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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


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?

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How do you know the map is surjective? – Rasmus Oct 13 '12 at 12:09
That was a part I was unsure of, but I eventually decided that it was because if I take any function $k:\mathbb{Q}\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $k(q)=q_{k}+\mathbb{Z}$ say, then it can just be written as $k=pg$ where $p:\mathbb{Q}\rightarrow{\mathbb{Q}/\mathbb{Z}}$ the natural map and $g:\mathbb{Q}\rightarrow\mathbb{Q}$ such that $g(q)=q_{k}$ @Rasmus – Paul Gilmartin Oct 13 '12 at 12:22
up vote 3 down vote accepted

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.

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Thanks a lot! I think I have just realised why my argument in my comment to @Rasmus was incorrect, the map $g$ I defined need not necessarily be a homomorphism. I really like this answer, but it has left me feeling that there must be another abelian group $B$ such that it is easier to prove $\operatorname{Ext}_{\mathbb{Z}}^{1}(\mathbb{Q},B)$ is non-zero, for the question was just the start of a long exercise in Rotman, and I'd be surprised if an answer this complex was the one in mind. A very nice answer nonetheless – Paul Gilmartin Oct 13 '12 at 13:33
I think (just) the non-surjectivity of $\hom_\Bbb Z(\Bbb Q,\Bbb Q)\rightarrow\hom_\Bbb Z(\Bbb Q,\Bbb Q/\Bbb Z)$ can be seen a bit easier: decomposing $\Bbb Q/\Bbb Z$ as direct sum of $p$-torsion factors shows that it has lots of automorphims; composing the natural $\Bbb Q\to\Bbb Q/\Bbb Z$ with such automorphisms gives lots of morphisms $\Bbb Q\to\Bbb Q/\Bbb Z$, all with kernel $\Bbb Z$, and only two of them come from $\hom_\Bbb Z(\Bbb Q,\Bbb Q)$. – Marc van Leeuwen Oct 13 '12 at 14:34

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