TITLE: Proof of Hilbert's nullstellensatz,
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.