# How does the coefficient ring influence the Euler characteristic?

Notation/background. Let $$C_\bullet$$ be a bounded chain complex of finitely generated abelian groups, with $$C_i\cong\mathbb{Z}^{a_i}$$. For any principal ideal domain $$R$$, let $$C_\bullet\otimes R$$ be the chain complex of finitely generated $$R$$-modules obtained by applying the functor $$(-)\otimes_\mathbb{Z}R\colon \mathbf{Ab}\to R\mathbf{Mod}$$ termwise. Furthermore, let $$\chi_R=\sum_{i\in\mathbb{Z}}(-1)^i\operatorname{rank} H_i(C_\bullet\otimes R)$$ be the Euler characteristic of $$C_\bullet\otimes R$$. It's a classical exercise in homological algebra (see e.g. Theorem 2.44 in Hatcher) to show that $$\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i {a_i}$$ whenever $$R$$ is a field and whenever $$R=\mathbb{Z}$$.

Question. What happens when $$R$$ is some other PID? Do we still get $$\chi_{R}=\sum_{i\in\mathbb{Z}} (-1)^i{a_i}$$?

Attempt/thoughts. Suppose that $$H_i(C_\bullet)\cong\mathbb{Z}^{b_i}\oplus T_i$$, where $$T_i$$ is the torsion subgroup of $$H_i(C_\bullet)$$. The universal coefficient theorem then tells us that $$H_i(C_\bullet\otimes R)\cong (H_i(C_\bullet)\otimes _{\mathbb{Z}}R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(H_{i-1}(C_\bullet),R)\\ \cong R^{b_i}\oplus (T_{i}\otimes_{\mathbb{Z}} R)\oplus \mathrm{Tor}_1^{\mathbb{Z}}(T_{i-1},R)\,.$$ When $$R$$ is a field (see this thread), then all free copies of $$R$$ obtained from the tensor products with $$T$$ are cancelled by free copies of $$R$$ obtained from the Tor-terms when we compute the alternating sum. It intuitively doesn't feel very likely that things would work out as nicely when $$R$$ is a general PID, but I find it hard to come up with a counterexample...

Does anyone have any ideas or suggestions about how to approach this?

• I think you might able to apply the universal coefficient theorem to pass to the field of fractions $F$ of $R$. We get $$H_i(C;F) \cong (H_i(C;R) \otimes_R F) \oplus \operatorname{Tor}_1^R(H_{i-1}(C;R),F)$$ and I think roughly the same analysis as in the thread you linked should show $\chi_R = \chi_F$. And we already know $\chi_F = \chi_{\mathbb{Z}}$! I haven't checked any of the details here, so take this with a grain of salt :) Commented Jan 10, 2021 at 19:24
• @diracdeltafunk I like the idea of using the field of fractions! Maybe one can even mimic the classical proof for $R=\mathbb{Z}$ (see e.g. Thm. 2.44 in Hatcher)? Idea: If $0\to A\to B\to C\to 0$ is a SES of f.g. $R$-modules, then the fact that $F\otimes_R\!-$ is exact (bc $F$ is torsion-free over $R$, and $R$ is a PID) implies that the induced sequence $0\to A\otimes F\to B\otimes F\to C\otimes F\to 0$ is exact. Since SESs over fields split, this gives $B\otimes F\cong (A\otimes F)\oplus (C\otimes F)$, which if I'm not mistaken gives $\operatorname{rk}B=\operatorname{rk}A+\operatorname{rk}C$. Commented Jan 10, 2021 at 20:42
• ...it might even work over a general integral domain $R$ with field of fractions $F$ (that this is a flat $R$-module follows from the more general fact that localization always is an exact functor), provided that we define $\operatorname{rk}A=\dim_F A\otimes_R F$. Commented Jan 10, 2021 at 21:21
• Oh right -- the fact that $F$ is flat is super useful! Commented Jan 10, 2021 at 21:22

I'll try to turn my comment into a proof. Let $$F$$ be the field of fractions of $$R$$. Since $$R$$ is a PID, the Universal Coefficient Theorem tells us that $$H_i(C;F) \cong (H_i(C;R) \otimes_R F) \oplus \operatorname{Tor}_1^R(H_{i-1}(C;R),F)$$ (this is an isomorphism of $$F$$-vector spaces). Since $$F$$ is flat over $$R$$, the $$\operatorname{Tor}$$ term is $$0$$. Now $$H_i(C;F) \cong H_i(C;R) \otimes_R F$$ implies $$\operatorname{rk}_F H_i(C;F) = \operatorname{dim}_F H_i(C;F) = \dim_F (H_i(C;R) \otimes_R F) = \operatorname{rk}_R H_i(C;R).$$ Thus, $$\chi_R = \chi_F = \chi_{\mathbb{Z}}$$.

Edit: To be extra clear; the thing I called $$H_i(C;F)$$ here is technically $$H_i((C_{\bullet} \otimes_{\mathbb{Z}} R) \otimes_R F)$$. The precise definition of $$H_i(C;F)$$ is $$H_i(C_{\bullet} \otimes_{\mathbb{Z}} F)$$. However, this is the same thing, since the map $$(A \otimes_{\mathbb{Z}} R) \otimes_R F \to A \otimes_{\mathbb{Z}} F$$ defined by $$(a \otimes r) \otimes f \mapsto a \otimes (rf)$$ gives a natural isomorphism $$({(-)} \otimes_{\mathbb{Z}} R) \otimes_R F \to {(-)} \otimes_{\mathbb{Z}} F$$. This is easy to see by thinking about these tensor products as pullbacks of sheaves on affine schemes.

Here is an attempt to combine the classical proof for the case $$R=\mathbb{Z}$$ (see e.g. Thm. 2.44 in Hatcher) with @diracdeltafunk's suggestion to work with the field of fractions.

Definition. Let $$R$$ be an integral domain, and let $$F=\operatorname{Frac}(R)$$ be its field of fractions. The rank of an $$R$$-module $$M$$ is then defined as $$\operatorname{rk} M=\dim_F (M\otimes_R F)$$.

Remark. If $$R$$ is a PID and $$M$$ is a finitely generated $$R$$-module with $$M\cong R^r\oplus R/(a_1)\oplus \cdots\oplus R/(a_m)$$, then $$M\otimes_R F\cong F^r$$ and we get $$\operatorname{rk} M=r$$. In other words: this definition of rank coincides with the usual definition of rank for finitely generated modules over a PID.

Lemma. Let $$R$$ be an integral domain, and let $$0\to A\to B\to C\to 0$$ be a short exact sequence of finitely generated $$R$$-modules. Then $$\operatorname{rk} B=\operatorname{rk} A+ \operatorname{rk} C$$.

Proof. Note that $$F$$ is a flat $$R$$-module (this follows for example from the more general fact that localization is an exact functor). Hence, the induced sequence $$0\to A\otimes_R F\to B\otimes_R F\to C\otimes_R F\to 0$$ of $$F$$-vector spaces is exact. Since every short exact sequence of vector spaces splits, this gives $$B\otimes_R F\cong (A\otimes_R F)\oplus(C\otimes_R F)$$. $$\:\square$$

Proposition. Let $$\chi_R$$ be defined as in the original post (but for an arbitrary integral domain $$R$$, with field of fractions $$F$$). Then $$\chi_R=\sum_{i\in\mathbb{Z}} (-1)^i\,a_i$$.

Proof. Let $$d_i^R=d_i\otimes \mathrm{id}_R$$ denote the differentials of the chain complex $$C_{\bullet}\otimes_{\mathbb{Z}} R$$. Note that we for every index $$i\in\mathbb{Z}$$ have short exact sequences $$0\to\operatorname{im}(d_{i+1}^R)\hookrightarrow \ker(d_{i}^R)\to H_i(C_\bullet\otimes_\mathbb{Z} R)\to 0\,,$$ $$0\to\operatorname{ker}(d_{i}^R)\hookrightarrow C_i\otimes R \to \operatorname{im}(d_{i}^R)\to 0\,.$$ By the lemma, this gives rise to equalities $$\operatorname{rk}\ker(d_{i}^R)=\operatorname{rk}\operatorname{im}(d_{i+1}^R) + \operatorname{rk} H_i(C_\bullet\otimes_\mathbb{Z} R)\,,$$ $$\operatorname{rk}(C_i\otimes_\mathbb{Z} R)=\operatorname{rk}\operatorname{ker}(d_{i}^R) + \operatorname{rk}\operatorname{im}(d_{i}^R)\,,$$ which implies $$a_i=\operatorname{rk}(C_i\otimes_\mathbb{Z} R)=\operatorname{rk} H_i(C_\bullet\otimes_\mathbb{Z} R)+\operatorname{rk}\operatorname{im}(d_{i+1}^R) + \operatorname{rk}\operatorname{im}(d_{i}^R)\,.$$ The desired result now follows by computing the alternating sum over all $$i\in\mathbb{Z}$$. $$\:\square$$

• This is very nice! I would next want to use the fact that every bounded chain complex of finitely generated abelian groups is quasi-isomorphic to a perfect complex (like $C_{\bullet}$), to weaken the assumption about $C_{\bullet}$. However, I think this requires $R$ to be flat over $\mathbb{Z}$. For example, the Euler characteristic of $0 \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$ changes when we tensor with $\mathbb{Z}/2$. Commented Jan 10, 2021 at 22:18