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I want to show that $\mathbb{Z_{6}}$ is an injective module over itself. I was thinking in using Baer's criterion but not sure how to apply it. So it suffices to look at non-trivial ideals, the non-trivial ideals of $\mathbb{Z_{6}}$ are:

(1) $I=\{0,3\}$

(2) $J=\{0,2,4\}$

So take a $\mathbb{Z_{6}}$-map $f: I \rightarrow \mathbb{Z_{6}}$. Since $f$ is a group homomorphism it must map generators to generators right? so $3 \mapsto 1$ and $0 \rightarrow 0$. Now can we say suppose $f(1)=k$ then define $g: \mathbb{Z_{6}} \rightarrow \mathbb{Z_{6}}$ by sending the remaining elements, (those distinct from 0 and 3), say n, to $nk$?

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Actually, $f$ can't map generators to generators, since the generator of $I$ is killed by $2$ but the generator of $\mathbb{Z}_6$ isn't. –  user29743 Oct 21 '12 at 20:08
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up vote 8 down vote accepted

It is true in general that $\mathbb Z_n$ is injective as a module over itself, and the argument in the general case is not much different to the particular case $n = 6$.

You are right that Baer's criterion is the right way to proceed, but you are wrong in your application of it.

There is no reason that a homomorphism of modules (or groups) should take generators to generators (think about any inclusion of a proper submodule or subgroup). Indeed, as countinghaus notes in a comment, it is not even possible in your context.

Why don't you think about the structure of $I$ (for example) as a $\mathbb Z_6$-module, and determine what the possible homomorphisms $f$ actually are. Then you should see that it is possible to extend such a homomorphism to $\mathbb Z_6$.

(The point is that $I$ is a cyclic $\mathbb Z_6$-module; it is generated by $3$. But it is not free: $3$ is annihlated by multplication by $2$, so its image in $\mathbb Z_6$ must also be annihilated by $2$. What are the possibilities?)


If you want to understand this problem more deeply, you may want to think about the general case. There is a certain amount of bookkeeping when thinking about the maps of ideals into $\mathbb Z_n$ and their possible extensions to $n$, which can be easier to keep track of in general by applying the Chinese Remainder Theorem. Indeed, already the CRT can help with the $\mathbb Z_6$ case, but if it isn't clear to you how to use it, you certainly don't need to use it.

Also, you might want to think about the case of $\mathbb Z$ as a module over itself, and why Baer's criterion fails in this case (so that $\mathbb Z$ is not injective as a module over itself).

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Thank you. Would this work? so f(3) must be 0 or 3. If it is equal to 0 then f is the zero map so extend it via the zero map again. It f(3) is equal to 3 then extend it via the identity morphism. –  user10 Oct 21 '12 at 21:37
    
@user10: Dear user, That sounds good. Cheers, –  Matt E Oct 21 '12 at 22:14
    
to argue the integers is not injective as a module over itself, can we simply say this follows because the rationals are indecomposable? –  user10 Oct 21 '12 at 22:34
    
@user10: Dear user, That's certainly a valid argument, but it's a good exercise to translate it into a direct counterexample to Baer's criterion. Regards, –  Matt E Oct 21 '12 at 23:17
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I found a solution for the 'general' case:

Let I be a ideal of $\mathbb{Z}/n\mathbb{Z}$ then we know that $I=\langle \overline{k} \rangle$ for some $k$ such that $k\mid n$.

If $f:I\rightarrow \mathbb{Z}/n \mathbb{Z}$ is a $\mathbb{Z}/n \mathbb{Z}$-morphism then $im f\subset I$. To show this we note that if $\overline{x}\in im f$ then there exist $\overline{c}=\overline{lk}$ with $\overline{l}\in\mathbb{Z}/n \mathbb{Z}$ s.t. $f(\overline{c})=\overline{x}$, but $n=ks$ for some $s$, so $\overline{0}=f(\overline{ln})=\overline{s}\cdot f(\overline{lk})=\overline{sx}$, then $sx=nt$ for some $t$, but again since $n=ks$, we get $x=tk$.

In particular for $\overline{k}\in I$ we have $f(\overline{k})=b\overline{k}$ for some $b$. So for any $x\in I$ we have $f(x)=bx$ then we just take the extension of $f$ to be the map $\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ where $x\mapsto ax$.

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