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Let $k$ be an algebraically closed field and $$K=\frac{k[x_1,\dots,x_n]}{m}$$ be a finitely generated $k$-algebra, where $m$ is a maximal ideal. $K$ is algebraic over $k$. Then why is $k$ isomorphic to $K$? Sorry if this is obvious.

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    $\begingroup$ It's very hard to tell from your question where your confusion lies. Are you asking why, if $K$ is algebraic over $k$, then $K$ is isomorphic to $k$? That is by definition of $k$ being algebraically closed. This is definitely the most basic fact of the ones you've stated in your question, so please let us know if that is not your question $\endgroup$ – Alex Mathers Jan 28 '17 at 7:09
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Hint: (Zariski) If a field $L$ is ring-finite over a subfield $K$, then $L$ is module-finite (and hence algebraic) over $K$.

So if $I$ is a maximal ideal of $k[x_1,x_2,\ldots,x_n]$, then $k[x_1,x_2,\ldots,x_n]/I$ is a field, and so an algebraic extension of $k$. Since $k$ is algebraically closed, we get $k[x_1,x_2,\ldots,x_n]/I$=k.

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  • $\begingroup$ But I am trying to prove that all maximal ideals are of that form specifically the weak form of Hilbert's nullstellensatz. $\endgroup$ – user385705 Jan 28 '17 at 6:32
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It is enough to show that $\mathfrak{m}=\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle$ for some $\alpha=(\alpha_1,\ldots,\alpha_n)\in\mathbb{A}_k^n$, where $\mathbb{A}_k^n$ is affine $n$-space.

Now, here is where we use the Nullstellensatz. Note that the Nullstellensatz gives us an order-reversing bijection between algebraic subsets of $\mathbb{A}_k^n$ and the radical ideals of $k[x_1,\ldots,x_n]$. Therefore, since any maximal ideal is prime and therefore radical, $\emptyset=Z(1)\subsetneq Z(\mathfrak{m})$ implies that $Z(\mathfrak{m})$ is nonempty, so if $\alpha\in Z(\mathfrak{m})$, then it is easy to check that $$Z(\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle)=\{\alpha\}\subset\mathbb{A}_k^n$$ and so if $Z(\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle)\subsetneq Z(\mathfrak{m})$ then we would have that $\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle\supsetneq\mathfrak{m}$ contradicting maximality, so that $Z(\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle)= Z(\mathfrak{m})$ and therefore $\mathfrak{m}=\langle x_1-\alpha_1,\ldots,x_n-\alpha_n\rangle$, since both ideals are prime and therefore radical.

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This answer from a book: Elementary algebtaic geometry, Klaus Hulek, p25:

Theorem: Let $k$ be a field with infinitely many elements, and let $A=k[a_1,...,a_n]$ be a finitely generated $k-$algebra. If $A$ is a field, then $A$ algebraic over $k$.

In proof of Hilbert's Nullstellensatz we have $K$ is a field and finitely generated $k-$ algebra, then by theorem implies that $K$ is algebraic over $k$ then we have $k \hookrightarrow k[x_1,...x_n] \to k[x_1,...x_n]/m =K $ then $K$ is an algebraic extension of $k$ so $K\cong k$ where $k$ is algebraic closed.

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