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I am not sure I am understanding a defintion and would like some input on if I am using it correctly.

Let $U_p$ be the sub-module of the $\mathbb{Z}$-module $\mathbb{Q}/\mathbb{Z}$ consisting of the classes mod $\mathbb{Z}$ of rational numbers of the form $k/p^n$ with $k \in \mathbb{Z} , n\in \mathbb{N}$ for some fixed prime p.

Let $E$ be the product $\mathbb{Z}$-module $M \times N$ where $M$ and $N$ are isomorphic to $U_p$ and $M$ and $N$ are canonically identified with sub-modules of $E$.

What does it mean for $M$ and $N$ are canonically identified with sub-modules of $E$.

I have in my notes there is a canonical identification $j: M \rightarrow M+N$ but I am confused because $M+N$ is not in the set $M\times N$

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Are you sure $M+N$ and $M\times N$ mean different things in this context? A $\mathbb Z$-module is nothing more nor less than an abelian group, except for the notation: modules use additive notation whereas (general) group theory tends to use multiplicative notation. And the "direct product of modules" is the natural generalization of "direct sum" of vector spaces. – Henning Makholm Sep 21 '11 at 13:48
up vote 2 down vote accepted

I don't understand your notes, but the canonical identification of $M$ with a submodule of $M\times N$ is the embedding $M\to M\times N$, $m\mapsto (m,0)$, which is obviously injective and a module homomorphism.

EDIT: I guess your notes should be an identification $M\to M\times 0$

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Since $M$ and $N$ are identified with submodules of $M\times N$ ($M$ is identified with the set of elements of the form $(m,0)$ for $m\in M$ and $N$ with the set of elements of the form $(0,n)$ with $n\in N$), you can form the sum $M+N$. This is the submodule of $M\times N$ consisting of sums $m+n$ with $m\in M$ and $n\in N$ (where you are identifying $m$ and $(m,0)$, and similarly $n$ with $(0,n)$). It's in fact clear that $M+N=M\times N$. In general, it only makes sense to talk about $M+N$ when both $M$ and $N$ are submodules of some larger module. In this case they are both submodules of $M\times N$.

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re: a couple of your flags. Please read all of this meta-thread (including the comments) for possible ideas. The moderators are not allowed to disclose personal information of any users of math.SE, so we cannot be of much help to you. But comparing handed-in solutions to ones given here, and making (potential) students aware that you are on to them might help stem their mis-use of the site. (Also, you can check responses to your flags by clicking on the number(s) next to the "helpful flags" item in your user profile.) – arjafi Sep 30 '13 at 2:54

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