Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that if $M$ is a direct sum of $M_1$ and $M_2$ then $M/M_1$ is isomorphic to $M_2$ and $M/M_2$ is isomorphic to $M_1$.

share|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Dec 31 '12 at 10:40

3 Answers 3

Well, if $M = M_1 \oplus M_2$, then every $m \in M$ can be written as $m=m_1+m_2$ where $m_1 \in M_1$ and $m_2 \in M_2$.

If you consider the linear map $\phi : M \to M_1$ which send every $m\in M$ to his projection over $M_1$, namely $m_1$, then the kernel of this map is exactly the set $M_2$, because $M_1\cap M_2 = \{0\}$.

However, this map is clearly onto, so by factorisation, $M/Ker\phi \cong Im\phi$, namely $M/M_2 \cong M_1.$

The other isomorphism can be deducted this way.

share|improve this answer

You can simply view the direct sum as the set $M:=\{(x,y)\mid x\in M_1, y\in M_2\}$ and component-wise addition and scalar multiplication, together with the obvious embeddings $i_1\colon M_1\to M, x\mapsto (x,0)$ and $i_2\colon M_2\to M, y\mapsto (0,y)$. In this setup, it is clear that $M/i_2(M_2)\to M_1, (x,y)+i_2(M_2)\mapsto x$ and $M/i_1(M_1)\to M_2, (x,y)+i_1(M_1)\mapsto y$ are isomorphisms.

share|improve this answer

I like to prove such things only using the universal properties..

Firstly a map out of the coproduct $M=M_1 \oplus M_2$ is the same as maps out of each guy individually. Moreover by the universal property of quotients maps out of $M/M_1$ are maps out of $M$ that vanish on $M_1$, so you directly see that $M_2$ is a quotient (recall that objects defined by a universal property, such as coproducts and quotients are best thought of as unique only up to unique isomorphism).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.