What are $\operatorname{Ext}(\mathbb Q, \oplus_k \mathbb Z/p^k)$ and $\operatorname{Ext}(\mathbb Z/p^\infty, \oplus_k \mathbb Z /p^k)$?

Question

Motivation: Let $p$ be a prime. I learned here that by a theorem of Kulikov, every abelian $p$-group $A$ is an extension of a divisible group by a direct sum of cyclic groups. That is, we have a short exact sequence

$$ 0 \to B \to A \to (\mathbb Z / p^\infty)^{(J)} \to 0$$

where $B = \oplus (\mathbb Z/p^k)^{(I_k)}$ is a sum of cyclic groups (I denote by $X^{(I)}$ an $I$-fold direct sum of copies of $X$). This motivates studying the groups

$$\operatorname{Ext}((\mathbb Z / p^\infty)^{(J)}, B) = \operatorname{Ext}(\mathbb Z / p^\infty, B)^J$$

for such $B$ as an approach to classifying $p$-groups. Moreover, the short exact sequence $0 \to \mathbb Z_{(p)} \to \mathbb Q \to \mathbb Z / p^\infty \to 0$ yields a short exact sequence

$$0 \to B \to \operatorname{Ext}(\mathbb Z/p^\infty, B) \to \operatorname{Ext}(\mathbb Q, B) \to 0$$

So it seems natural to start by studying $\operatorname{Ext}(\mathbb Q, B)$.

If $B$ is of bounded exponent, then both $\operatorname{Ext}$ groups vanish. So the simplest interesting case should be $B = \oplus_{k \in \mathbb N} \mathbb Z / p^k$. Thus I ask

Question 1: What is $\operatorname{Ext}(\mathbb Q, \oplus_{k \in \mathbb N} \mathbb Z / p^k)$?

I believe this is a divisible group, so it is of the form $\mathbb Q^{(I)} \oplus (\mathbb Z / p^\infty)^{(J)}$, so the question is to identify the cardinalities $I$ and $J$.

Everything can also be done $p$-adically, and I think I can see that $\operatorname{Ext}^1_{\mathbb Z}(\mathbb Q, B)$ is actually isomorphic to $\operatorname{Ext}^1_{\mathbb Z_p}(\mathbb Q_p, B)$. So a more refined question is

Question 2: What is $\operatorname{Ext}^1_{\mathbb Z_p}(\mathbb Q_p, \oplus_{k \in \mathbb N} \mathbb Z / p^k)$ (as a $\mathbb Z_p$-module)?

Similarly to before, it should be of the form $\mathbb Q_p^{(I')} \oplus \mathbb (\mathbb Z/p^\infty)^{(J)}$ so the question is to identify the cardinalities $I'$ and $J$.

Answer 1

Not a complete answer. I'm going to write $\text{Rlim}$ for $\lim^1$ because I can't figure out how to get $\lim^1$ to look nice. Writing $\mathbb{Z}/p^{\infty} \cong \text{colim}_n \, \mathbb{Z}/p^n$ gives a short exact sequence

$$0 \to \text{Rlim}_n \text{Hom}(\mathbb{Z}/p^n, B) \to \text{Ext}(\mathbb{Z}/p^{\infty}, B) \to \lim_n \text{Ext}(\mathbb{Z}/p^n, B) \to 0.$$

The rightmost term is just $\lim_n B/p^n B$, the $p$-completion of $B$. So I think this means $\text{Ext}(\mathbb{Z}/p^{\infty}, B)$ is what people call derived $p$-completion, and I guess the $\text{Rlim}$ term can be nontrivial in general. $\text{Ext}(\mathbb{Q}, B)$ is then the quotient of whatever this is by the image of $B$.

By functoriality, if $B$ is a $p$-group then $\text{Ext}(\mathbb{Q}, B)$ inherits both an action of $\mathbb{Q}$ and an action of $\mathbb{Z}_p$, so overall has an action of $\mathbb{Q} \otimes \mathbb{Z}_p \cong \mathbb{Q}_p$, and hence must be a vector space over $\mathbb{Q}_p$ of some dimension (and this determines its $\mathbb{Z}_p$-module structure also). As you say we also have $\text{Ext}(\mathbb{Q}, B) \cong \text{Ext}_{\mathbb{Z}_p}(\mathbb{Q}_p, B)$ via the short exact sequence

$$0 \to \mathbb{Z}_p \to \mathbb{Q}_p \to \mathbb{Q}_p/\mathbb{Z}_p \to 0$$

which makes the $\mathbb{Q}_p$-action a little easier to see.

When $B = \oplus_k \mathbb{Z}/p^k$ I believe (but haven't checked carefully) that the $p$-completion ought to be $\prod_k \mathbb{Z}/p^k$. Unfortunately the $\text{Rlim}$ term doesn't satisfy the Mittag-Leffler condition.

Edit: Okay, here's an alternative to try. We can instead start with the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z} \left[ \frac 1 p \right] \to \mathbb{Z}/p^{\infty} \to 0$, which produces another very similar short exact sequence

$$0 \to B \to \text{Ext}(\mathbb{Z}/p^{\infty}, B) \to \text{Ext}\left( \mathbb{Z} \left[ \frac 1 p \right], B \right) \to 0.$$

Writing $\mathbb{Z} \left[ \frac 1 p \right] \cong \text{colim} \left( \cdots \mathbb{Z} \xrightarrow{p} \mathbb{Z} \cdots \right)$ gives a short exact sequence

$$0 \to \text{Rlim } B \to \text{Ext}\left( \mathbb{Z} \left[ \frac 1 p \right], B \right) \to \lim B \to 0.$$

Now both the $\text{Rlim}$ term and the $\lim$ term are simpler. $\lim B$ is the limit of the sequence $\cdots B \xrightarrow{p} B \xrightarrow{p} B \cdots$ which is $\text{Hom} \left( \mathbb{Z} \left[ \frac 1 p \right], B \right) = 0$. This means we have an isomorphism $\text{Rlim } B \cong \text{Ext} \left( \mathbb{Z} \left[ \frac 1 p \right], B \right)$, where $\text{Rlim } B$ is the cokernel of the map

$$\prod_n B \xrightarrow{\partial} \prod_n B$$

where $\partial(\{ b_n \}) = \{ b_n - p b_{n+1} \}$ (and $\lim B$ is the kernel). This should be easier to understand although I admit I am not excited about the double indices necessary to understand this for $B = \oplus_k \mathbb{Z}/p^k$. Maybe this will help though.

Answer 2

$\DeclareMathOperator{\Ext}{Ext}$ $\DeclareMathOperator{\Hom}{Hom}$ $\newcommand{\Zp}{{\mathbb{Z}_p}}$ $\newcommand{\Qp}{{\mathbb{Q}_p}}$ $\newcommand{\div}{\mathrm{div}}$ $\newcommand{\tor}{\mathrm{tor}}$ Still not a complete answer, but here's some more progress, I think. Let $B$ be a direct sum of cyclic abelian $p$-groups. All $\Hom$'s, $\Ext$'s, $\otimes$'s, etc. are over $\Zp$.

Theorem 1: $\Ext(\Qp, B) \cong \Hom(\Qp, \hat B / B) \cong \Qp \otimes \widehat{\Zp^{(I)}} \oplus \Qp^{(J)}$ where

• $I = \dim_{\mathbb F_p}((\hat B / B)[p])$

• $J = \dim_{\Qp}((\hat B / B)/(\hat B / B)^\tor)$

Here $\hat B = \varprojlim_k B/p^k$ denotes the $p$-completion of $B$, $(-)^\tor$ denotes the torsion subgroup, and $[p]$ denotes the kernel of multiplication by $p$.

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We can be more concrete about this. Write $B'$ for the corresponding product of cyclic groups, i.e. if $B = \oplus_k (\Zp/p^k)^{(I_k)}$, then $B' = \prod_k (\Zp/p^k)^{I_k}$. Then we have

Proposition 2: The natural inclusion $B \to B'$ factors through $\hat B$, and

• $B = \{(x_{k,i}) \in B' \mid x_{k,i} \to 0\}$

• $\hat B = \{(x_{k,i}) \in B' \mid v_p(x_{k,i}) \to \infty\}$

• $(\hat B / B)[p] = \{(x_{k,i}) \in \hat B \mid k - v_p(x_{k,i}) \text{ is bounded by }1\} / B[p]$

• $(\hat B / B)^\tor = \{(x_{k,i}) \in \hat B \mid k - v_p(x_{k,i}) \text{ is bounded}\} / B$

In particular, the maps $B \to \hat B$ and $\hat B \to B'$ are monic. Moreover, $\hat B / B$ is divisible.

Here, $x_{k,i} \to 0$ means that $x_{k,i} = 0$ for all but finitely many $(k,i)$. Similarly, $v_p(x_{k,i}) \to \infty$ means that for each $n \in \mathbb N$, $v_p(x_{k,i}) \geq n$ for all but finitely many $(k,i)$, where $v_p$ is the $p$-adic valuation.

Proof: Straightforward.

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Lemma 3: $\Ext(\Qp,\hat B) = 0$.

Proof: The short exact sequence $0 \to \hat B \to B' \to B'/\hat B \to 0$ yields an exact sequence $\Hom(\Qp, B' / \hat B) \to \Ext(\Qp,\hat B) \to \Ext(\Qp,B')$. The last term vanishes because $\Ext(\Qp,\Zp/p^k) = 0$ and $\Ext(\Qp,-)$ commutes with products. To see that the first term vanishes, it suffices to show that the divisible part of $B'/\hat B$ is zero, which can be done using the explicit descriptions of Proposition 2.

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Lemma 4:

• If $D$ is a divisible $\Zp$-module, then $D \cong D^\tor \oplus D/D^\tor$. Moreover, $D^\tor \cong (\mathbb Z/p^\infty)^{(I)}$ where $I = \dim_{\mathbb F_p} D[p]$, while $D/D^\tor$ is a $\Qp$-vector space.

• $\Hom(\mathbb Z/p^\infty, (\mathbb Z/p^\infty)^{(I)}) \cong \widehat{\Zp^{(I)}}$ and $\Hom(\Qp, (\mathbb Z/p^\infty)^{(I)}) \cong \Qp \otimes \widehat{\Zp^{(I)}}$

Proof: The first bullet is clear. For the second, use the expression $\mathbb Z / p^\infty = \varinjlim_k \mathbb Z/p^k$, which is carried to a $\varprojlim$ by $\Hom$. Then observe that every homomorphism $\Qp \to (\mathbb Z/p^\infty)^{(I)}$ descends to $\mathbb Z / p^\infty \to (\mathbb Z/ p^\infty)^{(I)}$ after multiplication by some power of $p$.

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Proof of Theorem 1: The short exact sequence $0 \to B \to \hat B \to \hat B / B \to 0$ induces an exact sequence $\Hom(\Qp,\hat B) \to \Hom(\Qp,\hat B / B) \to \Ext(\Qp,B) \to \Ext(\Qp,\hat B)$. The last term vanishes by Lemma 3. To see that the first term vanishes, it suffices to observe that the divisible part of $\hat B$ is zero, which is obvious from the description of Proposition 2. So the middle map is an isomorphism. This establishes the first isomorphism of Theorem 1. The second follows from Lemma 4 and the divisibility of $\hat B / B$.

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Dimension count: Recall that we are considering $B = \oplus_k (\mathbb Z / p^k)^{(I_k)}$, where we are assuming the torsion is not bounded, i.e. there are infinitely many $k$ with $I_k \neq 0$.

1. $(\hat B / B)[p]$ is infinite-dimensional, so its $\mathbb F_p$-dimension is the same as its cardinality, which is clearly $I = \max(2^{\aleph_0}, \inf_n \prod_{k \geq n \mid I_k \geq \aleph_0} I_k)$. In particular, $(\hat B / B)[p]$ is at least continuum-dimensional. Therefore, $\Qp \otimes \widehat{\Zp^{(I)}}$ has dimension at least the continuum, and so its dimension is the same as its cardinality, which is the same as the cardinality of $\widehat{\Zp^{(I)}} = \varprojlim_k (\mathbb Z / p^k)^{(I)}$. To choose an element of this set, we first choose an element of $(\mathbb Z / p)^{(I)}$ ($I$ many choices), then we choose an element of $(p\mathbb Z / p^2)^{(I)}$ ($I$ many choices), and so forth, for a total cardinality of $I^{\aleph_0}$. Thus $\dim_{Qp}(\Qp \otimes \widehat{\Zp^{(I)}}) = I^{\aleph_0} \max(2^{\aleph_0}, \inf_n \prod_{k \geq n \mid I_k \geq \aleph_0} I_k^{\aleph_0})$.

2. A similar calculation shows that $J = I$.

Thus $\dim_\Qp(\Ext(\mathbb Z / p^\infty, B)) = I^{\aleph_0} = \max(2^{\aleph_0}, \inf_n \prod_{k \geq n \mid I_k \geq \aleph_0} I_k^{\aleph_0})$.