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I'm trying to solve this question. My TA told me that it was easy and the information/assumption given is useless.

Question

We have the following inclusions of $R$-modules $M\subseteq N \subseteq L$

Prove that if N and L are both essential extensions of M, then L is also an essential extension of N.

I have tried to solve it using his method: "Use the inclusion function".

Okay so I choose the injective function $i:N \hookrightarrow L$ given by the inclusion. Then I need to show, according to the defintion, that if I choose $S\subseteq L$ then $S\cap i(N) = S\cap N \neq \lbrace 0 \rbrace$.

So I need to show that there exists no submodule $S$ in $L$ such that $S \subseteq L-N$. How do I proceed with this technique?

postguest12
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1 Answers1

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You have to show, that for every submodule $S \subset L$ with $S \cap N = \{0\}$ it follows that $S = \{0\}$.

So let $S \subset L$ be a a submodule with $S \cap N = \{0\}$. Then $$\{0\} = (S \cap N) \cap M = S \cap (N \cap M) =S \cap M,$$ from what follows that $S$ must be $\{0\}$, since $L$ is an essential extension of $M$.

Hope this is the right argument. We didn't need that $N$ is an essential extension of $M$ though.

BIS HD
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  • Since $L$ is an essential extension of $M$ we know that there exists an injective function $\alpha: M \rightarrow L$ such that for each $S\subset L$ we have $S\cap \alpha(M)$. It appears to me that you use that the injective function in this case is the inclusion such that $\alpha ( M ) = M$ ? – postguest12 Nov 28 '13 at 22:26