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Let $k$ be an algebraically closed field. Let $\mathfrak{m}$ be a maximal ideal in $k[x_1,\ldots,x_n]$. Then $K:=k[x_1,\ldots,x_n]/\mathfrak{m}$ is a field. Moreover $K$ is also a finitely generated $k$-algebra, so by Zariski's Lemma $K$ is algebraic over $k$.

I can't prove what follows: since $k$ is algebraically closed, the natural map $$\phi:k\hookrightarrow k[x_1,\ldots,x_n]\twoheadrightarrow k[x_1,\ldots,x_n]/\mathfrak{m}=K$$ is an isomorphism between $k$ and $K$. I know that since $k$ is algebraically closed, every algebraic extension of $k$ equals $k$, but how can be proven that $\phi$ is a particular isomorphism?

bateman
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    Note that $\phi$ is one-to-one. As $K$ is algebraic over $k$ and $k$ is closed, it cannot be a proper injection. Hence its onto and therefore an isomorphism. – martini Oct 17 '13 at 19:00
  • @martini thanks for your comment. Honestly I can't see the injectivity. – bateman Oct 17 '13 at 19:06
  • @martini Well, I think one hsould simply note that $\phi$ is the identity on $k$. (A priori, it might be that $k$ is isomorphic to a subfield of itself) – Hagen von Eitzen Oct 17 '13 at 19:06
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    @bateman Injectivity is trivial as fields lack ideals. – Hagen von Eitzen Oct 17 '13 at 19:07
  • @HagenvonEitzen Oh yes I know that, but now I can't understand your further remark: $\phi$ is the identity on $k$ – bateman Oct 17 '13 at 19:14

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A homomorphism from a field to a nontrivial ring is injective (since the kernel is a proper ideal, hence zero). In particular, every homomorphism between fields is actually a field extension. Now use that an algebraically closed field has only one algebraic field extension, namely itsself.

Martin Brandenburg
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