I found the following assertion at page 62 in Geometry of schemes by Eisenbud and Harris:

Let $R$ be a local $K$-algebra of vector-space dimension $2$, where $K$ is an algebraically closed field, and let $\mathfrak m$ be its maximal ideal. Then $R/\mathfrak m\cong K$ and $\mathfrak m^2=0.$

I can prove the first claim, by Nullstellensatz and the fact that $K$ is algebraically closed, but I failed to see why the square of $\mathfrak m$ is $0$. The book says that we might use Nakayama's lemma, but I could not perceive how to use that lemma.
So any hint or help will be mostly appreciated, and thanks in advance.

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    $\begingroup$ It seems to me that this make sense. Evidently $\mathfrak{m}$ is a proper subspace of $R$, and so must either be $1$ or $0$ dimensional. If it's the latter, then $R$ is a field, and so can't be two dimensional over $K$ (since it's alg closed). So, it must be dimension one. Then, it's square is either zero dimensional, or equal to $\mathfrak{m}$ which would imply (by Nakayama) that $\mathfrak{m}=0$, which is, once again, impossible. Does that make sense to you? $\endgroup$ – Alex Youcis Aug 22 '13 at 4:46
  • $\begingroup$ @AlexYoucis Thanks for this clear and beautiful argument! Now I see it crystal clear. :) $\endgroup$ – awllower Aug 22 '13 at 7:12

The $K$-vector space $R$ admits a filtration $R\supseteq \mathfrak{m}\supseteq \mathfrak{m}^2\supseteq\cdots \supseteq\{0\}$ where each quotient (for $i\geq 0$) $\mathfrak{m}^i/\mathfrak{m}^{i+1}$ is an $R/\mathfrak{m}\cong K$-vector space.

The fact that $R$ has dimension $2$ as a $K$-vector space restricts the length of this chain. Indeed, if $\mathfrak{m}^2\subsetneq \mathfrak{m}$, then we're basically done. (Why?) In order to rule out the case $\mathfrak{m}^2=\mathfrak{m}$, use Nakayama's lemma.

I hope this helps! (I mentioned the filtration because it's a very useful thing to keep in mind.)

Exercise 1: Let $R$ be a local Artinian ring with unique maximal ideal $\mathfrak{m}$. Prove that $\mathfrak{m}$ is a nilpotent ideal, that is, $\mathfrak{m}^i=0$ for some positive integer $i$. (You can assume that $R$ is also Noetherian if you are not aware that this is a Corollary of the hypothesis that $R$ is Artinian.)

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  • $\begingroup$ I see! We can show the exercise by the finiteness of the filtration mentioned. And this answer is indeed very clear and excellent! Thanks very much. $\endgroup$ – awllower Aug 22 '13 at 7:15
  • $\begingroup$ You're welcome, @awllower! $\endgroup$ – Amitesh Datta Aug 22 '13 at 8:30

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