Why any short exact sequence of vector spaces may be seen as a direct sum?

Question

This is actually the first time I have worked with short exact sequences and I have no clue why the following assertion is true:

Any short exact sequence of vector spaces $$ 0 \longrightarrow U \overset{A}{\longrightarrow} V \overset{B}{\longrightarrow} W \longrightarrow 0 $$ reduces to a decomposition $V \simeq U \oplus W$.

Can anyone give a clear proof? I only know from the definition of short exact sequences that $A$ is injective, $B$ is surjective, and $\text{im}(A) = \ker(B)$

Yes sorry, V, U and W are finite vector spaces – Alonso Delfín Jan 15 '15 at 07:59 — Alonso Delfín, Jan 15 '15 at 07:59

Najib Idrissi · Accepted Answer · 2018-04-06T07:37:38.067

17

In general you cannot see $V$ as a direct sum $U \oplus W$ when these are modules over a general ring $R$. For example consider the short exact sequence: $$\newcommand\Z{\mathbb{Z}}0 \to \Z/2\Z \to \Z/4\Z \to \Z/2\Z \to 0,$$ it is not true that $\Z/4\Z$ is isomorphic to $\Z/2\Z \oplus \Z/2\Z$.

However when you are working with vector spaces over a field (and since you tagged this "differential topology" I assume you're actually interested in de Rham cohomology, which is over $\mathbb{R}$), then yes, it follows that $V$ is isomorphic to $U \oplus W$. A simple way to see that in finite dimension is through the rank-nullity theorem (and the fact that two vector spaces are isomorphic iff they have the same dimension).

In general though, there is still an isomorphism, even if $U$, $V$, $W$ have infinite dimension (and you assume the axiom of choice I guess). Every vector space is free, ie. has a basis. Choose a basis $\{w_i\}$ of $W$. Since $B$ is surjective, there are elements $v_i \in V$ such that $B v_i = w_i$. Define a linear map $C : W \to V$ by $Cw_i = v_i$. Since $\{w_i\}$ is a basis this determines a (unique) linear map $C$. Then it is a simple exercise to show that the following map is an isomorphism: $$\begin{align} U \oplus W & \to V\\ (u, w) & \mapsto (Au, Cw) \end{align}$$

edited Apr 06 '18 at 07:37

answered Jan 15 '15 at 07:53

Najib Idrissi

56,269

I apologize because I did not tell in the beginning that I was working with finite vector spaces. Thanks a lot for the answer. – Alonso Delfín Jan 15 '15 at 08:02
1

More generally, every short exact sequence splits is equivalent to every module being projective; i.e. the ring is finite direct product of matrix rings over division rings. – Jan 15 '15 at 19:14
By the way, this splitting condition is not enough when you consider exact sequence of groups that aren't necessarily abelian, see there. – Najib Idrissi Jan 16 '15 at 08:20
"It is a simple exercise...": or just apply the splitting lemma! – xuq01 Jan 26 '21 at 21:21
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@xuq01 The exercise is the proof of the splitting lemma. – Najib Idrissi Feb 01 '21 at 09:35

score 2 · Answer 2 · answered Jan 15 '15 at 07:50

2

The conclusion is false in general. Pick cyclic groups ${\mathbb Z}_p$, ${\mathbb Z}_q$, and ${\mathbb Z}_{pq}$. You can easily cook up a short exact sequence $$ 0\to{\mathbb Z}_p\to{\mathbb Z}_{pq}\to{\mathbb Z}_q\to0, $$ but the group in the center is not the direct product of the groups on its sides.

The conclusion holds if, e.g., you have a short exact sequence of vector spaces.

answered Jan 15 '15 at 07:50

Nicola

49

13

Actually if $p$ and $q$ are different primes then $\mathbb{Z}_{pq} \cong \mathbb{Z}_p \times \mathbb{Z}_q$ by the Chinese remainder theorem. – Najib Idrissi Jan 15 '15 at 07:53
Thanks for your answer, i forgot to write that I was working with finite vector spaces. – Alonso Delfín Jan 15 '15 at 08:04
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You can easily fix this by taking $q=p$, and maybe use the fact that $\Bbb{Z}_{p^2}$ has a normal subgroup of order $p$. So you can take the inclusion followed by the quotient. – D. Zack Garza Feb 06 '21 at 03:06

Why any short exact sequence of vector spaces may be seen as a direct sum?

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