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As I understand, if $0\rightarrow A\rightarrow X\rightarrow B\rightarrow 0$ is a short exact sequence of abelian groups, $\mbox{Ext }_{\mathbb{Z}}^{1}(B,A)$ gives all the isomorphism classes of what can come in as $X$. But when I consider $0\rightarrow \mathbb{Z}\rightarrow X\rightarrow \mathbb{Z}/(3)\rightarrow 0 $,$\ \ \ $ $\mbox{Ext } _{\mathbb{Z}}(\mathbb{Z}/(3),\mathbb{Z})=\mathbb{Z}/(3)\ \ $ but all I can think of for $X$ are only two, $\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}/(3)$. Am I missing something or am I not understanding the result of extension problem correctly?

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The isomorphisms should commute with the injection $A\to X$ and the surjection $X\to B$. This leaves less freedom for the isomorphisms, and thus we get more distinct $X$'es. – darij grinberg Jul 29 '10 at 13:19
Since this seems to be your point of confusion, I should add that the elements of $\mathrm{Ext}^1\left(B,A\right)$ are the isomorphism classes of EXACT SEQUENCES $0\to A\to X\to B\to 0$ and not just the isomorphism classes of those $X$ for which such exact sequences exist. An exact sequence $0\to A\to X\to B\to 0$ is not uniquely defined by $A$, $X$ and $B$; we also need to know the maps! For instance, in your case, there are four different exact sequences $0\to \mathbb Z\to\mathbb Z\to \mathbb Z\diagup\left(3\right)\to 0$, and they are isomorphic in pairs, but not all isomorphic to each other – darij grinberg Jul 29 '10 at 13:25
up vote 6 down vote accepted

$\mathrm{Ext} = \mathrm{Ext}^1$ does not classify the middle terms up to isomorphism. It classifies the short exact sequences up to equivalence, where the equivalence relation is generated by commutative diagrams having equalities on each end. It's possible for two inequivalent short exact sequences to have isomorphic middle terms. See Eisenbud's book, Exercises A3.26, A3.27, and especially A3.29 for more.

As for your particular example, consider the pullback of the exact sequence $0 \to \mathbb{Z} \rightarrow \mathbb{Z} \to \mathbb{Z}/(3) \to 0$ by the map given by multiplication by 2 on $\mathbb{Z}/(3)$. The middle term is isomorphic to $\mathbb{Z}$, but the short exact sequence is not equivalent to the original.

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1. Sorry, $\mbox{Ext }_{\mathbb{Z}}^{1}(\mathbb{Z}/(3),\mathbb{Z})$ being zero was my typo. I corrected it. 2. I guess Wikipedia article is wrong, then: I was comparing it with Eisenbud A3.26 and drew a wrong conclusion. But A3.29 clarified everything. Thanks! – ashpool Jul 29 '10 at 13:34
The wikipedia article says that "isomorphism classes of extensions" are classified by Ext$^1$. This doesn't seem to be wrong -- an extension is a short exact sequence, and the isom. classes of such are precisely the equivalence classes that Graham describes. I agree, though, that the article could have been clearer/more explicit. – George McNinch Jul 29 '10 at 14:08
You are absolutely right, George. Wikipedia article is correct and I was being very careless. – ashpool Jul 29 '10 at 14:12

The subtlety is in when two extensions are considered the same, in other words what are the morphisms of extension? There are different reasonable answers. For example a morphism could just be a morphism of short exact sequences (i.e. chain complexes).

However the statement that $Ext^1$ classifies extensions uses a very particular kind of morphism of extension: only those morphisms of short exact sequences which are the identity on A and B.

Let's look at extensions of $\mathbb{Z}/3$ by $\mathbb{Z}$, where the middle guy is a $\mathbb{Z}$. Suppose that the map from $f:\mathbb{Z} \to \mathbb{Z}$ is fixed and is multiplication by 3. Notice that there is a unique map $h: \mathbb{Z} \to \mathbb{Z}$ such that $f = hf$,
namely $h=id$. This means that for any such extension, after you've identified the map $f$ as multiplication by 3, there is no choice for morphisms of extension. A map of these extensions has to be the identity on all three terms.

In short, the extensions $\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/3$, up to isomorphism fixing the first and last group, are in bijection with homomorphisms from $\mathbb{Z} \to \mathbb{Z}/3$ which have kernel $3 \cdot \mathbb{Z}$. There are exactly two such homomorphisms, given by sending 1 to 1 or 1 to 2 in $\mathbb{Z}/3$. So there are two distinct non-spilt extensions.

However you can also use $Ext^1(B;A)$ to get extensions up to the weaker notion of equivalence described above. Since Ext is functorial in both variables, you have an action by the automorphism groups of both A and B. Extensions up to this weaker notion are in bijection with equivalence classes in the quotient of $Ext^1$ by these actions.

You can check in this case that the action by automorphisms of $\mathbb{Z}/3$ exchanges the two non-zero elements of $Ext^1(\mathbb{Z}/3, \mathbb{Z})$, and so there are exactly two extensions in this weaker sense: the spilt extension and the non-split one.

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