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The question is as in the title; I don't really have a more specific question, I am more interested in a "yes, here is an example of how it is useful," a "yes, but it really isn't useful," or a "no, here is why."

The context is we are covering homological algebra out of Lang for our algebra course. The problem is I still wasn't completely comfortable with the categorical properties of the Hom and Tensor functors before moving on to Ext and Tor, so I am trying to make some connections between the two notions to try and solidify my understanding.

I am comfortable with the adjoint relationship between Hom and Tensor; namely, $$Hom(Y\otimes X, Z)\cong Hom(Y, Hom(X,Z))$$

So potentially a candidate would be $$ Hom(Tor_n(Y,X),Z)\cong Hom(Y,Ext^n(X,Z))? $$

In my head I am thinking this could be useful as follows:

For any two abelian groups $A$ and $B$, $Tor_n^\mathbb{Z}(A,B)=0$ for $n\geq 2$. This is easy to see by taking a projective resolution $0\to ker(f)\to F\to A\to 0$, where $f:F\to A$ is a realization of $A$ as a quotient of a free group $F$.

From this, can we immediately conclude $Ext^n(A,B)=0$ for $n\geq 2$, without doing anymore work? (I am sure a direct proof would be equally straightforward, I am just interested in the hypothetical).

Thanks in advance

TomGrubb
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  • Leaving this as a comment, but I'm quite certain such a relationship (or something intimately similar) holds in the context of finite-dimensional algebras over an algebraically closed field. I have to check my notes. – Samantha Y Oct 27 '17 at 16:13
  • Ah, so the relationship I am thinking of is (letting $A$ be an algebra over $k$, and letting $k$ be an algebraically closed field), $Ext_A^n(M,Hom_k(N,k)) \cong Hom_k(Tor^A_n(M,N), k)$. – Samantha Y Oct 27 '17 at 16:18
  • Sorry for another comment, got timed out on the last one: IIRC, the proof boils down to the adjointness formula in the case $n=0$ and then the rest is just some standard homological algebra. – Samantha Y Oct 27 '17 at 16:36

1 Answers1

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No, here is why.

Recall that a right adjoint is left exact. So if $\operatorname{Ext}^n(X,.)$ were the right adjoint of some functor, it should preserve left exact sequence. But it doesn't ! Indeed, if $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence, then the long exact sequence $$...\rightarrow\operatorname{Ext}^{n-1}(X,C)\rightarrow\operatorname{Ext}^n(X,A)\rightarrow\operatorname{Ext}^n(X,B)\rightarrow\operatorname{Ext}^n(X,C)\rightarrow ...$$ shows that in general the map $\operatorname{Ext}^n(X,A)\rightarrow\operatorname{Ext}^n(X,B)$ is not into.

But the tensor-hom adjunction leads indeed to a derived version. For this, we need a more advanced tool : the derived category. (For simplicity, I assume that we are in $R$-Mod, moreover there are technical boundedness assumptions that I will not expand here). There are complexes $Y\otimes^L X$ and $R\operatorname{Hom}(X,Z)$, such that $$\operatorname{Hom}_D(Y\otimes^L X,Z)=\operatorname{Hom}_D(Y,R\operatorname{Hom}(X,Z))$$ with $H_n(Y\otimes^L X)=\operatorname{Tor}_n(Y,X)$ and $H^n(R\operatorname{Hom}(X,Z))=\operatorname{Ext}^n(X,Z)$.

I am not sure we can deduce formally the vanishing of $\operatorname{Ext}^2_\mathbb{Z}$ from the above derived adjunction (I will think about it). But if you know that you can compute $\operatorname{Ext}$ using a projective resolution of the first variable, the proof for $\operatorname{Tor}$ works for $\operatorname{Ext}$ and many other functors (namely every right exact covariant or left exact contravariant).

Roland
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