Commuting Matrices have a common eigenvector (using Hilbert's Nullstellensatz) I am aware that there exists elementary proof of the fact that commuting matrices have a common eigenvector. But recently I came across this following statement from the Wikipedia article "https://en.wikipedia.org/wiki/Triangular_matrix#Simultaneous_triangularisability" which goes as:
"The fact that commuting matrices have a common eigenvector can be interpreted as a result of Hilbert's Nullstellensatz: commuting matrices form a commutative algebra $k[A_1,A_2, \cdots A_n]$ over $K[x_{1}, x_{2}, \cdots x_n]$ which can be interpreted as a variety in k-dimensional affine space, and the existence of a (common) eigenvalue (and hence a common eigenvector) corresponds to this variety having a point (being non-empty), which is the content of the (weak) Nullstellensatz. In algebraic terms, these operators correspond to an algebra representation of the polynomial algebra in k variables."
I am having a hard time understanding the statements made here. In particular, I want to know
a)How can we interpret the algebra $k[A_1,A_2, \cdots A_n]$ as a variety?
b)What does the article mean by the statement "a common eigenvalue and hence a common eigenvector"?
We know that a common eigenvalue for two matrices does not mean they have the same eigenvector!
It will be nice if someone can help me out with this.
 A: Define $k[A_1,A_2,\dots, A_n]$ as the $k$ sub-algebra of $M_n(k)$ generated by $A_1,A_2,\dots ,A_n $.
Since $A_1,A_2,\dots ,A_n$ commute $k[A_1,A_2,\dots A_n ]$ is a commutative finitely generated $k$-algebra, you can consider the $k$-algebra homomorphism $$\psi : k[x_1,x_2,\dots ,x_n ]\rightarrow k[A_1,A_2,\dots,A_n ]$$ $$x_i\mapsto A_i$$ which is  a surjective morphism of commutative $k$-algebras. 
Assume $k$ is algebraically closed. Then any maximal ideal of $k[A_1,A_2,\dots ,A_n]$ is of the form $(A_1-\lambda_1I,A_2-\lambda_2 I,\dots ,A_n-\lambda_nI)$
If $p_i(x_i)$ is the characteristic polynomial of $A_i$, then $\psi(p_i)=0\implies p_i\in (x_1-\lambda_1,x_2-\lambda_2,\dots ,x_n-\lambda_n)$ In other words, $\lambda_i$ is an eigen-value of $A_i$. Let $V_i=\operatorname{Ker}(A_i-\lambda_iI)$. 
Suppose $A_1,A_2,\dots ,A_n$ have a common eigenvector say $v$ with $A_iv=\mu_i v$. Then $$p(A_1,A_2,\dots,A_n)=0\implies p(A_1,A_2,\dots ,A_n )v=0\implies p(\mu_1,\dots ,\mu_n)=0$$ $$ \text{ i.e. } \operatorname{Ker}\psi \subset (x_1-\mu_1,\dots ,x_n-\mu_n) $$In other words, the 'variety' contains a 'point'. 
Note: Strictly speaking, this only defines a closed $k$ sub-scheme of $\mathbb A^n_k$, namely $\operatorname{Spec}k[A_1,A_2,\dots,A_n ]$
Now comes the hard part of this answer. I shall slightly modify and spell out the details of this answer.
$\operatorname{Spec}k[A_1,A_2,\dots,A_n ]$ is $0$-dimensional. This is because $p_i(x_i)\in \operatorname{Ker}\psi \ \forall \ i$. In other words, $\operatorname{Spec}  k[A_1,A_2,\dots,A_n ]$ is just a finite collection of closed points and in particular has discrete topology.  $V=k^n$ is a quasi-coherent sheaf on $\operatorname{Spec}k[A_1,A_2,\dots,A_n ]$
Result: If $X=U\sqcup V$ is a dis-connected scheme, and $\mathcal F$ is a quasi-coherent sheaf on $X$, then $\mathcal F\cong (i_U)_*\mathcal F|_U\oplus (i_V)_*\mathcal F|_V$
 Using this, one splits $V$ as a direct sum of quasi-coherent sheaf over each point. Say $\operatorname{Spec}k[A_1,\dots, A_n ]=\{ m_1,m_2,\dots, m_N\}$. Then we have the stalk over each point is $V_{m_l}$. Since $V\neq 0$, by basic commutative algebra, we get $V_{m_l}\neq 0$ for some $l$. Let $\frac{v}{1}\in V_{m_l}$ and say $m_l=  (A_1-\lambda_1I,A_2-\lambda_2 I,\dots ,A_n-\lambda_nI) $,  For each $\alpha\in \{1,2,\dots, N \}\backslash \{ l\}$, exists $q_\alpha \in m_\alpha$ such that $q_\alpha(v)\neq 0$ So we get $\operatorname{Ann}(v)$ is $m_l$ -primary. So $k[A_1,A_2,\dots,A_n]v$ has a non-zero cyclic sub-module  say $k[A_1,\dots, A_n]v_0$ such that $\operatorname{Ann}v_0=m_l$. In other words $v_0$ is a common eigen-vector.
Update: To show $k[A_1,A_2,\dots,A_n ]$ is $0$-dimensional, observe that $k[A_1,A_2,\dots,A_n ]|_k$ is integral extension by Cayley-Hamilton theorem and $\dim k=0$.
