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I'm trying an old problem here: http://www.math.dartmouth.edu/archive/m111s09/public_html/homework-posted/hw1.pdf

Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?

My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?

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    In general, even if $k$ is a unit modulo $n$ you need not have that $k$ is a unit modulo $m$; e.g., $m=6$, $n=3$, $k=2$.2011-10-07
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    @Aturo, but 5 is a unit modulo 6, and it becomes 2 when mapped to $\mathbb Z_3$, so $k=2$ is still hit.2011-10-07
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    @Henning: Yes: the point is that Calvin seems to be trying to show that if $k$ is a unit modulo $n$, then it will have to be a unit modulo $m$ (at least, up to the point where he says he is "lost"), and that is false. Instead, he needs to show that there exists an $r$ such that $\gcd(k+nr,m)=1$; or else reduce to a case where it *is* true, like Rolando suggests.2011-10-07
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    @Arturo, I think the part leading up to "I'm lost here" is just meant to be "here are some random deductions I've made that turn out to lead nowhere" -- i.e., showing that he's done some work, which is a good thing and should be encouraged. They don't seem to aim anywhere in particular, but are just the most immediate conclusions one can draw from the available assumptions. Trying that is certainly a good initial plan in general. In particular: given the explicit wish to "find a unit which _maps to_ $k$", I don't see any implication that he expects $k$ itself to be that unit.2011-10-07
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    @Henning: I don't necessarily disagree; however, what my first comment is meant to show is that *that* line of attack *cannot* succeed, so that it was not merely that Calvin didn't get anywhere: that path doesn't lead anywhere. Perhaps I should have been clearer as to what my comment was meant to be doing (I was certainly not claiming the desired theorem is false).2011-10-08
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    @Arturo: OK; perhaps I was too thin-skinned. Sorry.2011-10-08

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