2.4 Field Homomorphisms

Section 2.3 looked at fields inside fields. This section looks at maps between fields. Two such maps have appeared already. Complex conjugation $z \mapsto \overline{z}$ turns sums into sums and products into products (Theorem 1.2.2 (a), (b)), and so does $a + b\sqrt 2 \mapsto a - b\sqrt 2$ on $\mathbb{Q}(\sqrt 2)$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Conjugation in $\mathbb{Q}(\sqrt 2)$ )). A map that respects both operations carries every equation that holds in one field to an equation in the other. That single observation decides where such a map can send an element, shows that it never collapses two elements, and rules out most maps between most pairs of fields.

Throughout, $F$ and $L$ are fields, and $n \cdot a$ for $n \in \mathbb{N}$ is the multiple of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Section 2.3 (“Multiples”): $0 \cdot a \mathrel{:=} 0$ , $(n + 1) \cdot a \mathrel{:=} n \cdot a + a$ . Powers in a field are defined by the same recursion with products: $a^0 \mathrel{:=} 1$ and $a^{n+1} \mathrel{:=} a^n a$ for $n \in \mathbb{N}$ , as Sets, Relations, and Cardinality, Definition 2.3.2 does for real numbers; the power laws $a^{m+n} = a^m a^n$ , $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$ of Sets, Relations, and Cardinality, Proposition 2.3.4 have proofs that use only (F1)–(F7), (F9) and induction, and hold in every field with the same proofs.

2.4.1. Maps that respect both operations

A homomorphism is a map that respects addition, multiplication and the number one.

Definition 2.4.1 (Field Homomorphism)

Let $F$ and $L$ be fields. A homomorphism (of fields) $\varphi \colon F \to L$ is a function such that, for all $a, b \in F$ , $\varphi(a + b) = \varphi(a) + \varphi(b) , \qquad \varphi(ab) = \varphi(a)\varphi(b) , \qquad \varphi(1) = 1 .$ A bijective (Sets, Relations, and Cardinality, Definition 1.4.5) homomorphism is an isomorphism; $F$ and $L$ are isomorphic, written $F \cong L$ , if there is an isomorphism $F \to L$ . An isomorphism $F \to F$ is an automorphism of $F$ .

In words: the first clause says that adding and then mapping gives the same as mapping and then adding; the second says the same for multiplying; the third pins down the image of $1$ , where $1$ on the left is $1_F$ and on the right $1_L$ . An isomorphism is a homomorphism that is also a one-to-one correspondence of elements, so $F$ and $L$ are then “the same field with the elements renamed”.

Examples.

The identity $\operatorname{id}_F \colon F \to F$ is an automorphism.
Conjugation $z \mapsto \overline{z}$ on $\mathbb{C}$ is a homomorphism by Theorem 1.2.2 (a), (b), since $\overline{1} = 1$ ; Example (Conjugation and Other Automorphisms) below shows that it is an automorphism.
The conjugation of $\mathbb{Q}(\sqrt 2)$ , $\sigma(a + b\sqrt 2) \mathrel{:=} a - b\sqrt 2$ , is an automorphism (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Conjugation in $\mathbb{Q}(\sqrt 2)$ )).

Why the clause $\varphi(1) = 1$ . The zero map $a \mapsto 0$ satisfies the first two clauses, since $0 = 0 + 0$ and $0 = 0 \cdot 0$ , and it carries no information about $F$ at all. The third clause excludes it, and the next proposition shows that the third clause is exactly what forces a homomorphism to remember every element.

Remark. The order-preserving field maps of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Section 6.1 (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 6.1.1), are the homomorphisms between ordered fields that also preserve $<$ . From here on, $F \cong L$ means isomorphic as fields, in the sense above; the relation of that chapter, an isomorphism that also preserves the order, is said in words, “isomorphic as ordered fields”. The two differ: the conjugation $\sigma$ of $\mathbb{Q}(\sqrt 2)$ is an automorphism of the field, and it is not order-preserving, since $0 < \sqrt 2$ while $\sigma(\sqrt 2) = -\sqrt 2 < 0 = \sigma(0)$ .

2.4.2. What a homomorphism must do

To find where a homomorphism sends an element, we take an equation that the element satisfies and apply $\varphi$ to both sides. The equation in $L$ that comes out often has only one solution.

Proposition 2.4.1 (Properties of Field Homomorphisms)

Let $\varphi \colon F \to L$ be a homomorphism of fields, and let $a, b \in F$ .

(a) $\varphi(0) = 0$ , $\varphi(-a) = -\varphi(a)$ and $\varphi(a - b) = \varphi(a) - \varphi(b)$ .
(b) If $a \ne 0$ , then $\varphi(a) \ne 0$ and $\varphi(a^{-1}) = \varphi(a)^{-1}$ .
(c) $\varphi$ is injective.
(d) The image $\varphi(F)$ is a subfield of $L$ , and $\varphi$ , with codomain $\varphi(F)$ , is an isomorphism $F \to \varphi(F)$ .
(e) If $\psi \colon L \to M$ is a homomorphism of fields, then $\psi \circ \varphi \colon F \to M$ is a homomorphism. If $\varphi$ is an isomorphism, its inverse function $\varphi^{-1} \colon L \to F$ is an isomorphism. The identity $\operatorname{id}_F$ is an automorphism.
(f) $\varphi(n \cdot a) = n \cdot \varphi(a)$ and $\varphi(a^n) = \varphi(a)^n$ for every $n \in \mathbb{N}$ .

Idea. For (a) and (b) the equations to push through $\varphi$ are $0 + 0 = 0$ , $a + (-a) = 0$ and $aa^{-1} = 1$ ; in $L$ the results are equations with a unique solution, by cancellation and the uniqueness of negatives and inverses. For (c): a homomorphism cannot send a non-zero element to $0$ , because a non-zero element has an inverse, and then $\varphi(a)\varphi(a^{-1}) = \varphi(1) = 1 \ne 0$ . Injectivity is that fact applied to $a - b$ . The rest is bookkeeping with (a)–(c), and induction for (f).

Before reading on: in the proof of (c), where is $\varphi(1) = 1$ used, and which map satisfies the other two clauses of Definition 2.4.1 but is not injective?

Proof.

(a) Since $0 + 0 = 0$ in $F$ , $\varphi(0) + \varphi(0) = \varphi(0 + 0) = \varphi(0) = \varphi(0) + 0 ,$ and additive cancellation in $L$ (Theorem 2.1.1 (b)) gives $\varphi(0) = 0$ . Next, $a + (-a) = 0$ gives $\varphi(a) + \varphi(-a) = \varphi(a + (-a)) = \varphi(0) = 0$ , so $\varphi(-a)$ is a negative of $\varphi(a)$ , and $\varphi(-a) = -\varphi(a)$ by the uniqueness of negatives (Theorem 2.1.1 (c)). Finally, $\varphi(a - b) = \varphi(a + (-b)) = \varphi(a) + \varphi(-b) = \varphi(a) - \varphi(b)$ .
(b) Let $a \ne 0$ . Since $aa^{-1} = 1$ , $\varphi(a)\varphi(a^{-1}) = \varphi(aa^{-1}) = \varphi(1) = 1 .$ If $\varphi(a) = 0$ , the left side would be $0 \cdot \varphi(a^{-1}) = 0$ (Theorem 2.1.1 (d)), and the display would read $0 = 1$ , against (F7) in $L$ . So $\varphi(a) \ne 0$ , and the display says that $\varphi(a^{-1})$ is an inverse of $\varphi(a)$ ; by the uniqueness of inverses (Theorem 2.1.1 (c)), $\varphi(a^{-1}) = \varphi(a)^{-1}$ .
(c) Let $\varphi(a) = \varphi(b)$ . By (a), $\varphi(a - b) = \varphi(a) - \varphi(b) = 0$ . By (b), read contrapositively, an element whose image is $0$ is $0$ ; so $a - b = 0$ , and $a = b$ (add $b$ to both sides, with (F2), (F1), (F4) and (F3)). So $\varphi$ is injective.
(d) We use Proposition 2.3.1 (c) for $\varphi(F) \subseteq L$ . First, $1 = \varphi(1) \in \varphi(F)$ . Next, two elements of $\varphi(F)$ are $\varphi(a)$ , $\varphi(b)$ with $a, b \in F$ (Sets, Relations, and Cardinality, Definition 1.6.1), and $\varphi(a) - \varphi(b) = \varphi(a - b) \in \varphi(F)$ by (a). If moreover $\varphi(b) \ne 0$ , then $b \ne 0$ , since $\varphi(0) = 0$ ; so by (b), $\varphi(a)\varphi(b)^{-1} = \varphi(a)\varphi(b^{-1}) = \varphi(ab^{-1}) \in \varphi(F)$ . Hence $\varphi(F)$ is a subfield of $L$ , a field with the operations of $L$ (Proposition 2.3.1 (a)). The map $\varphi \colon F \to \varphi(F)$ still satisfies the three clauses, since the operations of $\varphi(F)$ are those of $L$ and its one is $1_L$ ; it is surjective by the definition of the image, and injective by (c). So it is an isomorphism.
(e) For the composite, $(\psi \circ \varphi)(a + b) = \psi(\varphi(a) + \varphi(b)) = \psi(\varphi(a)) + \psi(\varphi(b))$ , likewise for products, and $\psi(\varphi(1)) = \psi(1) = 1$ (Sets, Relations, and Cardinality, Definition 1.4.2). Now let $\varphi$ be an isomorphism, with inverse function $\varphi^{-1} \colon L \to F$ (Sets, Relations, and Cardinality, Definition 1.5.2, which exists by Sets, Relations, and Cardinality, Theorem 1.5.4). For $x, y \in L$ , $\varphi\bigl(\varphi^{-1}(x) + \varphi^{-1}(y)\bigr) = \varphi(\varphi^{-1}(x)) + \varphi(\varphi^{-1}(y)) = x + y = \varphi\bigl(\varphi^{-1}(x + y)\bigr) ,$ so $\varphi^{-1}(x) + \varphi^{-1}(y) = \varphi^{-1}(x + y)$ , because $\varphi$ is injective. The same computation with products gives $\varphi^{-1}(x)\varphi^{-1}(y) = \varphi^{-1}(xy)$ . And $\varphi(1) = 1$ gives $\varphi^{-1}(1) = 1$ . The function $\varphi^{-1}$ is bijective, with inverse $\varphi$ (Sets, Relations, and Cardinality, Theorem 1.5.4); so it is an isomorphism. The identity satisfies the three clauses trivially and is bijective.
(f) By induction on $n$ (Logic and Proof, Axiom 4.2.1, read on $\mathbb{N} = \omega$ through Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Corollary 1.7.3). For $n = 0$ : $\varphi(0 \cdot a) = \varphi(0) = 0 = 0 \cdot \varphi(a)$ by (a), and $\varphi(a^0) = \varphi(1) = 1 = \varphi(a)^0$ . If the two formulas hold for $n$ , then $\begin{aligned} \varphi((n + 1) \cdot a) &= \varphi(n \cdot a + a) = n \cdot \varphi(a) + \varphi(a) = (n + 1) \cdot \varphi(a) , \\ \varphi(a^{n+1}) &= \varphi(a^n a) = \varphi(a)^n \varphi(a) = \varphi(a)^{n+1} . \end{aligned}$ This proves the proposition.

The answer to the prompt. In (c), $\varphi(1) = 1$ enters through (b): the display $\varphi(a)\varphi(a^{-1}) = 1$ is what makes $\varphi(a) = 0$ impossible. The zero map $F \to L$ satisfies the two other clauses, sends every element to $0$ , and is not injective. A map that respected sums and products but sent $1$ to $0$ would be the zero map: $\varphi(a) = \varphi(a \cdot 1) = \varphi(a)\varphi(1) = \varphi(a) \cdot 0 = 0$ , by (F7), the product clause, $\varphi(1) = 0$ , and Theorem 2.1.1 (d) with (F6) in $L$ .

We call the move of this proof push the equation through $\varphi$ : to learn $\varphi(x)$ , apply $\varphi$ to an equation that $x$ satisfies, and solve the resulting equation in $L$ . By (c) and (d), a homomorphism $F \to L$ is the same thing as a copy of $F$ inside $L$ : $F \cong \varphi(F)$ , a subfield of $L$ . By (e) and Sets, Relations, and Cardinality, Theorem 1.4.4 (c), composing isomorphisms gives isomorphisms, so $\cong$ is reflexive, symmetric and transitive on any set of fields.

Warning. A bijection that respects addition alone is not an isomorphism. The map $f \colon \mathbb{Q} \to \mathbb{Q}$ , $f(x) \mathrel{:=} 2x$ , is bijective (its inverse is $x \mapsto x/2$ ) and satisfies $f(a + b) = f(a) + f(b)$ , by (F9). But $f(1 \cdot 1) = 2$ , while $f(1)f(1) = 4$ ; and $f(1) = 2 \ne 1$ . All three clauses of Definition 2.4.1 must be checked.

Quick check. Is $a \mapsto -a$ a homomorphism $\mathbb{Q} \to \mathbb{Q}$ ?

Solution. No. It sends $1$ to $-1$ , and $-1 \ne 1$ in $\mathbb{Q}$ : if $-1 = 1$ , then $1 + 1 = 1 + (-1) = 0$ , while $1 + 1 = 2 \cdot 1 > 0$ in the ordered field $\mathbb{Q}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (h), Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 2.5.1). (In $\mathbb{F}_2$ , where $-1 = 1$ , the same map is the identity.) It also fails the product clause: $1 \cdot 1 = 1$ is sent to $-1$ , while the product of the images is $(-1)(-1) = 1$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.2 (c)). It does respect sums, by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.2 (c) and (F9): $-(a + b) = (-1)(a + b) = -a + (-b)$ .

2.4.3. Examples, and pairs of fields with no map between them

Example (Conjugation and Other Automorphisms)

(i) Show that conjugation $z \mapsto \overline{z}$ is an automorphism of $\mathbb{C}$ , that it is not the identity, and that $\overline{z} = z$ if and only if $z \in \mathbb{R}$ .
(ii) Let $K$ be a subfield of $L$ . Show that the inclusion $K \to L$ , $x \mapsto x$ , is a homomorphism, and that it is not surjective when $K \ne L$ .
(iii) Recall why $a + b\sqrt 2 \mapsto a - b\sqrt 2$ is an automorphism of $\mathbb{Q}(\sqrt 2)$ .
(iv) Which automorphisms does $\mathbb{R}$ have?

Solution.

(i) By Theorem 1.2.2 (a), (b), $\overline{z + w} = \overline{z} + \overline{w}$ and $\overline{zw} = \overline{z}\,\overline{w}$ ; and $\overline{1} = \overline{1 + 0i} = 1 - 0i = 1$ (Definition 1.2.1). So conjugation is a homomorphism. By Theorem 1.2.2 (a), $\overline{\overline{z}} = z$ , so conjugation is its own inverse, hence bijective (Sets, Relations, and Cardinality, Theorem 1.5.4): an automorphism. It is not the identity: $\overline{i} = -i$ , and $-i \ne i$ , since $i - (-i) = 2i = (0, 2) \ne (0, 0)$ .
For the fixed elements, let $z = a + bi$ with $a, b \in \mathbb{R}$ . Then $\overline{z} = a - bi$ (Definition 1.2.1), and comparing parts (Definition 1.1.1), $\overline{z} = z$ if and only if $-b = b$ , that is $2b = 0$ , that is $b = 0$ (as $2 = 2 \cdot 1 > 0$ in the ordered field $\mathbb{R}$ , Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.3.3, Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (h), so $2 \ne 0$ , and by Theorem 2.1.1 (f)). And $b = 0$ says $z = a + 0i = a \in \mathbb{R}$ (Proposition 1.1.1). This fact is not among the parts of Theorem 1.2.2; the two lines above prove it.
(ii) The operations of $K$ are those of $L$ , and $1_K = 1_L$ (Proposition 2.3.1 (a)). So $x \mapsto x$ satisfies the three clauses. If $K \ne L$ , some element of $L$ is not in $K$ , and it is not a value of the inclusion.
(iii) This is Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Conjugation in $\mathbb{Q}(\sqrt 2)$ ), recalled and not proved again: unique representation makes the formula a function, the product formula of Example (The Field Generated by the Square Root of Two) with $-b$ , $-d$ in place of $b$ , $d$ shows that products go to products, and the map is its own inverse.
(iv) Only the identity. Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Chapter 6, proved that a map $\sigma \colon \mathbb{R} \to \mathbb{R}$ with $\sigma(a + b) = \sigma(a) + \sigma(b)$ , $\sigma(ab) = \sigma(a)\sigma(b)$ and $\sigma(1) = 1$ , that is, a homomorphism $\mathbb{R} \to \mathbb{R}$ , is the identity, with no order condition assumed (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 6.3.4, with $\mathbb{R}$ complete by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.6.3). So $\mathbb{C}$ has a symmetry and $\mathbb{R}$ has none. Section 2.5 shows that $\mathbb{Q}$ and every $\mathbb{F}_p$ have no automorphism other than the identity either.

Pushing an equation through $\varphi$ also proves that no homomorphism exists: an equation that holds in $F$ and has no counterpart in $L$ leaves $\varphi$ nowhere to go.

Example (Pairs of Fields with No Homomorphism)

Show that there is no homomorphism (i) $\mathbb{C} \to \mathbb{R}$ ; (ii) $\mathbb{F}_2 \to \mathbb{Q}$ ; (iii) $\mathbb{Q} \to \mathbb{F}_p$ , for any prime $p$ . (iv) Which clause of Definition 2.4.1 does the zero map $\mathbb{Q} \to \mathbb{Q}$ fail?

Solution.

(i) Suppose $\varphi \colon \mathbb{C} \to \mathbb{R}$ is a homomorphism. In $\mathbb{C}$ , $i^2 = -1$ . Pushing this through $\varphi$ , $\varphi(i)^2 = \varphi(i^2) = \varphi(-1) = -\varphi(1) = -1 ,$ by (f) and (a) of Proposition 2.4.1. But $\mathbb{R}$ is an ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.3.3), in which every square is $\ge 0$ and $1 > 0$ , so $-1 < 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (e), (a)). So $\varphi(i)^2 = -1$ is impossible.
(ii) Suppose $\varphi \colon \mathbb{F}_2 \to \mathbb{Q}$ is a homomorphism. In $\mathbb{F}_2$ , $1 + 1 = 0$ (Example (Arithmetic on $\mathbb{Z}/n\mathbb{Z}$ and the Field $\mathbb{F}_2$ )). Pushing it through $\varphi$ , $1 + 1 = \varphi(1) + \varphi(1) = \varphi(1 + 1) = \varphi(0) = 0$ in $\mathbb{Q}$ , by (a). But $1 + 1 = 2 \cdot 1 > 0$ in the ordered field $\mathbb{Q}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (h), Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 2.5.1). So no such $\varphi$ exists.
(iii) Suppose $\varphi \colon \mathbb{Q} \to \mathbb{F}_p$ is a homomorphism. In $\mathbb{F}_p$ , $p \cdot 1 = 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 3.4.3 (d)). By Proposition 2.4.1 (f), $\varphi(p \cdot 1) = p \cdot \varphi(1) = p \cdot 1 = 0 = \varphi(0)$ , and $\varphi$ is injective by (c), so $p \cdot 1 = 0$ in $\mathbb{Q}$ . But $p \cdot 1 > 0$ in $\mathbb{Q}$ , as in (ii), since $p \ge 1$ . So no such $\varphi$ exists.
(iv) The zero map $a \mapsto 0$ respects sums and products, since $0 + 0 = 0$ and $0 \cdot 0 = 0$ , and it sends $1$ to $0 \ne 1$ . So it fails only $\varphi(1) = 1$ , and is not a homomorphism.

Section 2.5 explains (ii) and (iii) at once: both pairs of fields differ in the number of copies of $1$ that add up to $0$ .

2.4.4. Exercises

A. Check your understanding

Exercise (A1)

These parts are about Proposition 2.4.1, mainly its part (c).

(a) Where in the proof of (c) is the clause $\varphi(1) = 1$ used?
(b) Let $\zeta \colon \mathbb{Q} \to \mathbb{Q}$ be the zero map. Which clause of Definition 2.4.1 fails? Which of the conclusions (a)–(f) fail for $\zeta$ ?
(c) Run the proof of (c) with conjugation on $\mathbb{C}$ and the elements $a = 1 + i$ , $b = 1 - i$ : compute $a - b$ , its image, and conclude.
(d) State the idea of the proof of (c) in two sentences.

Solution.

(a) In (b), which (c) cites: $\varphi(a)\varphi(a^{-1}) = \varphi(aa^{-1}) = \varphi(1) = 1$ , and $1 \ne 0$ , so $\varphi(a) \ne 0$ for $a \ne 0$ .
(b) The clause $\zeta(1) = 1$ fails: $\zeta(1) = 0$ . Conclusion (a) holds ( $\zeta(0) = 0 = -0$ ). Conclusion (b) fails: $1 \ne 0$ and $\zeta(1) = 0$ . Conclusion (c) fails: $\zeta(1) = \zeta(2)$ . Conclusion (d) fails: the image $\{0\}$ does not contain $1$ , so it is not a subfield. Conclusion (f) holds for multiples (both sides are $0$ ) and fails for powers at $n = 0$ : $\zeta(a^0) = \zeta(1) = 0$ , while $\zeta(a)^0 = 1$ . Conclusion (e) fails in its first clause: for a homomorphism $\psi \colon \mathbb{Q} \to M$ (for example $\psi = \operatorname{id}_{\mathbb{Q}}$ ), $\psi \circ \zeta$ is the zero map, since $\psi(0) = 0$ by (a) for $\psi$ , and the zero map sends $1$ to $0$ , so it is not a homomorphism. The clause about inverses holds vacuously, since $\zeta$ is not bijective, and the clause about $\operatorname{id}_F$ does not involve $\zeta$ .
(c) $a - b = 2i \ne 0$ . Its image is $\overline{2i} = -2i \ne 0$ , as (b) predicts; so $\overline{a} - \overline{b} = \overline{a - b} \ne 0$ , and $\overline{a} \ne \overline{b}$ . Indeed $\overline{a} = 1 - i$ and $\overline{b} = 1 + i$ .
(d) A non-zero element has an inverse, so its image times another image is $\varphi(1) = 1 \ne 0$ ; hence a homomorphism sends no non-zero element to $0$ . Applied to a difference $a - b$ , this says that different elements have different images.

B. Practice

Exercise (B1: Images of quotients)

Let $\varphi \colon F \to L$ be a homomorphism, and let $a, b \in F$ with $b \ne 0$ . Prove that $\varphi(a/b) = \varphi(a)/\varphi(b)$ . Here is the start of a proof. Name the move, then complete it.

“Since $a = (a/b) \cdot b$ , applying $\varphi$ gives $\varphi(a) = \varphi(a/b)\,\varphi(b)$ . …”

Solution. The move is push the equation through $\varphi$ : the equation $a = (a/b)b$ , which holds since $(ab^{-1})b = a(b^{-1}b) = a$ by (F5), (F6), (F8) and (F7), is carried to $L$ and solved there.

Completion. Since $b \ne 0$ , $\varphi(b) \ne 0$ by Proposition 2.4.1 (b), so $\varphi(b)^{-1}$ exists. Multiplying $\varphi(a) = \varphi(a/b)\varphi(b)$ by $\varphi(b)^{-1}$ , with (F5), (F8) and (F7), $\varphi(a)\varphi(b)^{-1} = \varphi(a/b)\,\varphi(b)\varphi(b)^{-1} = \varphi(a/b) .$ That is, $\varphi(a/b) = \varphi(a)/\varphi(b)$ .

Exercise (B2: The automorphisms of the rational complex numbers)

Show that $\mathbb{Q}(i)$ (Example (Rational and Integer Complex Numbers)) has exactly two automorphisms: the identity and conjugation $a + bi \mapsto a - bi$ . Here is the first half of a proof. Name the move, then complete it.

“Let $\sigma$ be an automorphism of $\mathbb{Q}(i)$ . It fixes every rational: $\sigma(n) = \sigma(n \cdot 1) = n \cdot \sigma(1) = n$ for $n \in \mathbb{N}$ by Proposition 2.4.1 (f), $\sigma(-n) = -\sigma(n) = -n$ by (a), and $\sigma(m/n) = \sigma(m)/\sigma(n) = m/n$ for $m \in \mathbb{Z}$ , $n \in \mathbb{Z}_{>0}$ , by Exercise (B1: Images of quotients). Moreover $i^2 = -1$ . …”

Solution. The move is push the equation through $\sigma$ , applied to $i^2 = -1$ .

Completion. Every rational is $m/n$ with $m \in \mathbb{Z}$ , $n \in \mathbb{Z}_{>0}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.4 (b)), so $\sigma(q) = q$ for every $q \in \mathbb{Q}$ . Applying $\sigma$ to $i^2 = -1$ , $\sigma(i)^2 = \sigma(i^2) = \sigma(-1) = -1 = i^2 .$ By Corollary 2.1.2 in the field $\mathbb{Q}(i)$ , $\sigma(i) = i$ or $\sigma(i) = -i$ . For $a, b \in \mathbb{Q}$ , $\sigma(a + bi) = \sigma(a) + \sigma(b)\sigma(i) = a + b\,\sigma(i)$ . So either $\sigma(a + bi) = a + bi$ for all $a, b$ , and $\sigma$ is the identity, or $\sigma(a + bi) = a - bi$ for all $a, b$ , and $\sigma$ is conjugation. There are at most these two.

Both occur. The identity is an automorphism (Proposition 2.4.1 (e)). Conjugation sends $a + bi$ to $a + (-b)i \in \mathbb{Q}(i)$ , so it maps $\mathbb{Q}(i)$ into itself; it respects sums and products and sends $1$ to $1$ because conjugation on $\mathbb{C}$ does (Example (Conjugation and Other Automorphisms) (i)) and the operations of $\mathbb{Q}(i)$ are those of $\mathbb{C}$ ; and it is its own inverse on $\mathbb{Q}(i)$ , hence bijective. The two are different maps, since conjugation sends $i \in \mathbb{Q}(i)$ to $-i \ne i$ (Example (Conjugation and Other Automorphisms) (i)). So $\mathbb{Q}(i)$ has exactly two automorphisms.

Exercise (B3)

Let $F$ be the field of Exercise (B3: Which Are Fields?) (a): the set $\mathbb{Q}$ with the usual sum and the product $a \odot b \mathrel{:=} 2ab$ . Show that $f \colon F \to \mathbb{Q}$ , $f(x) \mathrel{:=} 2x$ , is an isomorphism of fields, so $F \cong \mathbb{Q}$ . Compare with the warning of this section.

Solution. The one of $F$ is $1/2$ : $a \odot \tfrac12 = 2a \cdot \tfrac12 = a$ . We check the three clauses, with $\mathbb{Q}$ ’s usual product on the right. Sums: $f(a + b) = 2a + 2b = f(a) + f(b)$ , by (F9). Products: $f(a \odot b) = 2(2ab) = (2a)(2b) = f(a)f(b)$ , by (F5) and (F6). One: $f(1/2) = 1$ . The map is bijective, with inverse $x \mapsto x/2$ : $2(x/2) = x$ and $(2x)/2 = x$ (Sets, Relations, and Cardinality, Theorem 1.5.4). So $f$ is an isomorphism.

The warning used the same formula $x \mapsto 2x$ as a map $\mathbb{Q} \to \mathbb{Q}$ , with the usual product on both sides, and it is not a homomorphism there. Whether a map is a homomorphism depends on the operations of the source as well as on the formula: on $F$ , the product $\odot$ contains the factor $2$ that the formula needs.

Exercise (B4: The fixed field)

Let $\sigma$ be an automorphism of a field $F$ . Show that its fixed set $\{ a \in F : \sigma(a) = a \}$ is a subfield of $F$ . Find the fixed set of conjugation on $\mathbb{C}$ and of conjugation on $\mathbb{Q}(\sqrt 2)$ .

Solution. Write $K$ for the fixed set, and use Proposition 2.3.1 (c). $\sigma(1) = 1$ , so $1 \in K$ . For $a, b \in K$ , $\sigma(a - b) = \sigma(a) - \sigma(b) = a - b$ by Proposition 2.4.1 (a), so $a - b \in K$ . If moreover $b \ne 0$ , then $\sigma(ab^{-1}) = \sigma(a)\sigma(b)^{-1} = ab^{-1}$ by (b), so $ab^{-1} \in K$ . Hence $K$ is a subfield. (Only the homomorphism property was used.)

For conjugation on $\mathbb{C}$ , $K = \mathbb{R}$ , by Example (Conjugation and Other Automorphisms) (i).

For conjugation on $\mathbb{Q}(\sqrt 2)$ , $a - b\sqrt 2 = a + b\sqrt 2$ holds if and only if $2b\sqrt 2 = 0$ , that is, $b = 0$ : in the ordered field $\mathbb{R}$ , $2 = 2 \cdot 1 > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (h), Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.3.3), so $2 \ne 0$ , and $\sqrt 2 \ne 0$ (since $0^2 = 0 \ne 2$ ), and a product of non-zero elements is non-zero (Theorem 2.1.1 (f)). So $K = \{ a + 0\sqrt 2 : a \in \mathbb{Q} \} = \mathbb{Q}$ .

C. Going deeper

Exercise (C1: Isomorphic fields have the same square roots of minus one)

(a) Let $F \cong L$ . Prove that $F$ has an element $x$ with $x^2 = -1$ if and only if $L$ has one.
(b) Deduce that $\mathbb{Q}(i) \not\cong \mathbb{Q}(\sqrt 2)$ and that $\mathbb{C} \not\cong \mathbb{R}$ .

Solution.

(a) Let $\varphi \colon F \to L$ be an isomorphism. If $x \in F$ has $x^2 = -1$ , then by Proposition 2.4.1 (f) and (a), $\varphi(x)^2 = \varphi(x^2) = \varphi(-1) = -\varphi(1) = -1 ,$ so $\varphi(x) \in L$ is such an element. Conversely, $\varphi^{-1} \colon L \to F$ is an isomorphism (Proposition 2.4.1 (e)), and the same computation carries an element $y \in L$ with $y^2 = -1$ to $\varphi^{-1}(y) \in F$ with $\varphi^{-1}(y)^2 = -1$ .
(b) In $\mathbb{Q}(i)$ , $i^2 = -1$ . In $\mathbb{Q}(\sqrt 2) \subseteq \mathbb{R}$ , every $y$ has $y^2 \ge 0 > -1$ , since $\mathbb{R}$ is an ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.3.3, Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.5 (e), (a)); so $\mathbb{Q}(\sqrt 2)$ has no such element, and by (a), $\mathbb{Q}(i) \not\cong \mathbb{Q}(\sqrt 2)$ . Likewise $i \in \mathbb{C}$ and no element of $\mathbb{R}$ squares to $-1$ , so $\mathbb{C} \not\cong \mathbb{R}$ .

The same argument works for any statement written with $+$ , $\cdot$ , $0$ , $1$ and elements: an isomorphism carries it in both directions. (For $\mathbb{C}$ and $\mathbb{R}$ , Example (Pairs of Fields with No Homomorphism) (i) shows more: there is no homomorphism $\mathbb{C} \to \mathbb{R}$ at all.)

Marking scheme. (10 points) - 3: pushes $x^2 = -1$ through $\varphi$ , with $\varphi(-1) = -1$ justified (by (a) and $\varphi(1) = 1$ ). - 3: the converse, by either route: through $\varphi^{-1}$ , with the fact that it is an isomorphism (or a homomorphism) justified (Proposition 2.4.1 (e)); or directly, writing $y = \varphi(x)$ by surjectivity, getting $\varphi(x^2) = \varphi(-1)$ , and concluding $x^2 = -1$ by injectivity. - 2: no element of $\mathbb{Q}(\sqrt 2)$ (or of $\mathbb{R}$ ) squares to $-1$ , from the order of $\mathbb{R}$ . - 2: both conclusions of (b) drawn from (a). Fatal errors (0 for the step): treating $\mathbb{Q}(\sqrt 2) \not\cong \mathbb{Q}(i)$ as evident because the sets “look different”; using $\varphi(-1) = -1$ without (a).

Exercise (C2: Find the error)

Here is an argument. Is its conclusion true? Does the argument prove it? If not, name the line that fails and why.

“Every homomorphism $\varphi \colon \mathbb{Q} \to \mathbb{Q}$ is the identity. (1) By definition, $\varphi(1) = 1$ , and $\varphi$ respects sums. (2) For $x \in \mathbb{Q}$ , $\varphi(x) = \varphi(x \cdot 1) = x\,\varphi(1)$ . (3) Hence $\varphi(x) = x \cdot 1 = x$ .”

Solution. The conclusion is true: the repair below proves it, and Section 2.5 proves it again by another route. The argument does not prove it.

Line (2) fails, at the second equality. “ $\varphi(x \cdot 1) = x\,\varphi(1)$ ” pulls the factor $x$ out of $\varphi$ , as if $\varphi$ were compatible with multiplication by a scalar $x$ ; that is not a clause of Definition 2.4.1, and nothing proved so far supplies it. So the line cites a general “fact” that is not available, entry (N9) of Logic and Proof, Example (What is not allowed). The clause that is available, multiplicativity, gives only $\varphi(x \cdot 1) = \varphi(x)\varphi(1) = \varphi(x)$ , which proves nothing.

Counterexample to the step (in its general form, “a homomorphism satisfies $\varphi(x \cdot 1) = x\,\varphi(1)$ ”): conjugation on $\mathbb{C}$ has $\overline{i \cdot 1} = -i$ , while $i \cdot \overline{1} = i$ .

Repair. For $n \in \mathbb{N}$ , $\varphi(n \cdot 1) = n \cdot \varphi(1) = n \cdot 1$ by Proposition 2.4.1 (f), and $n \cdot 1 = n$ in $\mathbb{Q}$ ; so $\varphi(n) = n$ . For $n \in \mathbb{N}$ , $\varphi(-n) = -\varphi(n) = -n$ by (a), so $\varphi(m) = m$ for every $m \in \mathbb{Z}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.2.3 (c)). Every rational is $m/n$ with $m \in \mathbb{Z}$ , $n \in \mathbb{Z}_{>0}$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.4 (b)), and $\varphi(m/n) = \varphi(m)/\varphi(n) = m/n$ by Exercise (B1: Images of quotients). So $\varphi = \operatorname{id}_{\mathbb{Q}}$ .