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If $d|m$ and $\overline{f} : \frac{\mathbb{Z}}{m\mathbb{Z}} \rightarrow \frac{\mathbb{Z}}{d\mathbb{Z}}$ is the map defined by $\overline{f}([a]_m)=[a]_d$ then show $\overline{f}$ maps $U_m\rightarrow U_d$.

This was both a homework problem and an exam problem for me and i got it wrong in both places :(. After I went to consult the professor he gave me the hint that $(a,m)=1 \rightarrow ax+my=1, m=dz$ but im not sure how this helps in the context of the problem.

I was under the impression that in a homomorphism units always get mapped to units, so why do we even need the stipulation that $d|m$?

Could someone please explain this step by step, because it seems like a very simple problem and im missing something fundamental.

Thanks in advanced for the help!

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The fact that $m=dz$ lets you conclude that $ax+dzy=1,$ which is only possible if $(a,d)=1.$ Then you know that $[a]_d$ is a unit. Also, what do you mean when you say $\overline f$ is a homomorphism, since $\Bbb Z/m$ and $\Bbb Z/d$ are not necessarily multiplicative groups (some elements are missing inverses)? – Andrew Nov 13 '12 at 22:35
$\bar f$ is not well defined if $d\not\mid m$. – lhf Nov 13 '12 at 23:13
up vote 3 down vote accepted

As mentioned by lhf:

$\bar f$ is not well defined if $d \nmid m$.

Consider for example $m = 3, d = 2$. Then we have $[3]_3 = [6]_3$, so if $\bar f$ is to be well defined we would need $[3]_2 = [6]_2$, which is obviously false.

So how can we see that $d \mid m$ is indeed the appropriate condition? If $[a]_m = [b]_m$, this means that $m \mid b - a$, i.e. there is an $x$ such that $b - a = mx$.

If we are to have $[a]_d = [b]_d$ (so that $\bar f$ is well defined), then also $b - a = dx'$ for some $x'$. The case $b = a + m$ then supplies us with the knowledge that $d \mid m$.

As to your supposition that a homomorphism always sends units to units, this is not true. Let $f: R \to R'$ be a homomorphism. Then we obviously have $f(1) f(r) = f(r) f(1) = f(r)$ for all $r \in R$, but this does not guarantee that $f(1) r' = r' f(1) = r'$ for $r'$ that are not in the image of $f$; so while the behaviour of $f(U_R)$ is very unit-like, it may not actually consist of units in $R'$ -- with a notable exception if $f$ is surjective.

As an explicit counterexample, consider the injection $\iota_1: \Bbb Z \to \Bbb Z \times \Bbb Z, \iota_1(n) = (n, 0)$.

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