Computing $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$

I'm trying to find an abelian group $B$ such that $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},B)$ is non-zero. My first guess was just to choose $B=\mathbb{Z}$. Using the following argument, I deduced that $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=\{0\}$, however I then read this question, which says in fact it is non-zero, so I'm assuming there's something wrong with my argument:

First, take the injective resolution $0\rightarrow{\mathbb{Z}}\rightarrow{\mathbb{Q}}\rightarrow{\mathbb{Q}/\mathbb{Z}}\rightarrow{0}$ of $\mathbb{Z}$, form the deleted resolution $0\rightarrow{\mathbb{Q}}\rightarrow{\mathbb{Q}/\mathbb{Z}}\rightarrow{0}$ (no longer exact) and apply the $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\bullet)$ functor to the deleted resolution to obtain the non-exact sequence

$$0\rightarrow{\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q})}\rightarrow{\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\rightarrow{0}}$$

So, as far as I can see, $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$ is the quotient of the kernal of the zero map from $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})$ by the image of the surjective map displayed in the functored sequence above, hence is zero. I'm guessing I have made a mistake either with the way I'm interpreting the functored sequence or with the definition of $\operatorname{Ext}$. Can anyone help me understand where I have gone wrong?

Solution 1:

There are indeed homomorphisms $\mathbb Q\to\mathbb Q/\mathbb Z$ that are not induced by a homomorphism $\mathbb Q\to\mathbb Q$.

For a prime $p$ we have the $p$-adic value on $\mathbb Q$ given by $\left |\pm\frac abp^k\right|_p=p^{-k}$ if $a,b$ are prime to $p$ and $k\in\mathbb Z$. For $n\in\mathbb N$ let $\mathbb Q_n\subset \mathbb Q$ be the set of rationals $x$ with $|x|_p\le \frac1n$ for all $p$. Then $\mathbb Q_n$ is a subgroup of $\mathbb Q$ and is cyclic with generator $g_n = (\prod_p p^{\lfloor \log_pn \rfloor})^{-1}$. A homomorphism $f\colon \mathbb Q\to \mathbb Q/\mathbb Z$ is determined by specifying values $f(g_n)$ such that $f(g_n)=\frac{g_n}{g_{n+1}}\cdot f(g_{n+1})$ always holds. Note that $k_n:=\frac{g_n}{g_{n+1}}$ is always an integer and if it is $>2$ (which happens infinitely often), we have several choices for $f(g_{n+1}) $, differing by $\frac1{k_n}$. At least one of the possible choices differs by at most $\frac1{2k_n}$ from $\frac12$, i.e. is in $[\frac13,\frac23]$. In the end we obtain a homomorphism $f\colon \mathbb Q\to \mathbb Q/\mathbb Z$ that cannot come from a homomorphism $\tilde f\colon \mathbb Q\to \mathbb Q$. Indeed, there would have do be infinitely many $m\in\mathbb N$ with $\tilde f(\frac1m)\ge\frac13$ or $\tilde f(\frac1m)\le-\frac13$. But then $|\tilde f(1)|=m\cdot|\tilde f(\frac1m)|\ge \frac m3$, which is absurd.