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Suppose $A$ is PID, $I\subset A$ is a nonzero ideal, show $A/I$ is an injective $A/I$-module.

I tried to prove this, but got stuck in my following argument.

To show $A/I$ is injective as $A/I$-module, by the equivalence definition of injective module, we only need to show that if $M$ is an $A/I$-module such that $A/I\subset M$, then $M$ decomposes as $M=A/I\oplus N$ for some submodule $N$.

First, $M$ can be considered as an $A$-module, and suppose $M$ is finitly generated,

then by the structure theorem of f.g. modules over PID, $M=A^n\oplus A/J_1\oplus \cdots\oplus A/J_k$, then since $M$ is an $A/I$-module, then we can assume $n=0, I\subset J_i, \forall 1\leq i\leq k$, and we are left to use $A/I\subset M$.

I think we can say there must some $J_i=I$, but how to deduce this?

For example, consider $k=1$, then it says $A/I\subset A/J, I\subset J$ implies $I=J$.

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Use the Baer criterion for injectvity: it'll make things simpler, I think. –  Mariano Suárez-Alvarez Nov 25 '12 at 20:44
    
still have no idea how to use Baer's criterion –  ougao Nov 26 '12 at 2:18
    
uncookedfalcon gives a nice related answer in this question math.stackexchange.com/questions/245335/… –  ougao Nov 27 '12 at 2:42

1 Answer 1

up vote 3 down vote accepted

A reformulation of Baer's criterion says that an $R$-module $Q$ is injective iff for any ideal $I\subset R$ and any $R$-module homomorphism $f:I\to Q$ there exists an element $x\in Q$ such that $f(a)=ax$ for all $a\in I$.

In your case $R=A/(a)$, $a\neq 0$. Take $I=(b)/(a)$ an arbitrary ideal of $A/(a)$ and a homomorphism $f:(b)/(a)\to R/(a)$. Set $\hat c=f(\hat b)$. Since $(a)\subset (b)$ there exists $\alpha\in A$ such that $a=\alpha b$. Then $\hat \alpha\hat c=\hat \alpha f(\hat b)=f(\hat \alpha \hat b)=f(\hat 0)=\hat 0$, so $\alpha c\in (a)$. As $\alpha\neq 0$ we get $c\in(b)$ and one can write $c=bx$. This shows that $f(\hat b)=\hat b\hat x$ and we are done.

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Thanks, I should have thought in depth. –  ougao Nov 26 '12 at 3:56

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