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In the answer to this question: Showing a homomorphism of a field algebraic over $\mathbb{Q}$ to itself is an isomorphism. I am curious why the surjectivity of the induced map on the finite set implies that the map $\varphi$ is itself surjective. If you restrict the domain of a function, and that function is surjective, does this imply that the function itself is surjective? Or am I missing something?

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  • $\begingroup$ This is because every element in $F$ belongs to such a finite set, since $F$ is an algebraic extension of $\mathbf Q$. $\endgroup$ – Bernard Feb 14 '18 at 17:31
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This is just a statement of linear algebra. Since the map is $\mathbb Q$ linear, it is a $\mathbb Q$ linear endomorphism, of a finite dimensional vector space. Apply rank nullity, and deduce that the kernel must be $0$ dimensional, so the image has full rank.

In the link you show, the proof is quite nice as well. Let $\alpha \in F$ be arbitrary. We wish to find some $\alpha^{\prime}$ so that $\phi(\alpha^{\prime})=\alpha$. Let $m_{\alpha}$ be the minimal polynomial for $\alpha$, and $S$ all of its roots. Since $\phi:S \to S$ is a bijective map from roots of $m_\alpha$ for each $\alpha \in F$, it follows that there exists some $\alpha^{\prime}$ that is a root of $m_\alpha$ so that $\phi(\alpha^{\prime})=\alpha$.

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  • $\begingroup$ How do we know that $\phi$ is a bijective map of the roots of $m_{\alpha}$? $\endgroup$ – ponchan Feb 14 '18 at 17:42
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    $\begingroup$ $\phi$ permutes roots of minimal polynomials, so the restriction is indeed a map $\phi:S \to S$. By assumption, this is injective. Since $S$ is finite, every injection $S \to S$ is in fact bijective. The last statement has nothing to do with algebra, it's just counting and various proofs can be found on this website. $\endgroup$ – Andres Mejia Feb 14 '18 at 17:44
  • $\begingroup$ I'm sorry, I don't see why $\phi$ permutes the roots of the minimal polynomial. $\endgroup$ – ponchan Feb 14 '18 at 17:47
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    $\begingroup$ if $p(x)=a_nx^n+\dots+a_0=0$, then $\phi(p(x))=a_n(\phi(x)^n)+\dots a_0=\phi(0)=0$. $\endgroup$ – Andres Mejia Feb 14 '18 at 17:49
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    $\begingroup$ since it fixes $a_0, \dots a_n \in k$, or the ground field. $\endgroup$ – Andres Mejia Feb 14 '18 at 17:49
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The point here is that $\varphi$ maps every element $\alpha \in F$ to some root of the minimal polynomial of $\alpha$. So $\varphi$ permutes these roots. If you pick any element of $F$, it will be the image of some root of its minimal polynomial (otherwise the map induced on $S$ would not be surjective), so $\varphi$ is surjective.

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