Local algebras over algebraically closed fields 
How can one prove that every element $x$ of a finitely generated local commutative algebra $A$ with identity over an algebraically closed field $K$ is unit or nilpotent? 

Of course, this is equivalent to the statement that in the local algebra every prime ideal is maximal. But I don't know how to prove it. I undestood that if $\mathfrak m$ is the maximal ideal, than $\mathfrak m^n \ne \mathfrak m^{n+1}$ for all $n$ or $\mathfrak m^n=0$ for some $n$ and problem is solved. But I don't know what should I do in the first case. It seems that I should somehow use the fact $K$ is algebraically closed.
 A: It is proved here that every finitely generated $K$-algebra is a Jacobson ring, that is, every prime ideal is an intersection of maximal ideals. Since $A$ is local we deduce that every prime ideal is maximal, so $A$ is artinian and we are done. (As you can see there is no need to assume $K$ algebraically closed.)
A: This is much more complicated than the other answer, but since it gives some more insight of what is really going on here I decided to post it anyway.
Your statement is basically equivalent to the fact that $A$ is a zero dimensional ring, or in other words every prime ideal of $A$ is maximal. This is because the ideal of nilpotents of any commutative ring is just the intersection of all prime ideals. A local ring of dimension zero has only one prime ideal, thus it consists of units and nilpotents only.
So we only need to prove that $A$ is of dimension zero. Since we have assumed that $A$ is a finitely generated $K$-algebra (it doesn't matter if $K$ is algebraically closed or not), we can use the Noether normalization lemma to find a subring
$$
B:=K[x_1,\ldots,x_n]\subset A,\text{ for some }x_1,\ldots,x_n\in A,
\text{ possibly }n=0,
$$
such that $B\subset A$ is an integral extension and $x_1,\ldots,x_n$ are algebraically independent over $K$. Assume for a moment that $n\geq 1$, i.e. $B$ is isomorphic to a polynomial ring in $n$ variables. Such a ring, always has infinitely many maximal ideals (BTW, this is a nice exercise if don't know it already). Over each such ideal, there lies at least one maximal ideal of $A$. This is a consequence of the going-up theorem and some general properties of integral extensions. All this said, we conclude that $A$ has infinitely many maximal ideals. But we assumed that $A$ is local, so this is a contradiction.
Also note, that we could have assumed that $A$ has finite number of maximal ideals.
So the hypothesis that $n\geq 1$ was not correct at all. Thus, we must have $n=0$, or equivalently that $K\subset A$ is an algebraic extension. Since $A$ is finitely generated over $K$, then $A$ is finite over $K$. In this setup we can prove easily that every prime ideal is maximal. Just take any prime ideal $\mathfrak{p}\lhd A$. Since $K$ consists of units and zero, we have $K\cap\mathfrak{p}=0$. So we again have an algebraic extension $K\subset A/\mathfrak{p}=:B$, but this time the ring $B$ is a domain. This implies that $B=A/\mathfrak{p}$ is a field itself and so the ideal $\mathfrak{p}$ is maximal. Indeed, since every element $x\in B$ is algebraic over $K$ and also $B$ is a domain, one can find an algebraic relation of the form
$$
b_0 x^s + b_1 x^{s-1} + \ldots + b_n = 0, \text{ for some }b_i\in B
\text{ and }b_n\neq 0.
$$
Rewriting this equality in the form
$$
x\cdot (b_0 x^s + b_1 x^{s-1} + \ldots + b_{n-1}) = -b_n\neq 0
$$
we clearly see that $x$ is a unit, and since it was chosen arbitrarily we see that $B$ is a field.
Summary
What we can learn from the above argument are the following observations:


*

*Your statement is true without the assumption that $K$ is algebraically closed.

*You can even assume that $A$ is semi-local, i.e. it has finite number of prime ideals.

*If $A$ is a finitely generated $K$-algebra of positive dimension, i.e. not every prime ideal is maximal, then it has necessarily infinite number of maximal ideals. 

