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Let $(A,m,k)$ be a local ring, and $A$ is a finitely generated $k$-algebra, and the maximal ideal $m$ is nilpotent. Let $x_1,\ldots,x_n \in m$ and their canonical images in $m/m^2$ generate this $k$-vector space.

How to show that $x_1,\ldots,x_n$ generate $A$ as $k$-algebra?

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When you put a hyphen between dollar-signs, so it's inside of $\TeX$, then it looks like a minus sign instead of a hyphen. Not only is a minus sign longer, but it is preceded and followed by spaces that don't happen with a hyphen. Observe: $k-$algebra versus $k$-algebra. –  Michael Hardy Mar 22 '12 at 16:35
    
Thank you, Michael. –  John Mar 23 '12 at 7:48

1 Answer 1

Well, your hypothesis is that $$ A=k[y_1,\dots,y_p] \text{ is local, }\mathfrak m^N=0, k=A/\mathfrak m \text{ and } \mathfrak m/\mathfrak m^2= k<\overline{x}_1,\dots,\overline{x}_n> $$ By Nakayama $\mathfrak m=(x_1,\dots,x_n)$. The chain $$ 0 \subseteq \mathfrak m^N \subseteq \mathfrak m^{N-1} \subseteq \mathfrak m^2 \subseteq \mathfrak m \subseteq A $$ then gives you the finite-dimensional $k$-vectorspaces $\mathfrak m^i/\mathfrak m^{i+1}$, so by induction it follows that $A$ itself is a finite-dimensional $k$-vectorspace [EDIT:] generated by $1$ and a finite set of monomials of the $x_i$

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Yes. Thank you,Blah. –  John Mar 23 '12 at 7:51

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