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Suppose $F$ is algebraic over $\mathbb{Q}$ and $\varphi \colon F \to F$ is a homomorphism. Prove that $\varphi$ is an isomorphism.

I came across this question here but I don't quite get why $F$ has to be algebraic over a field.

I know $\phi$ is injective since $\ker(\phi) = 0$. Shouldn't it then follow that $\phi$ is an isomorphism since $F$ and $F$ have the same cardinality?

Can anyone give a counterexample?

Edit: I also don't get why $\phi$ is surjective.

Let $\alpha$ be an element of $F$. Let $f(X)$ be the minimal polynomial of $\alpha$. Let $S$ be the set of all the roots of $f(X)$ in $F$. $\varphi$ induces an injective map $S\to S$. Since $S$ is a finite set, this map is surjective. Hence $\varphi$ is surjective.

Since $\phi$ maps $S$ onto $S$, how does that imply $\phi$ is surjective in the whole field?

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    $\begingroup$ Your proof only works when $F$ is finite (which is not the case). But notice that the actual proof of the statement looks at a finite set of roots of some minimal polynomial and then uses the fact that every injective endomorphism of a finite set is an isomorphism. $\endgroup$ – Martin Brandenburg Jun 8 '14 at 17:35
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Let $F=\mathbb Q(T)$. Then there is a homomorphism $F\to F$ given by $T\mapsto T^2$. It is not onto.

Edit: $\phi\colon F\to F$ is surjective because you can pick $\alpha\in F$ arbitrarily, then consider $S$ as in your quote, notice that $\phi$ is a permutation of the finite set $S$, hence there exists $\alpha'\in S$ with $\phi(\alpha')=\alpha$. In summary, for arbitrary $\alpha\in F$ there exists $\alpha'\in F$ such that $\phi(\alpha')=\alpha$.

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Let $\theta\in F$. Restriction $\psi$ of $\varphi$ on $\mathbb{Q}(\theta)$ is a homomorphism of finite extensions of $\mathbb{Q}$. You've mentioned that it's injective. But $\psi$ is a linear mapping of finite dimensional $\mathbb{Q}$-spaces and so has a determinant. It does not vanish due to injectivity. Then $\psi$ has an inverse. So $(\theta\psi^{-1})\varphi = \theta$

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