While reading about modules from Hilton & Stammbach's Homological algebra, I saw the following statement : $\Lambda$ is a ring.

$\Lambda$ modules are generalizations of vector spaces and abelian groups. If $\Lambda$ is a field then $\Lambda$ module is a vector space and if $\Lambda=\mathbb{Z}$ then $\Lambda$ module is an abelian group.

I understand this quite well. Here comes the next statement.

In theory of vector spaces there is no interest in the following question : Given vector spaces $A$ and $B$ over a field $K$, find all vector spaces $E$ over $K$ such that $E/B\cong A$ as any such $E$ is isomorphic to $A\oplus B$. However this question turns quite interesting in case of abelian groups $A,B,E$.

First of all i do not understand how does $E\cong A\oplus B$.. I can think of one map $E\rightarrow E/B\rightarrow A\rightarrow A\oplus B$ first map being the quotient map, second map being the isomorphism and third map being inclusion.. How ever i do not think this is a surjective map.

So, what is the isomorphism that they are talking about $E\cong A\oplus B$..

What is stopping them to do the same for abelian groups..


There isn't a unique isomorphism; if you're given a short exact sequence $0 \to A \to E \to B \to 0$ of vector spaces, to exhibit an isomorphism $E \cong A \oplus B$ from here you need to pick a splitting, which means a section $s : B \to E$ of the map $E \to B$. Given a splitting, the isomorphism $A \oplus B \to E$ has components the given map $A \to E$ and the splitting.

Splittings need not exist for general modules, and in particular don't exist for abelian groups in general: the smallest example is

$$0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0.$$

Splittings always exist for vector spaces (assuming the axiom of choice) because vector spaces have bases, and so are free.

  • $\begingroup$ Thanks for your help but i did not get what you are trying to tell... Are you saying writing the isomorphism $E\cong A\oplus B$ is not trivial? $\endgroup$ – kk lm Feb 1 '16 at 17:12
  • $\begingroup$ I mean, I guess it depends on what you mean by trivial. I'm saying there's no such thing as "the" isomorphism: you need to make choices to pick an isomorphism, and those choices in general involve an appeal to the axiom of choice (to find a basis of $B$, and then to use that basis to construct $s$). $\endgroup$ – Qiaochu Yuan Feb 1 '16 at 17:14
  • $\begingroup$ One thing that i understood is there is no splitting for $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$ as we do not have a group homomorphism $\mathbb{Z}_2\rightarrow \mathbb{Z}_4$ $\endgroup$ – kk lm Feb 1 '16 at 17:17
  • $\begingroup$ Well, we do. We just don't have one that's a section of the natural projection $\mathbb{Z}_4 \to \mathbb{Z}_2$. $\endgroup$ – Qiaochu Yuan Feb 1 '16 at 17:18
  • $\begingroup$ Ok Ok, we do have a homomorphsim $1\mapsto 2$ but that is not a section of natural projection $\endgroup$ – kk lm Feb 1 '16 at 17:20

If $E, B, A$ are vector spaces such that $E/B \cong A$, this means we have a surjective map $\pi \colon E \rightarrow A$ (the quotient map) whose kernel is precisely $B$. Now, in the category of vector spaces, given a vector space $E$ and a subspace $B \leq E$, we can always find another subspace $B' \leq E$ such that $B \oplus B' = E$ (this is an internal direct sum). To do that, we can (assuming the axiom of choice, if needed) choose a basis for $B$, extend it to a basis of $E$ and let $B'$ be the vector space spanned by the basis elements that do not belong to $B$.

Having found such a $B'$, we can see that $\pi|_{B'} \colon B' \rightarrow A$ is an isomorphism and so

$$E = B \oplus B' \cong B \oplus A = \ker(\pi) \oplus \mathrm{im}(\pi). $$

  • $\begingroup$ Thanks for sparing your time... This answer is useful... $\endgroup$ – kk lm Feb 1 '16 at 17:24
  • $\begingroup$ Just to clarify my understanding... We have surjective $\pi : E\rightarrow A$... We consider restriction $\pi' : B'\rightarrow A$... As kernel of $\pi$ is $B$, this map $\pi' $ is injective... given $a\in A$ we have $e\in E$ such that $\pi(e)=a$.. we have $e=b+b'$ unique and $\pi'(b')=\pi(b+b')=\pi(e)=a$.. Thus, $\pi'$ is surjective thus an isomorphism? $\endgroup$ – kk lm Feb 1 '16 at 17:54
  • $\begingroup$ @kklm Yep, your argument is correct. $\endgroup$ – levap Feb 1 '16 at 17:57
  • $\begingroup$ please see if you can help me with math.stackexchange.com/questions/1637839/… $\endgroup$ – kk lm Feb 2 '16 at 20:27

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