The following is exercise 3.5.3 from Weibel's Introduction to Homological Algebra:

Show that $\mathrm{Ext}^1(\mathbb{Z}[\frac{1}{p}],\mathbb{Z})= \hat{\mathbb{Z}}_p/\mathbb{Z}$ using $\mathbb{Z}[\frac{1}{p}]=\cup p^{-i} \mathbb{Z}$. Then show that $\mathrm{Ext}^1(\mathbb{Q},B)=(\Pi_p \hat{B}_p)/B$ for torsion-free $B$. (Here $\hat{B}_p=\varprojlim (B/p^iB)$ is the $p$-adic completion of $B$).

I can show the first statement using the short exact sequence $$0 \to {\varprojlim}^1 \mathrm{Ext}^{q-1} (A_i,B) \to \mathrm{Ext}^q(\cup A_i, B) \to \varprojlim \mathrm{Ext}^q(A_i,B) \to 0$$ Ideally the second statement would follow from the observation that $\mathbb{Q}$ is the union of $\mathbb{Z}[\frac{1}{2}] \subset \mathbb{Z}[\frac{1}{6}] \subset \mathbb{Z}[\frac{1}{30}] \subset\cdots$, and the computation $$\mathrm{Ext}^1(\mathbb{Q},B)=\mathrm{Ext}^1(\varinjlim \mathbb{Z}[\frac{1}{p_1\cdots p_r}],B)=\varprojlim \mathrm{Ext}^1(\mathbb{Z}[\frac{1}{p_1...p_r}],B)$$ $$= \varprojlim \hat{B}_{p_1\cdots p_r}/B=\varprojlim (\Pi \hat{B}_{p_i})/B=(\Pi_p \hat{B}_p)/B.$$ However the step where I take the $\varinjlim$ out of $\mathrm{Ext}^1$ is problematic, since to justify this would require that the category $Ab$ satisfy axiom (AB5*), i.e. filtered limits are exact, which is not the case.

I am unable to find another way to do this exercise, so I suspect there should be a reasoning showing that the passing of the (co)limit outside $\mathrm{Ext}^1$ is legitimate in this particular case. It would be appreciated if someone can point out how to patch this step, or even point out another way to do this exercise. Thank you in advance.

  • 3
    $\begingroup$ How do you know that $Ext^1(\mathbb{Z}[\frac{1}{p_1...p_r}],B)= \hat{B}_{p_1...p_r}/B$? I would think that whatever proof you have for that could be modified easily to give the result you want for $\mathbb{Q}$. $\endgroup$ Jun 25, 2017 at 23:45
  • $\begingroup$ @EricWofsey Yes you are right indeed! Thanks for pointing it out! $\endgroup$ Jun 26, 2017 at 0:43
  • $\begingroup$ Discussed here $\endgroup$
    – FShrike
    Dec 12, 2023 at 1:52

1 Answer 1


I think I might be close to finish that exercise, but maybe there is a mistake in the book.

The first part of this exercise was done in a previous exercise, and I'm not entirely sure how the hint is applied. The second part is what this question is about.

By Application 3.5.10, for $p$ prime, we have the short exact sequence \begin{equation*} 0 \to \varprojlim^1\mathrm{Hom}(\mathbb{Z}/p^i, B) \to \mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Z}_{p^{\infty}}, B) \to \hat{B}_p \to 0. \end{equation*} But $\mathrm{Hom}(\mathbb{Z}/p^i, B) = 0$ because $B$ is torsionfree. Therefore, the last short exact sequence implies that $\mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Z}_{p^{\infty}}, B) \cong \hat{B}_p$.

From Example 2.3.3, we know that $\mathbb{Q}/\mathbb{Z} \cong \bigoplus_p\mathbb{Z}_{p^{\infty}}$. By Proposition 3.3.4, $\mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, B) = \prod_p\hat{B}_p$.

We have a short exact sequence \begin{equation*} 0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0. \end{equation*} As $(\mathrm{Ext}^n(\bullet, \mathbb{Z}))_{n \in \mathbb{N}}$ is a universal cohomological $\delta$-functor, we have the following long exact sequence \begin{align*} 0 \to &\mathrm{Hom}(\mathbb{Q}/\mathbb{Z}, B) \to \mathrm{Hom}(\mathbb{Q}, \mathbb{Z}) \to \mathrm{Hom}(\mathbb{Z}, B) \to \\ &\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z}, B) \to \mathrm{Ext}^1(\mathbb{Q}, B) \to \mathrm{Ext}^1(\mathbb{Z}, B) \to \dots. \end{align*}

Since $B$ is torsionfree, we have $\mathrm{Hom}(\mathbb{Q}/\mathbb{Z}, B) = 0$, and since $\mathbb{Z}$ is proyective, we have $\mathrm{Ext}^1(\mathbb{Z}, B) = 0$.

In summary, we obtained the following exact sequence \begin{equation*} 0 \to \mathrm{Hom}(\mathbb{Q}, \mathbb{Z}) \to \mathrm{Hom}(\mathbb{Z}, B) \to \prod_p\hat{B}_p \to \mathrm{Ext}^1(\mathbb{Q}, B) \to 0. \end{equation*}

We know that $\mathrm{Hom}(\mathbb{Z}, B) = B$. However, I think $\mathrm{Hom}(\mathbb{Q}, \mathbb{Z})$ might not be zero. In certain situations, like $B = \mathbb{Z}$, it is indeed zero, but I'm not sure about the general torsionfree $B$.


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