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.


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.)

  • $\begingroup$ The fact that $A$ is local does not imply that every prime ideal is maximal. $\endgroup$ – Tomasz Lenarcik Nov 24 '13 at 0:45
  • $\begingroup$ @TomaszLenarcik In general no, but in this case, yes. $\endgroup$ – user89712 Nov 24 '13 at 0:46
  • $\begingroup$ The statement "every prime ideal is maximal" is (in this case) equivalent to the fact that $A$ is of finite dimension over $K$. But the latter does not follow from the assumptions. $\endgroup$ – Tomasz Lenarcik Nov 24 '13 at 0:50
  • $\begingroup$ @TomaszLenarcik If you have a local ring and you know that every prime ideal in that ring is the intersection of maximal ideals containing it, then what's your conclusion? $\endgroup$ – user89712 Nov 24 '13 at 0:53
  • $\begingroup$ Ok, now I see it! Local + f.g. indeed implies $\dim=0$. I'm not used to this kind of setup, since localizations of f.g. algebras usually tends not to be f.g. any more. Sorry for not reading carefully. $\endgroup$ – Tomasz Lenarcik Nov 24 '13 at 0:58

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.


What we can learn from the above argument are the following observations:

  1. Your statement is true without the assumption that $K$ is algebraically closed.
  2. You can even assume that $A$ is semi-local, i.e. it has finite number of prime ideals.
  3. 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.
  • 1
    $\begingroup$ 2. Certainly (and trivially) a semilocal Jacobson ring has Krull dimension $0$ and therefore a finite number of prime ideals. 3. follows immediately from 2. $\endgroup$ – user89712 Nov 24 '13 at 8:09

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