$[Modules] Show that $h_3$ is injective given comutative diagram$

Question

$[Modules] Show that $h_3$ is injective given comutative diagram$

Let $R$ be a ring with $1$. Consider the comutative diagram of $R-modules$ with exact lines attached.
Show that if $h_2$ and $h_4$ are injective and $h_1$ is surjective, then $h_3$ is injective.

Abstract Algebra

Ssvnormandysr1

Matchmaticians [Modules] Show that $h_3$ is injective given comutative diagram File #1

File #1 (png)

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

Suppose $h_3(x)=0$ for some $x\in M_3$. We have to show that $x=0$. We have $h_4(f_3(x))=g_3(h_3(x))=g_3(0)=0$ so $f_3(x)=0$ since $h_4$ is injective. Then there is $y\in M_2$ such that $x=f_2 (y)$ since the sequence is exact. Now $g_2 (h_2(y))=h_3 (f_2(y))=h_3(x)=0$. So there is $z\in N_1$ so that $h_2 (y)=g_1(z)$. Also there is $w\in M_1$ so that $z=h_1(w)$ since $h_1$ is surjective. Then $h_2(y)=g_1(z)=g_1(h_2(w))=h_2(f_1(w))$. So $y=f_1(w)$ since $h_2$ is injective. Then we get $$x=f_2(y)=f_2(f_1(w))=0$$since the sequence is exact.

Martin

Ssvnormandysr1

0

Hey, gareat job! Would you be able to answer the other question I asked about modules?
Martin

0

I think I have seen that question some time ago, but I cannot remember how to solve it now. Someone more proficient in algebra than me should take it.

$[Modules] Show that $h_3$ is injective given comutative diagram$

Answer

Related Questions

Search