Proof of Hilbert's nullstellensatz, 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.

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. 
