Let $k$ be an algebraically closed field, let $A$ be a finitely generated $k$-algebra. If $m\subset A$ is a maximal ideal, show that it is generated by $\operatorname{dim}_k m/m^2$ elements.
Consider the localization $A_m$; the proposition 2.8 (Atiyah-MacDonald) states that $mA_m$ is generated (as ideal) by $\operatorname{dim}_{A_m/mA_m}(m/m^2)A_m$ elements.
In general $A_m/mA_m\cong A/m$, but here $A/m\cong k$ by Zariski's lemma. So $mA_m$ is generated by $\operatorname{dim}_k(m/m^2)A_m$ elements.
Now, as $k$-modules $(m/m^2)A_m$ and $m/m^2$ are isomorphic; in fact we can view $m/m^2$ (and $k$) as an $A$-module, and recover the $k$-module structure as $k\otimes _A m/m^2$. Similarly, we can view $(m/m^2)A_m$ (and $k$) as an $A_m$-module, and recover the $k$-module structure as $k\otimes_{A_m}(m/m^2)A_m$. But also, we get the $A_m$-module structure of $(m/m^2)A_m$ as $A_m\otimes_Am/m^2$. Since $k\otimes_{A_m} A_m\cong k$, $$k\otimes _A m/m^2\cong k\otimes_{A_m} A_m\otimes_Am/m^2.$$
Thus $\operatorname{dim}_km/m^2=\operatorname{dim}_k(m/m^2)A_m$, and we are left to show that, if $n\in \mathbb N$ elements generate $mA_m$ as ideal, also $m$ is generated by $n$ elements in $A$. However this I don't understand if is true, i.e. from $n$ elements generating $(m/m^2)A_m$ I don't see how to get $n$ elements generating $m/m^2$. Thanks for any suggestion.
Best Answer
Let $k=\mathbb{C}$, $A=\mathbb{C}\times \mathbb{C}$ and $m=\langle (1,0)\rangle$.
Then $m/m^2=\{0\}$ but $m\neq 0$.