3

I am working on a problem that is asking me to compute Tor groups. I am trying to learn this material on my own, so I haven't had any formal education in this area.

Specifically, I am given the ring $R=\mathbb{Z}[X]/(X^n)$, an $R$-module $M$, and asked to calculate the groups $Tor_i^R(M, (x^m))$ where $x=X+(X^n)$ is the image of $X$ in $R$ and $0 \leq m \leq n-1$.

I know that in order to calculate these groups, I first need to construct a projective resolution for $(x^m)$ as an $R$-module, but I'm not sure where to begin. I am familiar with the proof that every module has a projective (in fact, free) resolution, but how do I actually construct the projective (free) modules and the maps between them? That is, how do I know what the canonical free modules look like concretely in this example?

Any help would be greatly appreciated.

Ethan
  • 148
  • You should be asking for flat resolutions, not projective. – 54321user Jul 20 '17 at 01:36
  • @MoarCake559 Not really. Projective resolutions work just fine to compute Tor functors. Flat resolutions work too because flat modules are acyclic for Tor, but that doesn't mean you have to use them. – Pedro Jul 20 '17 at 01:38
  • @PedroTamaroff But flat resolutions work in more domains. For example, you do not have enough projectives over $\mathbb{P}^1$. The general theory also focuses on flat resolutions for working with derived tensor products. – 54321user Jul 20 '17 at 01:43
  • @MoarCake559 That's not relevant in this situation, since we have projectives, which are flat. Why would the OP want to use a flat resolution? Care to exhibit a flat $\mathbb Z[x]/(x^3)$ module which is flat but not projective, anyways? – Pedro Jul 20 '17 at 02:05

1 Answers1

2

Let $A$ be your ring, and $I$ your ideal, where $m$ is as in your post. Begin by noting there is an epimorphism $A\to I$ obtained by multiplication by $x^m$. Its kernel is the ideal $(x^{n-m})$, and you can cover this by $A$ again by multiplication by $x^{n-m}$. You thus obtain a periodic resolution $$ \cdots\longrightarrow A\longrightarrow \cdots\stackrel{x^{n-m}}\longrightarrow A\stackrel{x^m}\longrightarrow I\to 0$$

from which you can compute Tor groups easily now, in particular you need only compute the first two Tor groups. Concretely, let $M$ be an $A$-module to the left. Tensoring the resolution above with $M$ one obtains

$$ \cdots\longrightarrow M\longrightarrow \cdots\stackrel{x^m}\longrightarrow M\stackrel{x^{m-n}}\longrightarrow M\to 0$$

where the maps are also multiplication by the indicated term. This gives that $$\operatorname{Tor}_i^A(I,M) = \begin{cases} \dfrac{\ker\{x^m : M \to M\} }{ x^{n-m} M } & \text{ if $i\neq 0$ is even}\\ \dfrac{\ker\{x^{n-m} : M \to M\} }{ x^mM} & \text{ if $i$ is odd}\\ \dfrac{M }{x^{n-m} M} & \text{ if $i=0$ } \end{cases}$$

Pedro
  • 122,002
  • Ah, thank you! That seems much easier and more straightforward than I anticipated. Is there a general rule of thumb for how to construct projective resolutions like this? It seems to me that there would have to be a fairly straightforward way to do it, since Tor and Ext groups don't seem like they should be too difficult to calculate. – Ethan Jul 20 '17 at 00:46
  • Hm. I seem to be getting that the Tor groups are all 0, aside from $Tor_0^R(M, (x^m)) \simeq M \otimes_R (x^m)$. Is this correct? If so, it seems like I should have been able to see that with less work, right? – Ethan Jul 20 '17 at 01:13
  • @Ethan No, the computation of projective resolutions is a discipline of its own, and so is the computation of Tor and Ext groups in general. This example was particularly simple because your ideal is principal and the resolution is periodic. But don't expect this to be a representative example if this endeavor. – Pedro Jul 20 '17 at 01:14
  • @Ethan The Tor groups are periodic, so if one is nonzero, there are many others that will be nonzero. Check your work. – Pedro Jul 20 '17 at 01:15
  • 1
    Yes, I had just realized that I could simplify things by noting that $M \otimes_R R \simeq M$. So if $i$ is even and nonzero, then we have $Tor_i^R(M, (x^m))=ann_M(x^m)/x^{n-m}M$, and if $i$ is odd we have $Tor_i^R(M, (x^m))=ann_M(x^{n-m})/x^mM$, where $ann_M(x^m)$ is the submodule in $M$ of elements annihilated by $x^m$. – Ethan Jul 20 '17 at 01:41