if $N$ is submodule of $M$ then $\pi_{1}\left(N\right)/\left(N\cap M_{1}\right)\cong\pi_{2}\left(N\right)/\left(N\cap M_{2}\right)$

Question

Let $M=M_{1}\oplus M_{2}$ ($M_{1},M_{2}$ are submodules of $M$). $\pi_{i}\left(i=1,2\right)$ is projection form $M$ to $M_{i}$. Prove that if $N$ is submodule of $M$ then $\pi_{1}\left(N\right)/\left(N\cap M_{1}\right)\cong\pi_{2}\left(N\right)/\left(N\cap M_{2}\right)$

Help me some hints.

awllower · Answer 1 · 2013-11-27T11:57:28.183

3

Denote $\pi_i(N)/(N\cap M_i)$ by $P_i, i=1, 2.$
Define a map $f: P_1\rightarrow P_2$ by sending $\pi_1(n)+(N\cap M_1)$ to $\pi_2(n)+(N\cap M_2).$ If $k\in N\cap M_1,$ then $\pi_2(k)=0,$ so $f$ is well-defined. And $f$ is evidently surjective, so it remains to show $f$ is injective. If $\pi_2(n)\in N\cap M_2,$ then, as $n=\pi_1(n)+\pi_2(n),$ we know that $\pi_1(n)\in N$ as well, i.e. $\pi_1(n)+(N\cap M_1)=0+(N\cap M_1),$ thus $f$ is injective. Since $f$ is also a homomorphism of modules, this shows that $f$ is an isomorphism.
Hope this helps.

P.S. Thanks to @jeanmfischer, for pointing this out, that, in fact $f$ is a zero-map, i.e. the two modules in question are zero, hence isomorphic. :P

edited Nov 27 '13 at 11:57

answered Nov 27 '13 at 10:54

awllower

16,536

+1 vote.:) May you explain more precise? I need more details – user109584 Nov 27 '13 at 10:58
Why if k∈N∩M1, then π2(k)=0? – Truong Nov 27 '13 at 11:09
@chuyenvien94 Because if $k \in N \cap M_1$ then $k \in M_1$ and $M_1 = ker (\pi_2)$. – jeanmfischer Nov 27 '13 at 11:13
Because $\pi_2$ sends an element in $M$ to its component in $M_2,$ while an element in $M_1$ has zero-component in $M_2.$ :) – awllower Nov 27 '13 at 11:13
@user109584 What kind of details do you need? I am glad I can help. :) – awllower Nov 27 '13 at 11:14
@awllower: Doesn't $\pi_1(N) \simeq N \cap M_1$ ? – jeanmfischer Nov 27 '13 at 11:38
So our $f$ is actually a zero-map! :P – awllower Nov 27 '13 at 11:50
1

Complicated formulas for simple things – jeanmfischer Nov 27 '13 at 12:14
Why $n=π{1}(n)+π{2}(n)$? – user109584 Nov 28 '13 at 03:59
Do you know the definitions of projections, and of direct sums? If so, then your question follows from the definitions really. – awllower Nov 28 '13 at 10:57

score 1 · Answer 2 · answered Nov 27 '13 at 12:22

1

Remark that $\pi_i(N) \simeq N \cap M_i$.

answered Nov 27 '13 at 12:22

jeanmfischer

1,779

Can you explain more precise? – user109584 Nov 27 '13 at 13:26
@user109584 : Do you want full explanations or just a hint ? – jeanmfischer Nov 27 '13 at 17:57
@user109584: Let $x \in N \cap M_1$. For $x \in M_1$, $\pi_1(x) = x$. Then because $x \in N$ and $\pi_1(x) = x$, $x \in \pi_1(N)$. So we have proved that $N \cap M_1 \subset \pi_1(N)$. – jeanmfischer Nov 27 '13 at 19:46
@user109584: that was the easy way. Now try to prove that $\pi_1(N) \subset N\cap M_1$. – jeanmfischer Nov 27 '13 at 19:58

if $N$ is submodule of $M$ then $\pi_{1}\left(N\right)/\left(N\cap M_{1}\right)\cong\pi_{2}\left(N\right)/\left(N\cap M_{2}\right)$

2 Answers2