# Is there a 'conjugation' on every algebraically closed field?

Let $K$ be an algebraically closed field. Then the polynomial $x^2+1\in K[x]$ has two distinct roots (when $K$ doesn't have characteristic 2). Let's suggestively call them $i$ and $-i$.

Does there exist an automorphism $\phi$ of $K$ such that $\phi(i)=-i$?

This is something I've been wondering about for a while. I'm asking this question here because I, unfortunately, have very little background in field theory and have no clue on how to approach this problem. Any help is appreciated.

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The second statement is false; if $K$ has characteristic $2$ we have $x^2 + 1 = (x + 1)^2$, so the polynomial has only a single root, namely $1 = -1$. – Qiaochu Yuan Jul 2 '14 at 1:56
If $K$ has characteristic $2$, then the two roots of $x^2+1$ are not distinct (namely, $1$ is a double root). – Henning Makholm Jul 2 '14 at 1:58
Oh, characteristic two. The algebraist's "you need to assume the limit exists". – Bryan Jul 2 '14 at 1:59

Not necessarily. For example, in the algebraic closure of $\mathbb F_5$, the two roots of $x^2+1$ are $1+1$ and $1+1+1$, and there's obviously no automorphism that interchanges them.

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Very nicely formulated! – Marc van Leeuwen Jul 2 '14 at 10:50

Henning's answer is correct. However, if we let $F$ be the smallest subfield of $K$ and $f(x) = x^2 + 1$ is irreducible in $F$, then your automorphism is legitimate. Here's why:

Let $L \subset K$ be its splitting field. Since $L$ is a splitting field of $f$, then it is by definition a Galois extension. At this point, we can consider the automorphism group $Aut(L/F)$. An element $\phi \in Aut(L/F)$ is a $F$-automorphism. That is to say, it is an automorphism of $L$ (that can be extended to $K$) that fixes $F$.

Next, it is a theorem that, since $L$ is a Galois extension $[L:F] = |Aut(L/F)|$. It is also a theorem that the elements of $Aut(L/F)$ are uniquely determined by their actions, as permutations, on the roots of the polynomial.

Obviously, the identity automorphism will fix the roots of $f$. Since $f$ is irreducible in $F$, then $[L:F] > 1$, and so there must exist at least one nontrivial automorphism. Since there are only two roots that can be permuted, then that nontrivial automorphism must swap the roots of $f$.

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I don't follow. I see that you've defined an automorphism of $L$. Are you claiming that it extends to an automorphism of $K$? I believe this if $K$ is algebraic over $L$ but it's not immediately obvious to me in the general case. – Qiaochu Yuan Jul 2 '14 at 2:31
Your objection has been addressed here, though I'm still not entirely satisfied: math.stackexchange.com/questions/219869/… – Kaj Hansen Jul 2 '14 at 3:28
Also, this argument is overkill; you don't need to appeal to so much Galois theory. You can just directly verify that $\sigma(i) = -i$ is an automorphism of $L$; the only thing you need is that $L$ has basis $\{ 1, i \}$ as a $K$-vector space. – Qiaochu Yuan Jul 2 '14 at 4:22
Ah, yes. That is easier! I really appreciate your comments on my posts @QiaochuYuan. I've learned a good bit from them now and in the past. – Kaj Hansen Jul 2 '14 at 4:31
While it was shown at mentioned other question that every automorphism of a subfield extends to an automorphism of the algebraically closed field, that extended automorphism may have infinite order while the original automorphism has finite order. I would hesitate to call an automorphism a "conjugation" if it does not have order$~2$. For instance over $\Bbb F_3$ the polynomial $x^2+1$ is irrededucible, and there is a (Frobenius) automorphism of order$~2$ of $\Bbb F_9$ that interchanges its roots, but the algebraic closure of $\Bbb F_3$ does not have any automorphism of order$~2$. – Marc van Leeuwen Jul 2 '14 at 11:42