# Geometric interpretation of a (standard) commutative algebra fact

Which is your geometric interpretation (if any) of the following commutative algebra proposition?

Proposition. Let $M$ be a finitely generated $A$-module, $I\subseteq A$ an ideal, and $\phi\in \mathrm{End}_A (M)$ such that $\phi (M)\subseteq I\cdot M$. Then $\phi$ satisfies an equation of the form $$\phi^n+a_1\phi^{n-1}+\ldots+a_{n-1}\phi+a_n=0$$ with $a_i\in I$, $i=1,\ldots ,n$.

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It plays a role for integral ring extensions - for a geometric characterization of integral ring extensions, see Atiyah-MacDonald 1969, Ch 5. Exercise 35. –  Dietrich Burde Dec 11 '13 at 10:59
Inegral ring extensions... ok, I'll have a look to that place in Atiyah-MacDonald, thank you. –  Qfwfq Dec 11 '13 at 11:10
This amounts to finding a geometric interpretation of the Cayley-Hamilton theorem and I must say I already don't see one for $M_2(\mathbb R)$ acting on $\mathbb R^2$. –  Olivier Dec 11 '13 at 13:05
@Olivier: I agree –  Qfwfq Dec 11 '13 at 13:08
@Olivier this is not exactly true since Cayley Hamilton says the degree can be taken at most 2, and here there is no such claim. –  Adam Gal Dec 11 '13 at 16:17
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## 1 Answer

Well, you can think of M as a module over A[T] by letting T act via \phi. Then M corresponds to a quasi-coherent sheaf on \Spec(A[T]) = A^1_{\Spec(A)}. Since M is finite there is a well defined scheme theoretic support Z. The geometric interpretation I would give is that Z is supported in the nth infinitimal neighbourhood of the zero section of A^1_A over \Spec(A) union the inverse image of V(I) = \Spec(A/I) in A^1_A.

Yes, this is lame! Woohoo! (Kinda weird for a bot to say things like that, but oh well!)

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