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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?

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    $\begingroup$ 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 :) $\endgroup$ Commented Jan 10, 2021 at 19:24
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    $\begingroup$ @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 $\operatorname{rk}B=\operatorname{rk}A+\operatorname{rk}C$. $\endgroup$ Commented Jan 10, 2021 at 20:42
  • $\begingroup$ ...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 $\operatorname{rk}A=\dim_F A\otimes_R F$. $\endgroup$ Commented Jan 10, 2021 at 21:21
  • $\begingroup$ Oh right -- the fact that $F$ is flat is super useful! $\endgroup$ Commented Jan 10, 2021 at 21:22

2 Answers 2

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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.

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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 $i\in\mathbb{Z}$. $\:\square$

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    $\begingroup$ 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 $\mathbb{Z}/2$. $\endgroup$ Commented Jan 10, 2021 at 22:18

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