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Let $R$ be a domain and $Q$ its field of fractions. Let $A$ be an $R$-module such that $rA=0$ for some $r \in R \backslash\{0\}$. Why $\mathrm{Ext}^n_R(Q,A)=0$ for all $n \geq 0$?

For $n=0$, $Ext^0_R(Q,A) \cong Hom_R(Q,A)$, if $f : Q \rightarrow A$, since $Q$ is a field must be injective, so if $a=f(1)$ then $0=ra=f(r)$, a contraddition. For $n >1$? The sequence $$ 0 \rightarrow A \rightarrow Ext^1_R(Q/R,A) \rightarrow Ext^1_R(Q,A) \rightarrow 0$$ is exact. Could the thesis follow from here?

user26857
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Rick88
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Let $\mathcal{A},\mathcal{B},\mathcal{C}$ be any $R$-linear categories and let $F : \mathcal{A} \times \mathcal{B} \to \mathcal{C}$ be any functor which is $R$-linear in each variable. (In your example, $\mathrm{Ext}^*_R : \mathbf{Mod}_R^{\mathrm{op}} \times \mathbf{Mod}_R \to \mathbf{Mod}_R$.) I claim that $F(A,B)=0$ if $A \in \mathcal{A}$ is any object such that $ r \cdot \mathrm{id}_A : A \to A$ is an isomorphism for all $r \in R \setminus \{0\}$ and $B \in \mathcal{B}$ is any object such that $r \cdot \mathrm{id}_B = 0$ for some $r \in R \setminus \{0\}$.

In fact, choose $r \in R \setminus \{0\}$ such that $r \cdot \mathrm{id}_B = 0$. Consider $r \cdot \mathrm{id}_{F(A,B)} : F(A,B) \to F(A,B)$. On the one hand, since $F$ is $R$-linear in the second variable, this is $F(\mathrm{id}_A,r \cdot \mathrm{id}_B)=0$. On the other hand, since $F$ is $R$-linear in the first variable, this is $F(r \cdot \mathrm{id}_A,\mathrm{id}_B)$, hence an isomorphism.

Martin Brandenburg
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  • The map $r:Q \rightarrow Q$ is the multiplication by $r$? – Rick88 Apr 18 '21 at 18:16
  • I'm not very familiar with this functorial language, which book do you recommend to start with to better understand these things? – Rick88 Apr 18 '21 at 18:20
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    Yes, $r : Q \to Q$ stands for mult. with $r$, but now I changed the notation and even generalized the proof. You only need the very basics here concerning functors, every textbook and even Wikipedia explains it. https://en.wikipedia.org/wiki/Functor + https://en.wikipedia.org/wiki/Preadditive_category#R-linear_categories + https://en.wikipedia.org/wiki/Ext_functor. The "joke" is that here the explicit construction of the Ext-functor is not necessary and only distracts from the easy categorical argument. If you don't like this general setting, just replace $F$ by $\mathrm{Ext}$ etc. – Martin Brandenburg Apr 18 '21 at 18:22
  • Ok, I understand it's very interesting, thank you very much. – Rick88 Apr 18 '21 at 21:19
  • Could a similar argument show that given two ideals $I, J$ of $R$ and $ r \in I $, $ Ext ^ 1_R (J / rJ, I) = 0 $? – Rick88 Apr 18 '21 at 21:26