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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?

Calvin Broadus
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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., $m=6$, $n=3$, $k=2$. – Arturo Magidin Oct 07 '11 at 21:34
  • @Aturo, but 5 is a unit modulo 6, and it becomes 2 when mapped to $\mathbb Z_3$, so $k=2$ is still hit. – hmakholm left over Monica Oct 07 '11 at 21:56
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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. – Arturo Magidin Oct 07 '11 at 22:05
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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. – hmakholm left over Monica Oct 07 '11 at 22:35
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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). – Arturo Magidin Oct 08 '11 at 02:46
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    @Arturo: OK; perhaps I was too thin-skinned. Sorry. – hmakholm left over Monica Oct 08 '11 at 02:50
  • See also https://mathoverflow.net/questions/32875/ – Watson Apr 30 '21 at 11:25

4 Answers4

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First reduce to the case where $n$ and $m$ are both powers of a prime $p$, using the Chinese remainder theorem. After this reduction the result is clear since the $p$-adic valuation of $n$ is lesser than the $p$-adic valuation of $m$ for any prime $p$.

KCd
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Louis La Brocante
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If you want a "pedestrian" solution without knowledge in group theory or number theory, you have to note that when $k$ is your unit, then $k+ln$ are also units, so you have several candidates for the unit in the bigger group and you only have to plug them into your equation and show that one of them gives a coefficient of $n$ divisible by $a$.

Phira
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This answer is just to observe that this is a special case of a very general result: if $R$ is an artinian ring and $I$ is a two-sided ideal of $R$, then the quotient map induces a surjection $R^\times \to (R/I)^\times$ (finite rings such as $\mathbf{Z}/n \mathbf{Z}$ are, of course, artinian). There is also closely related fact that for any ring $R$, the quotient map $R^\times \to (R/J)^\times$ is surjective, where $J$ is the Jacobson radical of $R$.

Stephen
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It's simply case $\,\gcd(\color{#0a0}{k,n})\!=\!1\,$ of below (which has an intuitive one-line high-school level proof).

$$\bbox[8px,border:1px solid #c00]{\gcd(\color{#0a0}{k,n},m)=1\,\Rightarrow\, \gcd (k\!+\!n\:\!x,\,m)\!=\!1\,\ \text{for some integer }x}\quad\ $$

Remark $ $ The above result implies that there are infinitely many coprimes to $\,m\,$ in every A.P. (Arithmetic Progression) $\,k+n\:\!x\,$ [except (obviously) when $\,k,n,m\,$ have a nontrivial common divisor]. It is a (much) simpler coprime form of Dirichlet's Theorem on primes in A.P. The linked post shows how to intuitively derive a simple one-line proof using the simple idea behind Stieltjes generalization of Euclid's proof of infinitely many primes (hint: coprimes to $\,m\neq 0\,$ arise by partitioning into $\rm\color{#c00}{two}\ \color{#0a0}{summands}$ all prime factors of $\,m,\,$ i.e. $\,\color{#c00}k+\color{#0a0}j\ $ is coprime to $\,m\,$ if every prime factor of $\,m\,$ divides exactly one of $\,k\,$ or $\,j).\,$ This method often yields simpler and more intuitive proofs vs. common alternative methods using CRT or $p$-adic valuations, e.g. here.

Bill Dubuque
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