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If we want to calculate $E(\mathbb{Z}/p\mathbb{Z},\mathbb{Z})$, i.e. equivalence classes of short exact sequences $\mathbb{Z}\rightarrow E\rightarrow\mathbb{Z}/p\mathbb{Z}$, we have $\mathbb{Z}/p\mathbb{Z}\cong\mathrm{Ext}(\mathbb{Z}/p\mathbb{Z},\mathbb{Z})\cong E(\mathbb{Z}/p\mathbb{Z},\mathbb{Z})$.

To get the explicit representative, we form the pushout as shown in the diagram: $$ \begin{array}{llllllllllll} \mathbb{Z}&\overset{\times p}{\rightarrow}&\mathbb{Z}&\rightarrow&\mathbb{Z}/p\mathbb{Z}\\ \downarrow\times a &&\downarrow &&{=}\\ \mathbb{Z}&\rightarrow&E&\rightarrow&\mathbb{Z}/p\mathbb{Z} \end{array} $$

The pushout is calculated as $\mathbb{Z}\oplus\mathbb{Z}/((a,p))$. When $(a,p)=1$, we have $ax+py=1$, so the invertible matrix $\begin{bmatrix}x & y\\-p & a\end{bmatrix}$ will take $(a,p)$ to $(1,0)$, which shows $E\cong\mathbb{Z}$. And $(1,0)$ is taken to $(x,-p)$, which is then equivalent to $-p$ in $E\cong\mathbb{Z}$, so the lower column is again a $\mathbb{Z}\overset{\times p}{\rightarrow}\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$ whenever $(a,p)=1$. But for different $a=1,2,...,p-1$, the sequence should be different, as $\times a$ are different elements in $\mathrm{Ext}(\mathbb{Z}/p\mathbb{Z},\mathbb{Z})$, which is a contradiction.

Surely I am wrong somewhere, but I just cannot find it out. Can anyone help me? Thanks.

Anonymous Coward
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    It is rather a feeling and I may mistake, but I suppose what changes is the projection $\pi_i \colon \mathbb Z \to \mathbb Z/(p)$. – Andrea Gagna Jun 29 '13 at 17:43
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    Oh yes. It really is. $(0,1)$ is taken to $(-p,a)$ and then $a$. I reminded that I was stuck on similar situations before... – Anonymous Coward Jun 29 '13 at 18:01

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They are actually different.

$(0,1)$ is taken to $(-p,a)$. So they are actually $\mathbb{Z}\overset{\times p}{\rightarrow}\mathbb{Z}\overset{(\times a) \mod p}{\rightarrow}\mathbb{Z}/p\mathbb{Z}$.

If we want an isomorphism: $$ \begin{array}{llllllllllll} \mathbb{Z}&\overset{\times p}{\rightarrow}&\mathbb{Z}&\overset{(\times a) \mod p}{\rightarrow}&\mathbb{Z}/p\mathbb{Z}\\ = &&\downarrow &&{=}\\ \mathbb{Z}&\overset{\times p}{\rightarrow}&\mathbb{Z}&\overset{(\times b) \mod p}{\rightarrow}&\mathbb{Z}/p\mathbb{Z} \end{array} $$ The unit of the upper and lower $\mathbb{Z}$ will be mapped to $a$ and $b$, respectively, which means $a\equiv b\mod p$.

Anonymous Coward
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