Let $R$ be a ring and $M$ an $R$-module with a finite free resolution (an “FFR module”). That is, there exists an exact sequence $0\to F_n\to F_{n-1}\to \cdots\to F_0\to M\to 0$ with each $F_i$ a finitely generated free $R$-module. If we denote by $r_i$ the rank of $F_i$, then the Euler characteristic of $M$ is defined to be $\chi(M) = r_0 – r_1 + \cdots + (-1)^nr_n$. One can easily prove that this is independent of the finite free resolution chosen and hence is a well-defined integer. In the post The Alternating Binomial Sum Vanishes we saw that it’s possible to prove that the alternating sum $\sum (-1)^i\binom{n}{i}$ vanishes by using some facts about Euler characteristics together with the Koszul complex.

Today’s problem shows that the Euler characteristic ins’t that interesting for ideals or homomorphic images of rings. First, here are some facts about the Euler characteristic, some of which might be useful for solving the problem:

  1. Any FFR module over a commutative ring has nonnegative Euler characteristic
  2. If an FFR module over a commutative ring has nonzero Euler characteristic, then the annihilator of $A$ is nil
  3. (Stallings) If $A$ is an FFR module with annihilator $I$ then $\chi(A)I = 0$.

We refer to these as Fact 1, Fact 2, and Fact 3.

Problem. If $I$ is a nonzero (!) ideal of a ring $R$ with a finite free resolution, show that $R/I$ also has a finite free resolution and that $\chi(I) = 1$ and $\chi(R/I) = 0$.

Solution. If $I$ has a finite free resolution then it can be pasted onto the the short exact sequence $0\to I\to R\to R/I\to 0$, so $R/I$ has an FFR too. Moreover, the exact sequence shows that $\chi(R) = 1 = \chi(I) + \chi(R/I)$. By Fact 1, we just have to show that $\chi(R/I) = 0$. By Fact 3, $\chi(R/I)$ annihilates the annihilator of $R/I$, so in particular it annihilates $I$. Since $I$ is nonzero, we see that $\chi(R/I)$ cannot be one. QED

In fact, as mentioned in the post on the “Alternative Binomial Sum”, the Euler characteristic of any module with nonzero annihilator must be zero (Vasconcelos’ theorem), but this might be a bit overkill. We also know that Fact 2 could have been used if we knew there were nonzero nilpotents in $I$.

Leave a Reply

Your email address will not be published. Required fields are marked *