# Equivalence between Ext and Hom

This is a question from Homology by Saunders Mac Lane. This is problem 5 page 76. I've been struggling to solve this problem for like more than a day, but still nothing valuable comes across my mind yet.

# Problem

For $p$ prime, and $C$ an Abelian group such that $pC = 0$, prove that:

$$\mbox{Ext}_\mathbb{Z}(C; G) \cong \mbox{Hom}(C; G/pG)$$

I've made a couple of attempts, but all of them fail.

• Firstly, I try to find some connections between the elements in $\mbox{Ext}_\mathbb{Z}(C; G)$, and the elements in $\mbox{Hom}(C; G/pG)$.

The elements of $\mbox{Ext}_\mathbb{Z}(C; G)$ is a short exact sequence of Abelian groups $0 \to G \to W \to C \to 0$, with $W$ varies. So given this exact sequence, how can I manage to find a homomorphism from $C$ to $G/pG$? And vice versa, given a homomorphism, how can I deduce a short exact sequence?

• Secondly, I try to make use of the fact that $pC = 0$. Although I know that $C$ can only be some group of orders $p^k$, can I deduce that $C = \bigoplus\mathbb{Z}_p$, or is it $C = \prod\mathbb{Z}_p$?

That's all I have in mind till now. :(

Is there some other way that to look at this problem?

Thanks very much,

And have a good day, :*

Edit (July, 03) I've managed to prove the problem for the simpliest case, i.e, for $C = \mathbb{Z}_p$, i.e, to prove: $$\mbox{Ext}_\mathbb{Z}(\mathbb{Z}_p; G) \cong \mbox{Hom}(\mathbb{Z}_p; G/pG)$$

I hope someone can have a quick look over the proof.

Consider the short exact sequence: $$0 \to p\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_p \to 0$$

Apply $\mbox{Ext}^n(-; G)$ to the above ses (? is this the right way to say it), gives the long one:

$$0 \to \mbox{Hom}(\mathbb{Z}_p; G) \to \mbox{Hom}(\mathbb{Z}; G) \xrightarrow{\chi} \mbox{Hom}(p\mathbb{Z}; G) \xrightarrow{\sigma} \mbox{Ext}(\mathbb{Z}_p; G) \to \mbox{Ext}(\mathbb{Z}; G) = 0$$

Hence $\sigma$ is epic, we'll have: $\mbox{Hom}(p\mathbb{Z}; G)/\mbox{Im} \chi \cong \mbox{Ext}(\mathbb{Z}_p; G)$.

Notice that a group homomorphism $f$ from $\mathbb{Z}$ to $G$ is completely determined if we know $f(1)$, similarly any $g: p\mathbb{Z} \to G$ will be determined by $g(p)$.

So, in fact, we'll have $\mbox{Im}(\chi) = \left\{ f: p\mathbb{Z} \to G \middle| f(p) \in pG\right \}$. And hence, 2 classes of $f$, and $g$ in $\mbox{Hom}(p\mathbb{Z}; G)/\mbox{Im}$ will be congruent iff $f - g \in \mbox{Hom}(p\mathbb{Z}, pG)$.

So: $\mbox{Hom}(p\mathbb{Z}; G)/\mbox{Im} \chi \cong \mbox{Hom}(p\mathbb{Z}; G/pG) \cong \mbox{Hom}(\mathbb{Z}; G/pG)$, since $\mathbb{Z} \cong p\mathbb{Z}$ as Abelian groups.

And moreover, one should notice that, for every $f \in \mbox{Hom}(\mathbb{Z}; G/pG)$, and for every $m, n \in \mathbb{Z}$, such that $(m - n) \vdots p$, then $f(m) = f(n)$. Which, in turns means that, $\mbox{Hom}(\mathbb{Z}; G/pG) \cong \mbox{Hom}(\mathbb{Z}_p; G/pG)$. Hence, it's done.

And I've found this pdf http://www.math.wichita.edu/~pparker/classes/handout/torext.pdf, which says that:

$$\mbox{Ext}\left( \bigoplus A_i; B\right) \cong \prod \mbox{Ext}\left( A_i; B\right)$$

So if it's true that $pC = 0 \Leftrightarrow C = \bigoplus \mathbb{Z}_p$, if so, then everything should be pretty clear.

# Questions

So some of my concerns left are:

• "Apply $\mbox{Ext}^n (-; G)$ to..." is this the right way to say it?
• Does the proof I gave for $\mbox{Ext}_\mathbb{Z}(\mathbb{Z}_p; G) \cong \mbox{Hom}(\mathbb{Z}_p; G/pG)$ look correct? Can it be shortened?
• I know that $\mbox{Ext}\left( \bigoplus A_i; B\right) \cong \prod \mbox{Ext}\left( A_i; B\right)$, but is it also true that $\mbox{Ext}\left( \prod A_i; B\right) \cong \bigoplus \mbox{Ext}\left( A_i; B\right)$? Can you guys guide me to some article, or some book, or website that have a proof to that? Or, can you guys just give me some hints on proving them?
• Is it true that $pC = 0 \Leftrightarrow \left[ \begin{array}{l} C = \oplus\mathbb{Z}_p \\ C = \prod \mathbb{Z}_p \end{array} \right.$?

Thank you very much,

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Whoops, I mis-typed it, I mean $pC = 0$. I'll correct it right away, thanks very much for pointing that out. – user49685 Jul 2 '13 at 17:25
You know that $0\to pG\to G\to G/pG\to 0$ is exact, do you know a long exact sequence which involves $\textrm{Hom}$ and $\textrm{Ext}$? – Thomas Andrews Jul 2 '13 at 17:28
Yup, of course, it's the sequence, $0 \to \mbox{Hom}(C; pG) \to \mbox{Hom}(C; G) \to \mbox{Hom}(C; G/pG) \xrightarrow{\chi} \mbox{Ext}(C; pG) \xrightarrow{\sigma} \mbox{Ext}(C; G) \to ...$. Should I prove that $\sigma \chi$ is iso? :-s – user49685 Jul 2 '13 at 17:32
Hmm, $\sigma\chi = 0$ implicitly, by exactness, sorry, I might have led you down a wrong path here. – Thomas Andrews Jul 2 '13 at 17:38
I have editted the post, can you please take a look to see if I make a mistake somewhere? And how can I proceed from there? Thanks so much, :* – user49685 Jul 3 '13 at 2:39

My answers to your four questions: (1) Yes, "Apply Ext$(-,G)$ to" is a standard way of saying it. (2) Your proof for the case $C=\mathbb Z/p$ looks OK. It might become a bit shorter and easier if you started with the exact sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/p$, where the map $\mathbb Z\to\mathbb Z$ is multiplication by $p$. That way, you never have to deal with $p\mathbb Z$ separately. (3) No, you can't reverse the $\bigoplus$ and $\prod$ in general. (In this respect Ext works just like Hom.) (4) Yes with $\bigoplus$; no with $\prod$. The point is that an abelian group $C$ with $pC=0$ amounts to a vector space over $\mathbb Z/p$, and every vector space has a basis, which gives you the $\bigoplus$ result. The $\prod$ version fails because a product of copies of $\mathbb Z/p$ is either finite (if there are only finitely many factors) or of cardinality at least $2^{\aleph_0}$ (if there are infinitely many factors), whereas $C$ could well be countably infinite.
And, can you give me a reference on proving $\mbox{Ext} (\bigoplus A_i; B) = \prod \mbox{Ext}(A_i; B)$? I searched Google, but I didn't come across any. :( – user49685 Jul 3 '13 at 4:37
Consider a resolution $\cdots\to \oplus P_{i,1}\to \oplus P_{i,0}\to \oplus A_i \to 0$. Applying $\mathrm{Hom}(-,B)$ to this resolution and applying the fact that $\mathrm{Hom}(\oplus A_i,B)\cong \prod\mathrm{Hom}(A_i,B)$, we have the desired result for Ext. If you need help figuring out why this is true for Hom, think about "a map from a coproduct is the same as...". – KReiser Jul 3 '13 at 7:57