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How should I show that a map $f(x) = x^{-1}$ for $x \neq 0$ and $f(0) = 0$ is not an automorphism for an infinite field?

Thanks for any hints.

Kuba

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Try to prove it is a homomorphism. Fail. Explain why you must fail. –  Hurkyl May 18 at 12:27

3 Answers 3

up vote 2 down vote accepted

Hint: Let $k$ be your field and fix a nonzero $y\in k$. If $f$ were an automorphism, then $f(x)+f(y)=f(x+y)$ would imply $x^2+xy+y^2$ is the zero polynomial in $k[x]$ because $k$ is infinite.

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Perhaps it could be mentioned that

the only fields $E$, finite or infinite, for which $f$ is an automorphism are the fields with $2$, $3$ and $4$ elements.

Clearly for the field with two and three elements $f$ is just the identity, whereas for the field with four elements $f$ is the Frobenius automorphism $x \mapsto x^{2}$. (This is because a nonzero element of the field with four elements satisfies $x^{3} = 1$, that is, $x^{2} = x^{-1}$.)

Suppose first that $E$ has characteristic $2$, and has more than two elements. As shown in the other answers, if $\alpha \ne 0, 1$ then $f(\alpha + 1) = f(\alpha) + f(1)$, implies $\alpha^{2} + \alpha + 1 = 0$. So $E$ has four elements, that is, $0, 1$ and the two roots of $x^{2} + x + 1$.

Now suppose the characteristic is not $2$, so $1 \ne -1$. If $\alpha \ne 0, -1$ is an arbitrary element, then $f(\alpha + 1) = f(\alpha) + f(1)$, implies $\alpha^{2} + \alpha + 1 = 0$. Set $\alpha = 1 \ne -1$ to get that $E$ has characteristic $3$. But then $0 = \alpha^{2} + \alpha + 1 = \alpha^{2} - 2 \alpha + 1= (\alpha - 1)^{2}$, so $\alpha \ne 0, -1$ can only be $1$, so that $E = \{ 0, 1, -1\}$ is the field with three elements.

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A shorter version of the last two paragraphs would be that if $\alpha \ne 0, 1$, then $\alpha$ is a root of $x^2 + x + 1$, so that $E$ has at most four elements, as claimed. –  Andreas Caranti May 19 at 11:41

An elemantary way;

Assume that $\phi$ is an automorphism of $F$ as you defined.Notice that if $\phi(x)=x$ then $x=1$ or $x=-1$ or $x=0$.

Now let $r$ be any nonzero elements of $F$ then set $x=r+\dfrac 1r$.

So we have, $\phi(x)=\phi(r+\dfrac 1r)=\phi(r)+\phi(\dfrac 1r)=\dfrac 1r+r=x$ which means that

$r$ must be a root of $$r+\dfrac 1r=0$$ $$r+\dfrac 1r=1$$ $$r+\dfrac 1r=-1$$

You have finite possible $r$ contradicting the fact that $F$ is infinite.

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