Let $R$ be a PID and $M$, $N$ be $R$-modules. I am trying to show that $$\forall n\ge 2~: \operatorname{Ext}_{R}^{n}(M,N)=0.$$

For example $\forall n\ge 2~: \operatorname{Ext}_{\mathbb Z}^{n}(M,N)=0$.

Here $\operatorname{Ext}^*_R$ is the derived functor of the functor of homomorphisms of $R$-modules $\operatorname{Hom}_R$. It's defined using projective/injective resolutions, and long exact sequences. On the other hand the definition of a PID involves ideals of the ring, and it's not immediately clear how to relate this condition to the definition of $\operatorname{Ext}$.


Over a PID, injective modules are the same as divisible modules. A module $M$ over a domain $R$ is divisible if, for every $x\in M$ and every $r\in R$, $r\ne0$, there is $y\in M$ such that $x=ry$.

If $M$ is injective, then $M$ is divisible: indeed, for $r\ne0$, the map $\mu_r\colon R\to R$ defined by $\mu_r(t)=rt$ is a monomorphism; if $x\in M$, the map $\mu_x\colon R\to M$ defined by $\mu_x(t)=tx$ is a homomorphism. By injectivity of $M$, there exists a homomorphism $g\colon R\to M$ such that $g\circ\mu_r=\mu_x$. Then $$ x=\mu_x(1)=g(\mu_r(1))=g(r)=rg(1) $$ and we can take $y=g(1)$.

Conversely, if $M$ is divisible, then $M$ is injective. By Baer's criterion, we just have to prove that every homomorphism $f\colon I\to M$, where $I$ is an ideal of $R$, can be extended to $R$. Since $R$ is a PID, we have $I=rR$. Let $x=f(r)$ and take $y\in M$ such that $x=ry$; then $\mu_y$ is the required extension.

It is obvious that any homomorphic image of a divisible module is divisible.

Let $M$ and $N$ be $R$-modules and embed $N$ in an injective module $E$. Then also $E/N$ is injective and applying $\operatorname{Hom}_R(M,-)$ to the exact sequence $0\to N\to E\to E/N\to 0$ gives the long exact sequence $$ \dots\to \operatorname{Ext}_R^{n-1}(M,E/N)\to \operatorname{Ext}_R^{n}(M,N)\to \operatorname{Ext}_R^{n}(M,E)\to\dotsb $$ If $n>1$, we have $\operatorname{Ext}_R^{n-1}(M,E/N)=0$ and $\operatorname{Ext}_R^{n}(M,E)=0$, because $E$ and $E/N$ are injective. Therefore $\operatorname{Ext}_R^{n}(M,N)=0$.

Of course one could go with a more direct method: if $N$ is an $R$-module, then, with the above notation, $0\to N\to E\to E/N\to 0\to 0\to\dotsb$ is an injective resolution for $N$ and this shows that the homology groups of the complex $0\to E\to E/N\to 0\to\dotsb$ are zero from the second on and this is preserved by any additive functor (such as $\operatorname{Hom}_R(M,-)$).

Another way for proving this is to recall that every submodule of a free module is again free. However, the proof of this fact is more complicated than the one about divisible modules.

Note that the fact that homomorphic images of injective modules are injective is equivalent to submodules of projective modules being projective.

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  • 3
    $\begingroup$ This isn't really the most direct or easy proof... $\endgroup$ – Najib Idrissi Nov 26 '15 at 8:34
  • $\begingroup$ Indeed. I have never seen anybody proven this with the notion of injectives, if not explicitly given that Ext is only defined as the right derived functors of $Hom(M,-)$. $\endgroup$ – MooS Nov 26 '15 at 8:38
  • $\begingroup$ @NajibIdrissi I added the “easier” proof. How Ext is defined by the OP is unknown. But I also added some other notes. I usually prefer the long exact sequence to lower level arguments using injective/projective resolutions. $\endgroup$ – egreg Nov 26 '15 at 8:42
  • $\begingroup$ It's not really a low-level argument as much as it is a basic fact following directly from the definition... $\endgroup$ – Najib Idrissi Nov 26 '15 at 8:47
  • $\begingroup$ @NajibIdrissi I agree, of course; take this as an introduction to “dimension shifting” methods. $\endgroup$ – egreg Nov 26 '15 at 8:56

Let $P_0 = \bigoplus_{m \in M} R_m$ be a direct sum of copies of $R$, one for each element of $M$ (the index is just here for bookkeeping reasons). This is a free $R$-module. This maps to $M$ through $\varepsilon : P_0 \to M$ by defining $\varepsilon_m : R_m \to M$, $x \mapsto x \cdot m$ and extending to the direct sum (coproduct).

The kernel $P_1 = \ker(P_0 \to M)$ is a submodule of the free module $P_0$, hence it is free as $R$ is a PID. Thus you get a free resolution (exact sequence): $$0 \to P_1 \to P_0 \to M \to 0.$$

Then $\operatorname{Ext}^n(M,N)$ is the homology of the chain complex you get from the exact sequence above after applying $\operatorname{Hom}_R(-,N)$, but this chain complex is zero in degrees $n \ge 2$, hence $\operatorname{Ext}^n(M,N) = 0$ for $n \ge 2$.

If you don't know that and only know the long exact sequence for $\operatorname{Ext}$, use the fact that $\operatorname{Ext}^n_R(P,N) = 0$ for $n \ge 1$ and $P$ projective (pretty much by definition), and the fact that a free module is projective. The long exact sequence derived from the short one above becomes (where $n \ge 2$): $$\dots \to \underbrace{\operatorname{Ext}^{n-1}_R(P_1,N)}_{= 0} \to \operatorname{Ext}^n_R(M,N) \to \underbrace{\operatorname{Ext}^n_R(P_0,N)}_{=0} \to \dots$$ and so $\operatorname{Ext}^n_R(M,N) = 0$.

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