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Is the function sending every element of a field to its multiplicative inverse (and 0 to 0) ever a field automorphism?

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closed as off-topic by Travis, GoodDeeds, Gabriel Romon, Kamil Jarosz, Bobson Dugnutt Mar 2 '16 at 22:43

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  • $\begingroup$ Did you try to see what would happen if it was? $\endgroup$ – Tobias Kildetoft Mar 2 '16 at 14:13
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    $\begingroup$ Maybe you should first try to show that a field $F$ for which the inversion is a field automorphism must be of characteristic $3$ or $2$ (Hint : for any field $F$ and any automorphism of $F$, the automorphism must be the identity on the primitive field). $\endgroup$ – Clément Guérin Mar 2 '16 at 14:21
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For $F_2$ and $F_3$ inversion is an automorphism, since it is the identity morphism. I'll prove that the field characteristic has to be $2$ or $3$ for iversion to be an automorphism.

Suppose that $\phi(x)={1 \over x}$ is an automophism. It means that

$$\frac{1}{x+y}=\frac{1}{x}+\frac{1}{y}$$ $$\frac{1}{xy}=\frac{1}{x}\frac{1}{y}$$

The first equation is an interesting one.

Suppose that the field has characteristic other than $2$, and pick $x \neq 0$.

$$\frac{1}{2x}=\frac{1}{x+x}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}$$ Multiplying by $x$ we arrive at $4=1$, or, equivalently, $3=0$, so the characteristic has to be $3$.

So, the field characteristic has to be $2$ or $3$.

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  • $\begingroup$ Alright, thanks for the necessary condition. Do you know if it's a sufficient one? $\endgroup$ – Vik78 Mar 2 '16 at 14:33
  • $\begingroup$ Vik... can you check the field with 4 elements and see if that works? $\endgroup$ – GEdgar Mar 2 '16 at 14:34
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    $\begingroup$ I like the word "iversion" (an i-version by Apple) :) it is a typo, of course. $\endgroup$ – Dietrich Burde Mar 2 '16 at 19:06
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Proposition. Inversion is an automorphism of a field $F$ if and and only if $F$ is one of $\mathbb F_2$, $\mathbb F_3$, or $\mathbb F_4$.

Proof: Let $F$ be a field and suppose that inversion is a field automorphism of $F$. If $x \in F \setminus \{0,1\}$, then $x$ and $1-x$ are both invertible, and hence, by our assumption, $$ 1=1^{-1}=(x+(1-x))^{-1}=x^{-1} + (1-x)^{-1}. $$ Multiplying out, it follows that $x(1-x)=(1-x) + x=1$, and hence $$ x^2-x+1=0. $$ Since a quadratic polynomial over a field has at most two roots, this means that there are at most two possibilities for $x$. Hence $F$ has at most $4$ elements.

If $F=\mathbb F_2$ or $F=\mathbb F_3$, then inversion is the identity map, and hence an automorphism. If $F=\mathbb F_4$, there is only one non-trivial field automorphism $\varphi$, namely the one which satisfies $\varphi(x)=x^2$. We know that multiplicative group $\mathbb F_4^\times$ is cyclic of order $3$. The field automorphism $\varphi$, when restricted to $\mathbb F_4^\times$, induces an automorphism of the multiplicative group $\mathbb F_4^\times$, which must also be non-trivial. However, the only non-trivial automorphism of a cyclic group of order $3$ is inversion. Hence $\varphi$ is the inversion automorphism. (Of course, one can also check this directly for $\mathbb F_4$, if one is more comfortable doing the explicit calculation.)

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