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seen Jul 19 at 21:00

Nov
10
awarded  Enthusiast
Nov
2
accepted Computing $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$
Nov
2
accepted Is the tensor product of two torsion-free modules always non-zero?
Nov
2
comment Is the tensor product of two torsion-free modules always non-zero?
Never mind, I think I know why this is true. I believe it follows from the fact that $\operatorname{Tor}_{1}^{R}(Q,M)=0$ for torsion free modules $M$, where $Q=K/M$. If there's a more elementary way to see this, please let me know.
Nov
2
comment Is the tensor product of two torsion-free modules always non-zero?
Thanks, but I'm unsure of exactly why $M$ is contained in $M\otimes_{R}{K}$. This isn't true in general so you must be using the torsion freeness of the modules, could you elaborate slightly please?
Nov
2
asked Is the tensor product of two torsion-free modules always non-zero?
Oct
22
accepted Global dimension of quasi Frobenius ring
Oct
22
asked Global dimension of quasi Frobenius ring
Oct
15
comment Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
It seems to be true (I think) that this proof only really needs the divisibility of $\mathbb{Q}/\mathbb{Z}$, which is equivalent to injectivity for abelian groups, so should hold for all injective abelian groups.
Oct
15
comment Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
Very nice, thanks!
Oct
15
awarded  Scholar
Oct
15
awarded  Supporter
Oct
15
accepted Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
Oct
15
comment Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
maybe that was the wrong word to use, basically it's from an exercise in a book that asks this question as part (a) and then asks to prove $\mathbb{Q}/\mathbb{Z}$ is injective later in the question, so I really mean without the use of any homological algebra. A proof using group theory or set theory (like the Zorn's lemma suggestion) is what I meant when I said elementary.
Oct
15
asked Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
Oct
13
comment Computing $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$
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
Oct
13
comment Computing $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$
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
Oct
13
awarded  Student
Oct
13
asked Computing $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$