## dim_k m/m^2 in a polynomial ring over an algebraically closed field (m being a maximal ideal)

I followed a course of commutative algebra during one semester and now I'm trying to enlarge a bit my knowledge. I want to show that given $A=k[X_1,\cdots,X_n]$ ($k$ algebraically closed) and $\mathfrak{m} \subset A$ a maximal ideal, that $dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 = n$. First of all, is this true ?

I know that $A/\mathfrak{m} = k$.

In fact, I think I've got ideas but small problems in the two ineaqualities. First it is clear that this is an $A/\mathfrak{m}$ vector space (universal properties and so on). Concerning its dimension :

1. First the "easy part" (at least it should be...) for $dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 \leq n$ : I start with the Nullstellensatz, taking $X_1-a_1,\cdots,X_n-a_n$ generators of $\mathfrak{m}$ as an $A$-module. Their classes also generate $\mathfrak{m}/\mathfrak{m}^2$ as an $A$-module. But I don't see how they could generate it with only $A/\mathfrak{m}$ coefficients. I mean, I can write $\overline{m} = \overline{P_1}\times\overline{X_1-a_1} + \cdots + \overline{P_n}\times\overline{X_n-a_n}$ (the bar means modulo $\mathfrak{m}^2$) but the $P_i$ are a priori in $A$, so why could I have some $\overline{P_i}$ in $A/\mathfrak{m}$ ? But I know I must be missing something evident here...

2. I think $dim_{A/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2 \geq n$ requires more work. In fact I read some things about algebraic geometry and tried to adapt it : if $\mathfrak{m}/\mathfrak{m}^2$ is generated by $r$ elements $f_1,\cdots,f_r$, then $\mathfrak{m} = \langle f_1,\cdots,f_r \rangle + \mathfrak{m}^2$. Then I wanted to use Nakayama's lemma but I needed $\mathfrak{m}$ to be in the Jacobson radical of $A$, which is not the case. So I had the idea of localising $A$ on $A\smallsetminus \mathfrak{m}$ and trying again. I'm a beginner with localisation, so maybe I made mistakes but I think that $\mathfrak{m}A_{\mathfrak{m}}/\mathfrak{m}A_{\mathfrak{m}}^2$ is isomorphic to $\mathfrak{m}/\mathfrak{m}^2$. Since $A_{\mathfrak{m}}$ is local I can use Nakayama's lemma and so $\mathfrak{m}A_{\mathfrak{m}} = \langle \frac{f_1}{1},\cdots,\frac{f_r}{1} \rangle$. But knowing that $\mathfrak{m}A_{\mathfrak{m}} = \langle \frac{X_1-a_1}{1},\cdots,\frac{X_n-a_n}{1} \rangle$ and using Krull's height theorem (thanks to Pete L. Clark : http://mathoverflow.net/questions/25875/generators-of-a-maximal-ideal-of-kx1-xn) I get $r \geq n$. Now because of the isomorphism, I can conclude.

My questions are : is this good, am I on the good way ? If not, what is the problem and what can I do ? (I'm not asking for any solution or demonstration, just for hints or advices, since I've spent several hours on this problem now...)

For the first part, rethink the fact that $\mathfrak{m}/\mathfrak{m}^2$ is a vector space over $A/\mathfrak{m}$. It's not because of any "universal properties and so on"; it's because it's annihilated by $\mathfrak{m}$. In particular, for any element $P \in A$ and any element $\bar{f} \in \mathfrak{m}$, $P \cdot \bar{f} = \bar{P} \cdot \bar{f}$. In other words, the action of $A$ on $A/\mathfrak{m}$ is via the homomorphism $A \to A/\mathfrak{m}$. – Graham Leuschke May 26 2010 at 14:55
By applying an isomorphism ("translation") you can assume all the $a_i$'s are zero. Then you know a basis of $\mathfrak m$ and of $\mathfrak m^2$ (composed both of monomials), and you can immediately compute the quotient. – Mariano Suárez-Alvarez May 26 2010 at 14:55
Graham > yes I know, I mean I used a universal property, using the fact that $\mathfrak{m}$ annihilates my module, so I can get a morphism that makes the module an $A/\mathfrak{m}$ module. Mariano > ow thanks... If I'm not wrong, this directly shows that the quotient only consists of linear combinations over $k$ of $X_1,\cdots,X_n$ and solves immediatly my problem ?! Ariyan > interesting, thanks for that. – Laurent May 26 2010 at 15:13