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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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  • $\begingroup$ It plays a role for integral ring extensions - for a geometric characterization of integral ring extensions, see Atiyah-MacDonald 1969, Ch 5. Exercise 35. $\endgroup$
    – Dietrich Burde
    Dec 11, 2013 at 10:59
  • $\begingroup$ Inegral ring extensions... ok, I'll have a look to that place in Atiyah-MacDonald, thank you. $\endgroup$
    – Qfwfq
    Dec 11, 2013 at 11:10
  • $\begingroup$ 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 $\mathbb R^2$. $\endgroup$
    – Olivier
    Dec 11, 2013 at 13:05
  • $\begingroup$ @Olivier: I agree $\endgroup$
    – Qfwfq
    Dec 11, 2013 at 13:08
  • $\begingroup$ @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. $\endgroup$
    – Adam Gal
    Dec 11, 2013 at 16:17

1 Answer 1

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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 $\textrm{Spec}(A[T]) = A^1_{\textrm{Spec}(A)}$. 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 $n$-th infinitimal neighbourhood of the zero section of $A^1_A$ over $\textrm{Spec}(A)$ union the inverse image of $V(I) = \textrm{Spec}(A/I)$ 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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