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I have that $R$ is a Noetherian local ring with maximal ideal $M$, and I want to show that $M/M^2$ is a finite dimensional vector space over the field $R/M$.

I think I've proved this (though I may be wrong), but I didn't use the fact that $R$ was local, so I'm wondering if it's necessary:

Since $R$ is Noetherian, $M$ is finitely generated, say by $x_1,\ldots, x_n$. Now, let $m+M^2\in M/M^2$. First of all, we can write $m=r_1x_1+\cdots+r_nx_n$. I now claim that $m+M^2=(r_1+M)x_1+\cdots+(r_n+M)x_n+M^2$, which I think is equivalent to saying that for any $m'\in M, m-r_1x_1-\cdots-r_nx_n-m'(x_1+\cdots+x_n)\in M^2$, which is true because everything besides the last term cancel to $0$, and the last term is the product of two elements of $M$. However, I have a feeling this proof is not complete/correct, since I didn't use the fact that $R$ is a local ring. Is that fact required?

user26857
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Nishant
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  • Notice that we have $$\text{$R$ commutative ring, $K$ field which is an $R$-module, $M$ finitely generated $R$-module} \\implies\ \text{$\dim_K(K \otimes_R M)<\infty$}$$ Apply this to $K=R/m,; K\otimes_R m \cong m/m^2$. – Watson Jan 24 '17 at 12:47
  • @Watson Can you give a reference of the result you mentioned. – Babai May 15 '17 at 05:17
  • @Babai : if $M$ is generated over $R$ by $m_1, \dots, m_n$, then you could prove that $1_K \otimes m_1, \dots, 1_K \otimes m_n$ generates $K \otimes_R M$ as $K$-vector space. – Watson May 15 '17 at 16:15

1 Answers1

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Your proof is basically correct, but perhaps it's helpful to look at a more general situation (indeed it's irrelevant that $R$ is local, keeping in mind that, over a field, a module being finitely generated is the same as it being finite-dimensional). Let $R$ be a ring and $M$ a finitely generated $R$-module. Then for any ideal $I$ of $R$, $M/IM$ is a finitely generated $R/I$-module. The proof is as you suggest. Take a set of generators $x_1,\ldots,x_n$ for $M$ as an $R$-module. Then they also generate $M/IM$ over $R/I$: if $x\in M$, we write $x=\sum_i r_ix_i$, and then, looking modulo $IM$, we have $x+IM=\sum_i (r_i+I)(x_i+IM)$. This equality holds because the definition of the $R/I$-module structure on $M/IM$ is $(r+I)(x+IM)=rx+IM$.

Keenan Kidwell
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  • Oh, okay. I guess the fact that it's a local ring is only used in the subsequent parts of the problem. – Nishant Jul 04 '14 at 17:15