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Let $R$ be a local Noetherian ring (maximal ideal $m$, residue field $k$). Suppose $\{x_{1}, \ldots, x_{n}\}$ generate $m$. Is it true that dim$_{k}(m/m^2) \leq n$?

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At the very least, @greeft, the title should satisfy the minimal rules of grammar, and «generates of an ideal» does not! Master of the English language is not a requirement, but accepting corrections is immensely helpful. –  Mariano Suárez-Alvarez Feb 19 '13 at 22:33
    
Please pardon both my typo in the subject and my accidental undoing of your corrections. –  greeft Feb 20 '13 at 14:32

2 Answers 2

up vote 5 down vote accepted

Consider the map $\phi:R^n\to\def\m{\mathfrak m}\m$ which sends the $i$th standard basis element $e_i$ of $R^n$ to $x_i$. The hypothesis that $M$ is generated by the $x_i$ means that $\phi$ is surjective.

Whenever we have a morphism of $R$-modules $f:M\to N$, we have an induced map $\bar f:M/\m M\to N/\m N$, and if $f$ is surjective, $\bar f$ is also surjective, as one can easily check.

Applying this to $\phi$ we get a surjective map $\bar\phi:R^n/\m R^n\to\m/\m^2$. It is immediate that $\bar\phi$ is a map of $R/\m$-vector spaces, and that $\dim_{R/\m}R^n/\m R^n=n$. Usual linear algebra then implies that $\dim_{R/\m}\m/\m^2\leq n$.

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+1 Nice answer Mariano. –  user38268 Feb 19 '13 at 22:13
    
Thanks :-) ${}{}$ –  Mariano Suárez-Alvarez Feb 19 '13 at 22:13

I rephrase Mariano's answer in elementary terms: $x_1,\dotsc,x_n$ generate $\mathfrak{m}$ over $R$, hence $\overline{x_1},\dotsc,\overline{x_n}$ generate $\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$, since $x = r_1 \cdot x_1 + \dotsc + r_n \cdot x_n$ implies $\overline{x} = \overline{r_1} \cdot \overline{x_1} + \dotsc + \overline{r_n} \cdot \overline{x_n}$. Hence the dimension is $\leq n$.

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Heh. I started out writing more or less what you wrote.But then I wanted to actually prove it, and then one sees that one has to pretty much do what I wrote. (If you do use tensor products, then I'd say this is not elemetary: people tend to hate $\otimes$ :-) ) –  Mariano Suárez-Alvarez Feb 19 '13 at 22:31

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