Automorphisms of Fields

We describe some rigidity (or lack thereof) results for field automorphisms. This is based on another problem from my girlfriend’s algebra exam.

It is worthwhile to acknowledge the enemy, which is that one expects large fields to have many field automorphisms. Here is the standard example.

Proposition. Let \(k\) be a field, and let \(K\) be a normal extension of \(k\). Then any automorphism of \(k\) extends to \(K\).

Proof. This is a standard application of Zorn’s lemma. (Note that we did not assume that \(K/k\) is finite!) Consider the partially ordered set \(\mathcal P\) of pairs \((L,\sigma_L)\), where \(L\) is an intermediate extension, and \(\sigma_L\) is an extension of \(\sigma\); the ordering is given by \((L,\sigma_L)\le(L’,\sigma_{L’})\) if and only if \(L\subseteq L’\) and \(\sigma_{L’}|_L=\sigma_L\). Our argument now has two steps.

• We use Zorn’s lemma to show that \(\mathcal P\) has a maximal element. Indeed, \(\mathcal P\) is nonempty because it has \((k,\sigma)\). Then for any nonempty chain \(\{(L_i,\sigma_i)\}\), we can define an upper bound by setting \(L\) to be the union of the fields in the chain \(\bigcup_iL_i\), and we can define \(\sigma_L\) on \(L\) as the union of the \(\sigma_i\)s, which is well-defined because \(\sigma_j|_{L_i}=\sigma_i\) whenever \(i\le j\). Then \((L,\sigma)\) is the required intermediate element.

• Let \((L,\sigma_L)\) be a maximal pair. We claim that \(L=K\), which will complete the proof. Well, for any \(x\in K\), we should show that \(x\in L\), from which \(K\subseteq L\) and hence equality will follow. To show that \(x\in L\), we let \(L’\) be the normal closure of \(L(x)\), and we claim that \((L,\sigma_L)\) can be extended to \((L’,\sigma_{L’})\), from which \(L=L’\) will follow by maximality.

  We now have to do some algebra. If \(x\) is a transcendental element over \(L\), then \(\sigma_L\) extends to \(L’=L(x)\) by sending \(x\mapsto x\). Otherwise, \(x\) is algebraic, so \(L’\) is generated over \(L\) by the roots of the minimal polynomial \(f\) of \(x\) over \(L\). Thus, we may extend \(\sigma_L\) to \(L’\) by sending the roots of \(f\) to the roots of \(\sigma_L(f)\) (which are in \(L’\) by its normality!). (Technically speaking, one should extend one root at a time.) \(\blacksquare\)

Example. Because \(\mathbb C\) is algebraically closed, it is a normal extension of any subfield. Thus, for any finite extension \(K\) of \(\mathbb Q\), we see that any automorphism of \(K\) will extend to \(\mathbb C\). For example, \(K=\mathbb Q(\sqrt[3]2,\zeta_3)\) has an automorphism sending \(\sqrt[3]2\mapsto\zeta_3\sqrt[3]2\); extending this to \(\mathbb C\) reveals that \(\mathbb C\) has an automorphism \(\sigma\) such that \(\sigma(\mathbb R)\not\subseteq\mathbb R\).

Here is the prototypical result we are interested in.

Proposition. Let \(\sigma\colon\mathbb R\to\mathbb R\) be a field automorphism. Then \(\sigma=\operatorname{id}_{\mathbb R}\).

Proof. While there are many “wild” additive functions \(\mathbb R\to\mathbb R\), preserving multiplication adds some rigidity because it will introduce topology into the picture. We proceed in steps.

1. Because \(\sigma\) is a field automorphism, it fixes \(1\), so it fixes \(\mathbb Z\) by additivity, and it fixes \(\mathbb Q\) by multiplicativity.
2. The main claim is that \(\sigma\) preserves nonnegativity; this is where some topology enters the picture. Indeed, \(x\in\mathbb R\) is nonnegative if and only if it is a square, say \(x=y^2\), but then \[\sigma(x)=\sigma(y)^2,\] so \(\sigma(x)\ge0\) as well.
3. We now claim that \(\sigma\) is increasing. Indeed, \(x\le y\) implies that \(y-x\) is nonnegative, so \(\sigma(y-x)\) is nonnegaative by the previous step, so \(\sigma(x)\le\sigma(y)\).
4. We complete the proof. For any real number \(x\in\mathbb R\), we would like to show that \(\sigma(x)=x\). We will show that \(\sigma(x)\ge x\); the other inequality is analogous. The point is that \(\sigma\) already fixes \(\mathbb Q\), so we note that \(\sigma(x)\ge\sigma(q)=q\) for any rational \(q\) less than \(x\). By the dentity of \(\mathbb Q\) in \(\mathbb R\), it follows that \[\sigma(x)\ge\sup\{q\in\mathbb Q:q\le x\}=x.\] The proof of \(\sigma(x)\le x\) is fully symmetric. \(\blacksquare\)

The remark about topology in the above proof has a more obvious implementation as follows.

Proposition. The only continuous field automorphisms of \(\mathbb C\) are the identity and complex multiplication.

Proof. It is not so hard to check that the identity and complex conjugation are continuous field automorphisms in \(\operatorname{Gal}(\mathbb C/\mathbb R)\), so we will focus on showing that these are the only ones.

Let \(\sigma\) be a continuous field automorphism. As in the proof of the previous proposition, we see that \(\sigma\) fixes \(\mathbb Q\). Because \(\mathbb Q\) is dense in \(\mathbb R\), it follows that \(\sigma\) also fixes \(\mathbb R\)! (Here is where we used the continuity of \(\sigma\)!)

Now, any complex number can be written in the form \(a+bi\) where \(a\) and \(b\) are real numbers. Thus, \(\sigma(a+bi)=a+b\sigma(i)\), so it remains to pin down \(\sigma(i)\). Well, \[\sigma(i)^2+1=\sigma\left(i^2+1\right)=0,\] so \(\sigma(i)\) is a root of the polynomial \(x^2+1\). Thus, \(\sigma(i)\in\{\pm i\}\). We now see that \(\sigma(i)=i\) corresponds to the identity, and \(\sigma(i)=-i\) corresponds to complex conjugation. \(\blacksquare\)

Because I can’t resist, we remark that there are also no nontrivial field automorphisms for \(\mathbb Q_p\).

Proposition. Let \(\sigma\colon\mathbb Q_p\to\mathbb Q_p\) be a field automorphism. Then \(\sigma=\operatorname{id}_{\mathbb Q_p}\).

Proof. We proceed similarly to the proof for \(\mathbb R\). Our argument follows Keith Conrad’s.

1. As above, we see that \(\sigma\) fixes \(\mathbb Q\).
2. The main claim is that \(\sigma\) preserves \(1+p\mathbb Z_p\). Indeed, we claim that \(a\in 1+p\mathbb Z_p\) if and only if \(a\) is nonzero and an \(n\)th power for all positive integers \(n\) coprime to \(p\); this proves the claim because this latter conclusion is preserved by the multiplicative map \(\sigma\). Anyway, we show the two directions separately.
   • In one direction, note that \(a\in 1+p\mathbb Z_p\) implies that \(x^n-a\) has a root whenever \(p\nmid n\) (which is \(1\pmod p\)) by Hensel’s lemma.
   • In the other direction, if \(a\) is always an \(n\)th power, then we start by writing \(a=p^\nu b\) for \(\nu\in\mathbb Z\) and \(b\in\mathbb Z_p^\times\). Whenever \(a\) is an \(n\)th power, we see that \(n\mid\nu\) by taking valuations; by taking \(n\to\infty\), we conclude that \(\nu=0\). Thus, \(a\in\mathbb Z_p^\times\). To check that \(a\equiv1\pmod p\), we note that \(a\) is a \((p-1)\)st power, so we find \(a=c^{p-1}\equiv1\pmod p\) upon noting that any such \(c\) must also be in \(\mathbb Z_p^\times\).
3. We claim that \(\left|\sigma(a)-\sigma(b)\right|_p=\left|a-b\right|_p\). By setting \(c\) to be the difference, it remains to check that \(\left|\sigma(c)\right|_p=\left|c\right|_p\). Because \(\sigma\) already fixes \(p\), we may assume that \(c\in\mathbb Z_p^\times\). We would like to appeal to the previous step, but this requires us to work with elements in \(\mathbb Z_p^\times\) which are not \(1\pmod p\). Well, note that \(x^{p-1}-1\) splits into linear factors; in fact, by Hensel’s lemma, we see that there is a root \(\zeta_c\) such that \(c\equiv\zeta_c\pmod p\). The point is that \(\sigma(\zeta_c)^{p-1}=1\) implies that \(\left|\sigma(\zeta_c)\right|_p=1\) already, but then we see that \(\left|\sigma(c/\zeta_c)\right|_p=1\) by the previous step as well, so the claim follows.
4. We complete the proof. The previous step shows that \(\sigma\) is Lipschitz continuous, so it is continuous, so fixing the dense subset \(\mathbb Q\subseteq\mathbb Q_p\) implies that \(\sigma\) fixes all of \(\mathbb Q_p\). \(\blacksquare\)