝑦$ , then $𝑥 = 𝑦$ by Corollary 1.1.14, which applies because $𝑎 \neq 0$ in a field. So the products are $𝑝 - 1$ distinct elements of the $(𝑝 - 1)$ -element set ${1, \dots, 𝑝 - 1}$ , hence all of it.

(b) Let $𝑃 = 1 \cdot 2 \dots (𝑝 - 1)$ , which is nonzero by Proposition 1.1.17. By (a), the product of $𝑎 \cdot 1, \dots, 𝑎 \cdot (𝑝 - 1)$ is the product of $1, \dots, 𝑝 - 1$ in another order, which is $𝑃$ . Reordering factors, the same product is $𝑎^{𝑝 - 1} 𝑃$ . So $𝑎^{𝑝 - 1} 𝑃 = 𝑃 = 1 \cdot 𝑃$ , and cancelling $𝑃$ (Corollary 1.1.14) gives $𝑎^{𝑝 - 1} = 1$ .

Contrast this use of cancellation with Exercise 1.1.45: here we already know that $𝔽_{𝑝}$ is a field.

Exercise 1.1.52.standard Let

𝑝

be an odd prime. How many elements of

𝔽_{𝑝}

are squares, that is, of the form

𝑥^{2}

for some

𝑥 \in 𝔽_{𝑝}

Solution.

Uses. Exercise 1.1.40, Proposition 1.1.17 (No zero divisors), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime)

Hint 1.

0

is a square. For the nonzero squares, ask when two nonzero elements have the same square.

Hint 2. By Exercise 1.1.40,

𝑥^{2} = 𝑦^{2}

if and only if

𝑦 = \pm 𝑥

. When

𝑝

is odd,

𝑥 \neq - 𝑥

for

𝑥 \neq 0

Hint 3. So squaring is two-to-one on the

𝑝 - 1

nonzero elements.

Sketch. The nonzero elements split into

(𝑝 - 1) / 2

pairs

{𝑥, - 𝑥}

with

𝑥 \neq - 𝑥

. Elements of a pair have the same square, different pairs have different squares, and squares of nonzero elements are nonzero. With

0

, that makes

(𝑝 + 1) / 2

squares.

There are $(𝑝 + 1) / 2$ squares.

First, $𝑥 \neq - 𝑥$ for $𝑥 \neq 0$ : otherwise $(1 + 1) 𝑥 = 𝑥 + 𝑥 = 0$ , so $1 + 1 = 0$ by Proposition 1.1.17, but $1 + 1 = 2 \neq 0$ in $𝔽_{𝑝}$ because $𝑝$ is odd. So the $𝑝 - 1$ nonzero elements split into $(𝑝 - 1) / 2$ disjoint pairs ${𝑥, - 𝑥}$ .

By Exercise 1.1.40, two nonzero elements have the same square if and only if they lie in the same pair. Also $𝑥^{2} \neq 0$ for $𝑥 \neq 0$ by Proposition 1.1.17. So the nonzero squares are in bijection with the pairs, and there are $(𝑝 - 1) / 2$ of them. Adding the square $0 = 0^{2}$ gives $(𝑝 + 1) / 2$ .

For example, the squares in $𝔽_{7}$ are $0$ , $1$ , $2 = 3^{2}$ , and $4 = 2^{2}$ , which is $(7 + 1) / 2 = 4$ elements.

Exercise 1.1.53.hard Prove or disprove: the set

𝐾 =

{𝑎 + 𝑏 \sqrt[3]{2} | 𝑎, 𝑏 \in ℚ}

is a subfield of

ℝ

Solution.

Uses. Definition 1.1.32 (Subfield), $\sqrt[3]{2}$ is irrational

Hint 1.

𝐾

contains

0

and

1

and is closed under addition and negation. Try products.

Hint 2. Is

{(\sqrt[3]{2})}^{2} = \sqrt[3]{4}

𝐾

? Suppose

\sqrt[3]{4} = 𝑎 + 𝑏 \sqrt[3]{2}

and multiply both sides by

\sqrt[3]{2}

Hint 3. You get

2 = 𝑎 𝑏 + (𝑎 + 𝑏^{2}) \sqrt[3]{2}

. Use the irrationality of

\sqrt[3]{2}

, twice.

Sketch. False:

𝐾

is not closed under multiplication, because

\sqrt[3]{4} \notin 𝐾

. Writing

\sqrt[3]{4} = 𝑎 + 𝑏 \sqrt[3]{2}

and multiplying by

\sqrt[3]{2}

gives

2 = 𝑎 𝑏 + (𝑎 + 𝑏^{2}) \sqrt[3]{2}

; irrationality of

\sqrt[3]{2}

forces

𝑎 = - 𝑏^{2}

and then

𝑏^{3} = - 2

, which is impossible for rational

𝑏

The claim is false: $𝐾$ is not closed under multiplication. Write $𝑡 = \sqrt[3]{2}$ , so $𝑡 \in 𝐾$ and $𝑡^{3} = 2$ . We will show that $𝑡^{2} \notin 𝐾$ .

First, $𝑡$ is irrational. If $𝑡 = 𝑟 / 𝑠$ with integers $𝑟, 𝑠$ , $𝑠 > 0$ , and $gcd (𝑟, 𝑠) = 1$ , then $𝑟^{3} = 2 𝑠^{3}$ , so $𝑟$ is even. Writing $𝑟 = 2 𝑘$ gives $8 𝑘^{3} = 2 𝑠^{3}$ , so $𝑠^{3} = 4 𝑘^{3}$ and $𝑠$ is even too, a contradiction.

Suppose $𝑡^{2} = 𝑎 + 𝑏 𝑡$ with $𝑎, 𝑏 \in ℚ$ . Multiplying by $𝑡$ ,

2 = 𝑡^{3} = 𝑎 𝑡 + 𝑏 𝑡^{2} = 𝑎 𝑡 + 𝑏 (𝑎 + 𝑏 𝑡) = 𝑎 𝑏 + (𝑎 + 𝑏^{2}) 𝑡 .

If $𝑎 + 𝑏^{2} \neq 0$ , then $𝑡 =$ $(2 - 𝑎 𝑏) / (𝑎 + 𝑏^{2})$ is rational, which is false. So $𝑎 = - 𝑏^{2}$ , and then $2 = 𝑎 𝑏 = - 𝑏^{3}$ , so ${(- 𝑏)}^{3} = 2$ with $- 𝑏 \in ℚ$ , again contradicting the irrationality of $𝑡$ . Hence $𝑡^{2} \notin 𝐾$ , although $𝑡 \in 𝐾$ , so $𝐾$ is not a subfield of $ℝ$ .

Exercise 1.1.54.hard Fix $𝑐 \in ℝ$ and give $ℝ^{2}$ the componentwise addition and the multiplication

(𝑎, 𝑏) (𝑎^{'}, 𝑏^{'}) = (𝑎 𝑎^{'} + 𝑐 𝑏 𝑏^{'}, 𝑎 𝑏^{'} + 𝑏 𝑎^{'}) .

(For $𝑐 = - 1$ this is $ℂ$ .) For which $𝑐$ is $ℝ^{2}$ with these operations a field?

Solution.

Uses. Proposition 1.1.30 ( $ℂ$ is a field), Proposition 1.1.17 (No zero divisors), Definition 1.1.20 (Commutative ring)

Hint 1. Show that every axiom except multiplicative inverses holds for every

𝑐

; for associativity, expand both sides. So the question is when nonzero elements are invertible.

Hint 2. For

𝑐 < 0

imitate

ℂ

: multiply

(𝑎, 𝑏)

(𝑎, - 𝑏)

Hint 3. For

𝑐 \geq 0

, write

𝑐 = 𝑡^{2}

with

𝑡 \geq 0

, and look for zero divisors among

(𝑡, 1)

and

(- 𝑡, 1)

Sketch. For every

𝑐

the structure is a commutative ring with identity

(1, 0)

. If

𝑐 < 0

, then

(𝑎, 𝑏) (𝑎, - 𝑏) =

(𝑎^{2} - 𝑐 𝑏^{2}, 0)

with

𝑎^{2} - 𝑐 𝑏^{2} > 0

for

(𝑎, 𝑏) \neq (0, 0)

, which gives inverses. If

𝑐 = 𝑡^{2} \geq 0

, then

(𝑡, 1) (- 𝑡, 1) =

(0, 0)

, so there are zero divisors. Answer: exactly when

𝑐 < 0

The answer is: exactly when $𝑐 < 0$ .

The other axioms hold for every $𝑐$ . Addition is componentwise, so its axioms hold as in $ℝ^{2}$ . The multiplication is commutative, because the formula is symmetric under swapping $(𝑎, 𝑏)$ and $(𝑎^{'}, 𝑏^{'})$ . The element $(1, 0)$ is an identity, $(𝑎, 𝑏) (1, 0) =$ $(𝑎, 𝑏)$ , and $(1, 0) \neq (0, 0)$ . Distributivity holds because each coordinate of the product is a sum of terms each involving exactly one coordinate of the second factor, so it is additive in the second factor. For associativity, with $(𝑎, 𝑏),$ $(𝑎^{'}, 𝑏^{'}),$ $(𝑎^{″}, 𝑏^{″})$ ,

\begin{aligned} ((𝑎, 𝑏) (𝑎^{'}, 𝑏^{'})) (𝑎^{″}, 𝑏^{″}) & = ((𝑎 𝑎^{'} + 𝑐 𝑏 𝑏^{'}) 𝑎^{″} + 𝑐 (𝑎 𝑏^{'} + 𝑏 𝑎^{'}) 𝑏^{″}, (𝑎 𝑎^{'} + 𝑐 𝑏 𝑏^{'}) 𝑏^{″} + (𝑎 𝑏^{'} + 𝑏 𝑎^{'}) 𝑎^{″}), \\ (𝑎, 𝑏) ((𝑎^{'}, 𝑏^{'}) (𝑎^{″}, 𝑏^{″})) & = (𝑎 (𝑎^{'} 𝑎^{″} + 𝑐 𝑏^{'} 𝑏^{″}) + 𝑐 𝑏 (𝑎^{'} 𝑏^{″} + 𝑏^{'} 𝑎^{″}), 𝑎 (𝑎^{'} 𝑏^{″} + 𝑏^{'} 𝑎^{″}) + 𝑏 (𝑎^{'} 𝑎^{″} + 𝑐 𝑏^{'} 𝑏^{″})), \end{aligned}

and both first coordinates equal $𝑎 𝑎^{'} 𝑎^{″} +$ $𝑐 (𝑎 𝑏^{'} 𝑏^{″} + 𝑏 𝑎^{'} 𝑏^{″} + 𝑏 𝑏^{'} 𝑎^{″})$ , while both second coordinates equal $𝑎 𝑎^{'} 𝑏^{″} +$ $𝑎 𝑏^{'} 𝑎^{″} +$ $𝑏 𝑎^{'} 𝑎^{″} +$ $𝑐 𝑏 𝑏^{'} 𝑏^{″}$ . So for every $𝑐$ we have a commutative ring, and it is a field exactly when every nonzero element has a multiplicative inverse.

If $𝑐 < 0$ . For $(𝑎, 𝑏) \neq (0, 0)$ , $𝑞 =$ $𝑎^{2} -$ $𝑐 𝑏^{2} =$ $𝑎^{2} +$ $| 𝑐 | 𝑏^{2} >$ $0$ . Since $(𝑎, 𝑏) (𝑎, - 𝑏) =$ $(𝑎^{2} - 𝑐 𝑏^{2}, - 𝑎 𝑏 + 𝑏 𝑎) =$ $(𝑞, 0)$ , the element $(𝑎 / 𝑞, - 𝑏 / 𝑞)$ is an inverse of $(𝑎, 𝑏)$ . So the structure is a field.

If $𝑐 \geq 0$ . Let $𝑡 = \sqrt{𝑐}$ . Then

(𝑡, 1) (- 𝑡, 1) = (- 𝑡^{2} + 𝑐, 𝑡 - 𝑡) = (0, 0),

while $(𝑡, 1)$ and $(- 𝑡, 1)$ are nonzero. By Proposition 1.1.17 the structure is not a field.

Exercise 1.1.55.hard Construct a set with operations

+

and

\cdot

that satisfy every field axiom except distributivity, in which

0 \cdot 0 \neq 0

. (So distributivity cannot be removed from the proof of Proposition 1.1.15.)

Solution.

Uses. Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ ), Proposition 1.1.15 (Multiplying by zero), the division algorithm: $𝑥 mod 𝑛 = 𝑦 mod 𝑛$ if and only if $𝑛 ∣ 𝑥 - 𝑦$

Hint 1. Without distributivity,

+

and

\cdot

are unrelated. Keep the addition of

ℤ / 3 ℤ

and look for a separate multiplication on

{0, 1, 2}

Hint 2. The axioms about multiplication alone ask for a commutative, associative operation with identity

1

in which every nonzero element has an inverse. The addition of

ℤ / 3 ℤ

has all these properties, but its identity is

0

. Shift it so that the identity becomes

1

Hint 3. Try

𝑎 * 𝑏 =

(𝑎 + 𝑏 - 1) mod 3

Sketch. On

{0, 1, 2}

take the addition of

ℤ / 3 ℤ

and the multiplication

𝑎 * 𝑏 =

(𝑎 + 𝑏 - 1) mod 3

. This multiplication is commutative and associative with identity

1

, and every element has an inverse, but

0 * 0 = 2

Let $𝐹 = {0, 1, 2}$ with the addition $+_{3}$ of $ℤ / 3 ℤ$ , and define

𝑎 * 𝑏 = (𝑎 + 𝑏 - 1) mod 3 .

The additive axioms hold by Lemma 1.1.21. For the multiplication we use that $𝑥 mod 3 = 𝑦 mod 3$ whenever $3 ∣ 𝑥 - 𝑦$ . The operation $*$ is commutative since $𝑎 + 𝑏 - 1 = 𝑏 + 𝑎 - 1$ . It is associative: $(𝑎 + 𝑏 - 1) mod 3$ differs from $𝑎 + 𝑏 - 1$ by a multiple of $3$ , so

(𝑎 * 𝑏) * 𝑐 = ((𝑎 + 𝑏 - 1) mod 3 + 𝑐 - 1) mod 3 = (𝑎 + 𝑏 + 𝑐 - 2) mod 3,

and in the same way $𝑎 * (𝑏 * 𝑐) =$ $(𝑎 + 𝑏 + 𝑐 - 2) mod 3$ . The element $1$ is an identity, since $𝑎 * 1 = 𝑎 mod 3 = 𝑎$ , and $1 \neq 0$ . Every element has an inverse: $(2 - 𝑎) mod 3$ differs from $2 - 𝑎$ by a multiple of $3$ , so $𝑎 * ((2 - 𝑎) mod 3) =$ $(𝑎 + 2 - 𝑎 - 1) mod 3 =$ $1$ . So every field axiom except distributivity holds. However,

0 * 0 = (- 1) mod 3 = 2 \neq 0 .

By Proposition 1.1.15, distributivity must therefore fail. Indeed $0 * (0 + 0) = 2$ , while $0 * 0 + 0 * 0 = 2 +_{3} 2 = 1$ .

Exercise 1.1.56.hardBeyond 21-242 $ℤ / 4 ℤ$ is not a field, but there is a field with four elements. Let $𝔽_{4} = {0, 1, 𝛼, 𝛽}$ , four distinct symbols, with the following operations. Addition: $0$ is an identity, $𝑥 + 𝑥 = 0$ for every $𝑥$ , and the sum of two different elements of ${1, 𝛼, 𝛽}$ is the third. Multiplication: $0 𝑥 = 𝑥 0 = 0$ , $1 𝑥 = 𝑥 1 = 𝑥$ , $𝛼^{2} = 𝛽$ , $𝛽^{2} = 𝛼$ , and $𝛼 𝛽 = 𝛽 𝛼 = 1$ .

(a)Show that $𝔽_{4}$ is a field.
(b)Show that $char 𝔽_{4} = 2$ , and explain why $𝔽_{4}$ is not $ℤ / 4 ℤ$ with its elements renamed.

Solution.

Uses. Definition 1.1.1 (Field), Lemma 1.1.21 (Arithmetic in $ℤ / 𝑛 ℤ$ ), Proposition 1.1.24 ( $ℤ / 𝑛 ℤ$ is a field if and only if $𝑛$ is prime), Definition 1.1.28 (Characteristic)

Hint 1. Do not check all

64

triples. Identify the elements with pairs of elements of

𝔽_{2}

0 = (0, 0)

1 = (1, 0)

𝛼 = (0, 1)

𝛽 = (1, 1)

. What does addition become?

Hint 2. The nonzero elements are

1

𝛼

𝛼^{2} = 𝛽

, and

𝛼^{3} = 1

. How does multiplication of nonzero elements act on exponents?

Hint 3. For distributivity, write multiplication by

𝛼

as a map on pairs and check that it respects addition. Multiplication by

𝛽

is that map applied twice.

Sketch. As pairs over

𝔽_{2}

, addition is coordinatewise, which gives the additive axioms. Nonzero elements are

𝛼^{0}, 𝛼^{1}, 𝛼^{2}

with

𝛼^{𝑖} 𝛼^{𝑗} = 𝛼^{𝑖 +_{3} 𝑗}

, so multiplication of nonzero elements is addition in

ℤ / 3 ℤ

, which gives the multiplicative axioms. Multiplication by

𝛼

(𝑢, 𝑣) \mapsto

(𝑣, 𝑢 + 𝑣)

, which respects addition, and multiplication by

𝛽

is its square, which gives distributivity. Since

1 + 1 = 0

, the characteristic is

2

, whereas

ℤ / 4 ℤ

is not even a field.

(a) We check the axioms in groups.

Addition. Identify $0, 1, 𝛼, 𝛽$ with the pairs $(0, 0),$ $(1, 0),$ $(0, 1),$ $(1, 1)$ of elements of $𝔽_{2}$ . Under this identification the given addition is coordinatewise addition in $𝔽_{2}$ : each pair added to itself gives $(0, 0)$ , and the sum of two different nonzero pairs is the third, for example $(1, 0) + (0, 1) = (1, 1)$ . Coordinatewise addition is commutative and associative with identity $(0, 0)$ , because addition in $𝔽_{2}$ is, and every element is its own additive inverse.

Multiplication. If some factor is $0$ , both sides of the commutative and associative laws are $0$ . The nonzero elements are $𝛼^{0} = 1$ , $𝛼^{1} = 𝛼$ , and $𝛼^{2} = 𝛽$ , and $𝛼^{3} = 𝛼 𝛽 = 1$ . Checking the table, $𝛼^{𝑖} 𝛼^{𝑗} = 𝛼^{𝑖 +_{3} 𝑗}$ for all $𝑖, 𝑗 \in {0, 1, 2}$ ; for example $𝛽 𝛽 =$ $𝛼^{2 +_{3} 2} =$ $𝛼^{1} =$ $𝛼$ . So multiplying nonzero elements amounts to adding exponents in $ℤ / 3 ℤ$ , and multiplication of nonzero elements is commutative and associative because addition in $ℤ / 3 ℤ$ is (Lemma 1.1.21). The identity is $𝛼^{0} = 1 \neq 0$ , and $𝛼^{𝑖}$ has inverse $𝛼^{3 - 𝑖}$ for $𝑖 = 1, 2$ .

Distributivity. We show $𝑥 (𝑦 + 𝑧) =$ $𝑥 𝑦 +$ $𝑥 𝑧$ . For $𝑥 = 0$ and $𝑥 = 1$ this is clear. Multiplication by $𝛼$ sends $0 \mapsto 0$ , $1 \mapsto 𝛼$ , $𝛼 \mapsto 𝛽$ , $𝛽 \mapsto 1$ , which in pairs is

(𝑢, 𝑣) \mapsto (𝑣, 𝑢 + 𝑣) .

This map respects addition: $(𝑢 + 𝑢^{'}, 𝑣 + 𝑣^{'})$ goes to $(𝑣 + 𝑣^{'}, 𝑢 + 𝑢^{'} + 𝑣 + 𝑣^{'}) =$ $(𝑣, 𝑢 + 𝑣) +$ $(𝑣^{'}, 𝑢^{'} + 𝑣^{'})$ . So $𝛼 (𝑦 + 𝑧) =$ $𝛼 𝑦 +$ $𝛼 𝑧$ . Since $𝛽 = 𝛼 𝛼$ , associativity gives $𝛽 𝑦 = 𝛼 (𝛼 𝑦)$ , and therefore

𝛽 (𝑦 + 𝑧) = 𝛼 (𝛼 𝑦 + 𝛼 𝑧) = 𝛼 (𝛼 𝑦) + 𝛼 (𝛼 𝑧) = 𝛽 𝑦 + 𝛽 𝑧 .

So $𝔽_{4}$ is a field.

(b) $1 + 1 = 0$ and $1 \neq 0$ , so $char 𝔽_{4} = 2$ . Renaming the elements of $ℤ / 4 ℤ$ cannot change whether it satisfies the field axioms, and $ℤ / 4 ℤ$ is not a field by Proposition 1.1.24 (in it $2 \cdot 2 = 0$ ). So $𝔽_{4}$ is not a renamed $ℤ / 4 ℤ$ . Another way to see it: $𝑥 + 𝑥 = 0$ for every $𝑥 \in 𝔽_{4}$ , while $1 + 1 = 2 \neq 0$ in $ℤ / 4 ℤ$ .

Exercise 1.1.57.hard Let

𝐹

be a finite field. Show that the sum of all the elements of

𝐹

0

, unless

𝐹

has exactly two elements.

Solution.

Uses. Proposition 1.1.13 (Solving $𝑎 + 𝑥 = 𝑏$ and $𝑎 𝑥 = 𝑏$ ), Proposition 1.1.17 (No zero divisors)

Hint 1. Call the sum

𝑆

. Multiplying every element of

𝐹

by a fixed nonzero

𝑐

permutes

𝐹

. What does that say about

𝑐 𝑆

Hint 2. You get

𝑐 𝑆 = 𝑆

. To conclude

𝑆 = 0

you need such a

𝑐

with

𝑐 \neq 1

. When does one exist?

Sketch. If

| 𝐹 | > 2

, pick

𝑐 \notin {0, 1}

. Since

𝑥 \mapsto 𝑐 𝑥

is a bijection of

𝐹

𝑐 𝑆 = 𝑆

, so

(𝑐 - 1) 𝑆 = 0

with

𝑐 - 1 \neq 0

, and

𝑆 = 0

. If

𝐹 = {0, 1}

, then

𝑆 = 1

Let $𝑆$ be the sum of all elements of $𝐹$ . If $𝐹$ has exactly two elements, then $𝐹 = {0, 1}$ and $𝑆 = 1 \neq 0$ , which is the exception.

Suppose $𝐹$ has more than two elements, and pick $𝑐 \in 𝐹$ with $𝑐 \neq 0$ and $𝑐 \neq 1$ . By part (b) of Proposition 1.1.13, for each $𝑦 \in 𝐹$ the equation $𝑐 𝑥 = 𝑦$ has exactly one solution, so $𝑥 \mapsto 𝑐 𝑥$ is a bijection $𝐹 \to 𝐹$ . Hence the elements $𝑐 𝑥$ , for $𝑥 \in 𝐹$ , are the elements of $𝐹$ in another order, and their sum is $𝑆$ . By distributivity their sum is also $𝑐 𝑆$ . So $𝑐 𝑆 = 𝑆$ , that is, $(𝑐 - 1) 𝑆 = 0$ . Since $𝑐 - 1 \neq 0$ , Proposition 1.1.17 gives $𝑆 = 0$ .

Exercise 1.1.58.challenge Show that commutativity of addition follows from the other field axioms. That is, suppose

𝐹

has operations

+

and

\cdot

satisfying every field axiom except possibly

𝑎 + 𝑏 = 𝑏 + 𝑎

, and show that

𝑎 + 𝑏 = 𝑏 + 𝑎

for all

𝑎, 𝑏 \in 𝐹

Solution.

Uses. Definition 1.1.1 (Field)

Techniques. Proving from axioms (Technique 1.1.7)

Hint 1. Expand

(1 + 1) (𝑎 + 𝑏)

in two different ways using distributivity.

Hint 2. You get

𝑎 +

𝑎 +

𝑏 +

𝑏 =

𝑎 +

𝑏 +

𝑎 +

𝑏

. To finish you want to cancel

𝑎

on the left and

𝑏

on the right, but our proof of cancellation used commutativity of addition.

Hint 3. Re-prove what you need from the remaining axioms. First show that if

𝑎 + 𝑏 = 0

then

𝑏 + 𝑎 = 0

, and that

0 + 𝑎 = 𝑎

. Then both cancellation laws follow.

Sketch. Without commutativity of

+

, the right inverse

𝑏

𝑎

is also a left inverse, and

0

is also a left identity; this gives cancellation on both sides. Distributivity on each side (multiplication is still commutative) expands

(1 + 1) (𝑎 + 𝑏)

as both

𝑎 + 𝑎 + 𝑏 + 𝑏

and

𝑎 + 𝑏 + 𝑎 + 𝑏

. Cancelling

𝑎

on the left and

𝑏

on the right gives

𝑎 + 𝑏 = 𝑏 + 𝑎

We may still use commutativity of multiplication, so both distributive laws hold and $1 𝑥 = 𝑥 1 = 𝑥$ . Associativity of $+$ lets us drop parentheses in sums (the remark in Section 1.1.2 does not need commutativity for this).

Inverses work on both sides. Let $𝑎 + 𝑏 = 0$ , and choose $𝑐$ with $𝑏 + 𝑐 = 0$ . Then

𝑏 + 𝑎 = (𝑏 + 𝑎) + 0 = (𝑏 + 𝑎) + (𝑏 + 𝑐) = 𝑏 + (𝑎 + 𝑏) + 𝑐 = 𝑏 + 0 + 𝑐 = 𝑏 + 𝑐 = 0 .

$0$ is a left identity. With $𝑎 + 𝑏 = 0$ as before, $0 +$ $𝑎 =$ $(𝑎 + 𝑏) +$ $𝑎 =$ $𝑎 +$ $(𝑏 + 𝑎) =$ $𝑎 +$ $0 =$ $𝑎$ .

Cancellation on both sides. Suppose $𝑥 + 𝑦 = 𝑥 + 𝑧$ , and let $𝑥 + 𝑤 = 0$ , so $𝑤 + 𝑥 = 0$ as well. Adding $𝑤$ on the left, $𝑦 =$ $0 +$ $𝑦 =$ $(𝑤 + 𝑥) +$ $𝑦 =$ $𝑤 +$ $(𝑥 + 𝑦) =$ $𝑤 +$ $(𝑥 + 𝑧) =$ $(𝑤 + 𝑥) +$ $𝑧 =$ $0 +$ $𝑧 =$ $𝑧$ . Similarly, if $𝑦 + 𝑥 = 𝑧 + 𝑥$ , adding $𝑤$ on the right gives $𝑦 =$ $𝑦 +$ $(𝑥 + 𝑤) =$ $(𝑦 + 𝑥) +$ $𝑤 =$ $(𝑧 + 𝑥) +$ $𝑤 =$ $𝑧$ .

The main computation. By distributivity, applied in two different ways,

\begin{aligned} (1 + 1) (𝑎 + 𝑏) & = (1 + 1) 𝑎 + (1 + 1) 𝑏 = 𝑎 + 𝑎 + 𝑏 + 𝑏, \\ (1 + 1) (𝑎 + 𝑏) & = 1 (𝑎 + 𝑏) + 1 (𝑎 + 𝑏) = 𝑎 + 𝑏 + 𝑎 + 𝑏 . \end{aligned}

So $𝑎 +$ $(𝑎 + 𝑏) +$ $𝑏 =$ $𝑎 +$ $(𝑏 + 𝑎) +$ $𝑏$ . Cancelling $𝑎$ on the left and $𝑏$ on the right gives $𝑎 + 𝑏 = 𝑏 + 𝑎$ , as desired.

More practice

The practice bank has 261 exercises on this section’s topics.

$\cdot$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$

$\cdot$	$0$	$1$	$2$	$3$	$4$	$5$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$	$5$
$2$	$0$	$2$	$4$	$0$	$2$	$4$
$3$	$0$	$3$	$0$	$3$	$0$	$3$
$4$	$0$	$4$	$2$	$0$	$4$	$2$
$5$	$0$	$5$	$4$	$3$	$2$	$1$

$\cdot$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$

$\cdot$	$0$	$1$	$2$	$3$	$4$	$5$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$	$5$
$2$	$0$	$2$	$4$	$0$	$2$	$4$
$3$	$0$	$3$	$0$	$3$	$0$	$3$
$4$	$0$	$4$	$2$	$0$	$4$	$2$
$5$	$0$	$5$	$4$	$3$	$2$	$1$

$\cdot$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$

$\cdot$	$0$	$1$	$2$	$3$	$4$	$5$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$	$5$
$2$	$0$	$2$	$4$	$0$	$2$	$4$
$3$	$0$	$3$	$0$	$3$	$0$	$3$
$4$	$0$	$4$	$2$	$0$	$4$	$2$
$5$	$0$	$5$	$4$	$3$	$2$	$1$