Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm attempting to understand Eisenbud's proof that:

If $x_1,x_2,\ldots,x_i$ is an $M$-sequence, then $H^i(M\otimes K(x_1,...,x_n))=((x_1,\ldots,x_i)M:(x_1,\ldots,x_n))/(x_1,\ldots,x_i)M$.

Here $M$-sequence means $M$-regular, i.e. $(x_1,\ldots,x_i)M\neq M$ and for each $j \in {1,\ldots,i}, x_j$ is not a zero divisor on $M/(x_1,\ldots,x_{j-1})M$, and $K$ is the functor taking an ideal of $A$ to its Koszul complex.

The proof proceeds by induction on both $i$ and $n$, and in the middle we apply the induction hypothesis to find that $H^{i-1}(M\otimes K(x_1,\ldots,x_n))=0,$ from which Eisenbud claims we can use the long exact sequence of homology $$\cdots\rightarrow H^{i-1}(M\otimes K(x_1,\ldots,x_n))=0\rightarrow H^{i-1}(M\otimes K(x_1,\ldots,x_{n-1})\overset{x_n}{\rightarrow} H^{i-1}(M\otimes K(x_1,\ldots,x_{n-1})\rightarrow H^{i}(M\otimes K(x_1,\ldots,x_n))\rightarrow \cdots$$ to see that $H^i(M\otimes K(x_1,\ldots,x_n)$ must be the kernel of the next boundary map after the ellipsis.

But then the kernel of the last arrow shown should be 0, and thus the same for the second-to-last arrow's image, which means the boundary map, multiplication by $x_n$, I show must be surjective. That's where I fall off the tracks. All I can see is that it's injective!

Thanks for your help.

---EDIT--- I'm sorry if my writing's unclear! I really doubt this is very tricky, if I can just get hold of one passionate diagram-chaser. Here's my version of the whole proof, minus the bit I remain stuck on.

Show that for a finitely generated module $M$ over a Noetherian ring $A$ with an ideal $I=(x)=(x_1,\ldots,x_n)$, if $x_1,\ldots,x_i$ is $M$-regular: $$H^l(M\otimes K(x))=((x_1,\ldots,x_i)M:I)/(x_1,\ldots,x_i)M$$ Deduce $H^k(M\otimes K(x))=0$ for $k<i$

We'll first do the deduction, assuming the main result. Then I claim $((x_1,\ldots,x_k)M:I)=(x_1,\ldots,x_k)M$ because if some $m\in M\setminus (x_1,\ldots,x_k)M$ is sent to $(x_1,\ldots,x_k)M$ by $a\in I, a$ is a zero divisor mod $(x_1,\ldots,x_k)M$. But we assumed $(x)$ was regular all the way up to $x_i,$ so in particular $x_{k+1}m \neq 0$ in $M/(x_1,\ldots,x_k)M$, and thus $m \notin ((x_1,\ldots,x_k)M : I)$.

We perform double induction on $i$ then $n$. Begin with $i=0$, when the claim just says that the zeroth homology is $(0 : I)$. This is immediate from the definition of the Koszul complex: $a(x_1,\ldots,x_n)$ is zero just if each $ax_i$ is.

Now fix $i$ and vary $n$. Defining $x'=(x_1,\ldots,x_{n-1}),$ we can find the module we want as the kernel of the boundary map $x_n: H^i(M\otimes K(x'))\rightarrow H^i(M\otimes K(x'))$, for by the induction hypothesis regarding $n$ we have $ H^i(M\otimes K(x'))=((x_1,\ldots,x_i)M:(x_1,\ldots,x_{n-1}))/(x_1,.\ldots,x_i)M$. The elements of this homology are the parts of $M$ that go into $(x_1,\ldots,x_i)M$ under multiplication by any element of the ideal $(x_1,\ldots,x_{n-1})$; if we take one such, $m$, and multiply it by $x_n,$ and find it's in $(x_1,\ldots,x_i)M,$ then we've gotten that all $n$ generators of $I$ send $m$ into $(x_1,\ldots,x_i)M$. That's precisely the definition of $((x_1,\ldots,x_i)M: I))/(x_1,\ldots,x_i)$, which is what we want $H^i(M\otimes K(x))$ to be.

Another bit of info we have from induction, now on $i$, is that $H^{i-1}(M\otimes K(x))=0,$ by the same argument from above showing this works for every fragment of an initial sequence. From than we get that the boundary map at level $H^{i-1}$ is injective, so that the image of $H^{i-1}(M\otimes K(x'))$ in $H^i(M\otimes K(x))$ is a quotient of $H^{i-1}(M\otimes K(x'))$ by an isomorphic subset of itself. But I cannot see why $H^{i-1}(M\otimes K(x'))$ should be finite, or any other reason why this quotient should be zero.

share|improve this question
    
Try asking at MathOverflow if you don't get any good answers in a few days. –  Grumpy Parsnip May 14 '12 at 1:45
add comment

1 Answer

up vote 2 down vote accepted

OK, I've worked it out-we just needed to apply the induction hypothesis to see that the homology is zero all the way up to the module we're interested in-it applies in the very same way to the two terms of the long sequence after the one I had already shown is 0. Then the map I couldn't quite get to be injective becomes so, and we're done.

Incidentally, this documents an error in the first printing of Eisenbud's Commutative Algebra with a View Towards Algebraic Geometry. I imagine it's corrected in more recent versions, but it's not mentioned in the errata on his site.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.