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Let $A$ be an abelian group.

I know that $Ext_\mathbb{Z}^1(\mathbb{Z}/p,A)=A/pA$.

Are there any similar formula about $Ext_\mathbb{Z}^1(A,\mathbb{Z}/p)$?

I know that $Ext_R^n(A,B)\neq Ext_R^n(B,A)$ generally.

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3 Answers 3

up vote 3 down vote accepted

We need an injective resolution for $\mathbb Z/p$ in the second variable. An Abelian group is injective if and only if it is divisible.

So, the first step will be to embed $\mathbb Z/p$ into a divisible Abelian group. For this, we can use the Prüfer $p$-group $\mathbb Z(p^\infty)$.

The cokernel of this embedding is again the Prüfer $p$-group, so that we get the injective resolution $\mathbb Z(p^\infty)\to\mathbb Z(p^\infty)$ given by multiplication with $p$.

Hence we get $$ Ext_\mathbb{Z}^1(A,\mathbb{Z}/p)=\frac{Hom_\mathbb{Z}(A,\mathbb{Z}(p^\infty))}{p\cdot Hom_\mathbb{Z}(A,\mathbb{Z}(p^\infty))}. $$

I don't think that it is easy to understand homomorphisms into $\mathbb{Z}(p^\infty)$ because it is a direct/injective limit. (Edit: But please read Matt E's comment below.)

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Aha, then, if $A$ is torsion abelian group, then $Ext_\mathbb{Z}^1(A,\mathbb{Z}/p)$ is trivial since $Hom_\mathbb{Z}(A,\mathbb{Z})=0$. Right? But, I remember that $Ext_\mathbb{Z}^1(\mathbb{Z}/p,\mathbb{Z}/p)=\mathbb{Z}/p$. Isn't it? –  John Smith Jun 11 '11 at 20:10
    
Right, that follows from the observation is your question. Let met think for a moment. –  Rasmus Jun 11 '11 at 20:14
    
Now, your answer seems to be perfect. Thank you for your effort. –  John Smith Jun 11 '11 at 20:34
    
Thanks, that's nice to hear. –  Rasmus Jun 11 '11 at 20:36
1  
Dear Rasmus, Note that Homs into $\mathbb Z(p^{\infty})$ are a version of the Pontrjagin dual of $A$, and so not so strange or necessarily difficult to compute (at least for e.g. finitely generated $A$). (See my answer for a little more explanation.) Best wishes, –  Matt E Jun 12 '11 at 0:01

Form the injective resolution $$0 \to \dfrac{1}{p}\mathbb Z/\mathbb Z \to \mathbb Q/\mathbb Z \buildrel \cdot p \over \to \mathbb Q/\mathbb Z \to 0.$$ Applying $\mathrm{Hom}(A,\text{--})$ to this, we find that $$\mathrm{Ext}^1(A, \mathbb Z/p) = \mathrm{Ext}^1(A,\dfrac{1}{p}\mathbb Z/\mathbb Z) = {\mathrm Hom}(A,\mathbb Q/\mathbb Z)/p.$$

Note that $\mathrm{Hom}(A,\mathbb Q/\mathbb Z)$ is (speaking somewhat informally; see my comment addressed to Rasmus below) a version of the Pontrajagin dual of $A$. (In particular, if $A$ is torsion, then it is precisely the Pontrjagin dual.) In fact, we could replace $\mathbb Q/\mathbb Z$ by the circle group $S^1$ in the above injective resolution, and then we would find that $$\mathrm{Ext}^1(A,\mathbb Z/p) = \hat{A}/p\hat{A},$$ where $\hat{A} = \mathrm{Hom}(A,S^1)$ is the Pontrjagin dual of the abelian group $A$ (with its discrete topology).

[Note: I began writing this answer some hours ago, but then got distracted. It is essentially the same as that of Rasmus, although, rather than using either $\mathbb Q/\mathbb Z$ or $S^1$, he uses $\mathbb Z(p^{\infty})$, which is the $p$-power torsion part of $\mathbb Q/\mathbb Z$. (If you like, you can write it as $\mathbb Q_p/\mathbb Z_p$.) These different descriptions reflect the flexibility available in choosing an injective resolution of $\mathbb Z/p$.]

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The maximally useful is the one using $S^1$, though, because the result looks like it might have a meaning :) –  Mariano Suárez-Alvarez Jun 12 '11 at 0:04
    
What does "a version of the Pontrjagin dual" mean? –  Rasmus Jun 12 '11 at 9:52
    
@Rasmus: Dear Rasmus, Strictly speaking, Pontrjagin duality means Homs into the circle group $S^1$. But for a torsion group this is the same as Homs into $\mathbb Q/\mathbb Z_p$, and it is sometimes useful (at least for me) to think of taking Homs into $\mathbb Q/\mathbb Z$, or even $\mathbb Q_p/\mathbb Z_p$ if one is primarily focussed on $p$-power torsion modules, as a kind of variant of Pontrjagin duality. I didn't mean more anything more than that. Regards, –  Matt E Jun 13 '11 at 8:15
    
Dear Matt, thank you for the explanation. Best, –  Rasmus Jun 13 '11 at 8:32
    
@Rasmus: Dear Rasmus, You're welcome. Also, you probably noticed a minor typo in my previous comment: $\mathbb Q/\mathbb Z_p$ should read $\mathbb Q/\mathbb Z$. Regards, –  Matt E Jun 13 '11 at 8:34

I would just add that $\operatorname{Ext}(A,G)$ is fully computable when $A$ and $G$ are finitely generated abelian groups.

You can find the proofs in any decent book on homological algebra:

1) If $A$ is free abelian, then $\operatorname{Ext}(A,G)=0$

2) If $G$ is divisible $\operatorname{Ext}(A,G)=0$

3) $\operatorname{Ext}(A,\prod G_j)=\prod \operatorname{Ext}(A,G_j)$

4) $\operatorname{Ext}(\sum A_j,G)=\prod \operatorname{Ext}(A_j,G)$

5) $\operatorname{Ext}(\mathbb{Z}/m\mathbb{Z},G)=G/mG$ (as noted)

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