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In the Wikipedia article on Hilbert's Nullstensatz, http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz the following application of the Weak Nullstensatz is mentioned:

Commuting matrices

The fact that commuting matrices have a common eigenvector – and hence by induction stabilize a common flag and are simultaneously triangularizable – can be interpreted as a result of the weak Nullstellensatz, as follows: commuting matrices form a commutative algebra $K[A_1,\ldots,A_k]$ over $K[x_1,\ldots,x_k];$ the matrices satisfy various polynomials such as their minimal polynomials, which form a proper ideal (because they are not all zero, in which case the result is trivial); one might call this the '''characteristic ideal''', by analogy with the characteristic polynomial.

One then defines an eigenvector for a commutative algebra as a vector $v$ such that for all $x \in A$ one has $x(v) = \lambda(x)\cdot v$ for a linear functional $\lambda\colon A \to K.$ This simply linearizes the definition of an eigenvalue, and is the correct definition for a common eigenvector, as if $v$ is a common eigenvector, meaning $A_i(v)=\lambda_i v,$ then the functional is defined as $\textstyle{\lambda(c_0 I + c_1 A_1 + \cdots c_k A_k) := c_0 + \sum c_i \lambda_i}$ (treating scalars as multiples of the identity matrix $A_0 := I$, which has eigenvalue 1 for all vectors), and conversely an eigenvector for such a functional $\lambda$ is a common eigenvector. Geometrically, the eigenvalue corresponds to the point in affine $k$-space with coordinates $(\lambda_1,\ldots,\lambda_k)$ with respect to the basis given by $A_i.$

Then the existence of an eigenvalue $\lambda$ is equivalent to the ideal generated by (the relations satisfied by) $A_i$ being non-empty, which exactly generalizes the usual proof of existence of an eigenvalue existing for a single matrix over an algebraically closed field by showing that the characteristic polynomial has a zero.

I am somewhat struggling to make sense of that. The Weak Nullstellensatz says that I find a functional $\lambda \colon K[A_1, \dots, A_k] \to K$, and $\lambda(A_i)$ is an eigenvalue of $A_i$. But how do I conclude that a common eigenvector exists?

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If someone does manage to rescue this proof, please make sure to mention it on mathoverflow.net/questions/42512/… –  darij grinberg Oct 23 '10 at 15:45
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up vote 3 down vote accepted

My "solution" is a bit strange, but I hope it is correct.

We regard $V$ as a module over $A$. $X = {\rm Spec}\ A$ is zero-dimensional. Then (as a sheaf on $X$) $V$ decomposes into a direct sum of sheaves over the finitely many points of $X$. This corresponds to a decomposition $V$ into subspaces corresponding to different eigenvalues in the sense you described. We must have a point $p$ such that the piece $V_p$ above it (which is the localization of $V$ at the maximal ideal $\mathfrak{m}$ of $A$ corresponding to $p$) is nonzero. Localizing at this ideal and restricting ourselves to the subspace $V_p$, we may assume that $A$ is local. If there would be no common eigenvector in $V_p$, then we would have $\mathfrak{m}V_p = V_p$, so by Nakayama's lemma $V_p = 0$, a contradiction.

darij: Does it deserve to be "awfully sophisticated" by your standards?

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To elaborate: By Nakayma, $V_p \supsetneq \mathfrak{m} V_p \supsetneq \mathfrak{m}^2 V_p \supsetneq \cdots$, with strict inequality until we reach the zero module. So, there is some $n$ such that $\mathfrak{m}^n V_p \neq (0)$ but $\mathfrak{m}^{n+1} V_p=0$. For $v \in \mathfrak{m}^n V_p \setminus \{ 0 \}$, the vector $v$ is annihilated by $\mathfrak{m}$. –  David Speyer Oct 23 '10 at 17:04
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Just remember that the standard proof that commuting matrices have a common eigenvector is trivial. If $A, B$ commute and $V_\alpha$ is the $\alpha$-eigenspace of $A$ (non-zero for some $\alpha$), then $\alpha Bv=BAv=A(Bv)$ for every $v\in V$, so $V_\alpha$ is invariant with respect to $B$. Now if $V_\alpha$ is the whole space, we can forget about $A$ and reduce the number of matrices. If $V_\alpha$ is not the whole space, we can reduce the dimension. No Nakayama or Hilbert are needed. –  Mark Sapir Oct 23 '10 at 17:44
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