Showing that simple modules over finitely generated commutative algebras are 1-dimensional and isomorphic to... 
Let $A= \Bbb{C}[x_1, ... ,x_n]$.
a) Show that every simple $A$-module is 1-dimensional and is isomorphic to $\Bbb{C}[x_1, ... , x_n]/(x_1 - a_1, ... , x_n - a_n)$ for some $a_1, ... , a_n \in \Bbb{C}$.
b) Extend this result to finitely generated commutative algebras.

I could do part (a), but I'm not really sure about part (b). I would be glad if anybody could give me a hint so I can get started...
Thanks in advance
 A: In great generality: if $M \ne 0$ is a simple module over a ring $R$, then $M \cong R/m$, for some maximal ideal $m \subseteq R$.
Proof: $M$ simple means that $M$ has no nontrivial proper submodules. Now any nonzero element $x \in M$ generates a nonzero submodule $Rx \subseteq M$, hence $Rx = M$ is cyclic, so $M \cong R/I$ for some $R$-ideal $I$ (in fact $I = \text{ann}_R(x)$). Since ideals of $R/I$ are the $R$-submodules of $R/I$, and correspond to ideals in $R$ containing $I$, we must have that $I = m$ is maximal, and indeed $R/m$ is a nonzero simple $R$-module.
Now (in the commutative setting), Hilbert's Nullstellensatz implies that a finitely generated $\mathbb{C}$-algebra that is a field, must be $\mathbb{C}$ itself. Thus if $A$ is any finitely generated $\mathbb{C}$-algebra, then for any maximal ideal $m$ of $A$, $A/m$ is a finitely generated $\mathbb{C}$-algebra that is a field, hence must be $\mathbb{C}$. Thus the simple $A$-modules are precisely $\{A/m \mid m \subseteq R \text{ maximal} \}$, and these are all isomorphic to $\mathbb{C}$.
A: I think all you need to do is to notice that if you have any family of matrices $\{B_\alpha\}$ such that they pairwise commute, then you can simultaneously diagonalize all $B_\alpha$ (this is a standard fact of linear algebra).
Since any representaion of any commutative algebra is just picking a family of pairwise commuting matrices (possibly satisfying some relations), you see that irreducible representations must be $1$-dimensional. Otherwise, you can pick a basis where all the matrices are diagonal, and take span of the first basis vector, which will give you a subrepresentation.
Having now a $1$-dimensional (irreducible) $A$-module $V\simeq \mathbb{C}$, it is by definition a homomorphism (surjective, since $1\mapsto id$) $f\colon A\to End(V)\simeq \mathbb{C}$. By definition, $\ker f$ is a maximal ideal in $A$, since $A/ker f\simeq \mathbb{C}$ (the last is just the first isomorphism theorem).
A: Here is a hint. A finitely generated $\mathbf{C}$-algebra is isomorphic to $\mathbf{C}[x_1,\ldots,x_n]/I$ for some ideal $I$, and the maximal ideals of such a quotient are in natural bijection with the maximal ideals of $\mathbf{C}[x_1,\ldots,x_n]$ containing $I$, and moreover, the quotients are canonically isomorphic. That is, if $\mathfrak{m}$ is a maximal ideal of $\mathbf{C}[x_1,\ldots,x_n]$ containing $I$, then $\mathbf{C}[x_1,\ldots,x_n]/\mathfrak{m}\cong (\mathbf{C}[x_1,\ldots,x_n]/I)/(\mathfrak{m}/I)$.
