Question

The following might be a very trivial question. If so, I don't mind it being closed, but would appreciate a reference where I could read about it.

Let $R$ and $S$ be commutative rings and let $R^\times$ and $S^\times$ denote their respective multiplicative groups of units. Let $f:R \to S$ be a ring homomorphism and let $f^\times : R^\times \to S^\times$ denote the induced group homomorphism. Finally, suppose that $f$ is surjective.

Under what conditions (if any) will $f^\times$ be surjective?

Thanks in advance!

Answer 1

If $R$ is a local ring and $S$ is its residue field then the map is onto, but that's too easy, isn't it?

I don't think this is a trivial question at all! For example, consider the ring ${\mathbf Z}[\sqrt{2}]$, which has infinitely many units ($\pm 1$ times powers of $1+\sqrt{2}$). For any nonzero prime ideal $(\pi)$ (the ring is a PID so the ideal is principal, not that it matters), we can reduce mod $\pi$ and get a map ${\mathbf Z}[\sqrt{2}] \rightarrow {\mathbf Z}[\sqrt{2}]/(\pi)$. This is onto and the target ring is a finite field, so its unit group is cyclic. Asking whether the map of unit groups is onto is essentially equivalent to asking if $1 + \sqrt{2}$ is a generator of the units mod $\pi$. This doesn't always happen (e.g., when $\pi = 5$ the ring ${\mathbf Z}[\sqrt{2}]/(5)$ is a field of size 25,
$1+\sqrt{2} \bmod 5$ has order 12, and $(1+\sqrt{2})^{6} \equiv -1 \bmod 5$, so the whole unit group of ${\mathbf Z}[\sqrt{2}]$ maps onto only half the units mod 5). However, it is conjectured that there are infinitely many prime ideals $(\pi)$ such that $1+\sqrt{2} \bmod \pi$ is a generator of the units. This is still an open problem, although it is known to follow from suitable versions of the Generalized Riemann Hypothesis.

This is a generalization of Artin's primitive root conjecture, which says that any nonzero integer $a$ other than $\pm 1$ or a perfect square should be a generator of the units mod $p$ for infinitely many primes $p$. For example, $10 \bmod p$ should be a generator for infinitely many $p$. (Concretely, this says there should be infinitely many $p$ such that $1/p$ has decimal period $p-1$, which is the longest it could conceivably be for any $p$.) Artin's original conjecture may not seem like it fits your specific question, since ${\mathbf Z}$ has only two units, but it is straightforward to make Artin's problem fit your question, e.g., use ${\mathbf Z}[1/10]$ instead of ${\mathbf Z}$ and its unit group is $\pm 10^{\mathbf Z}$. Artin's conjecture for $a=10$ amounts to saying the unit group of ${\mathbf Z}[1/10]$ maps onto the unit group of its reduction modulo infinitely many primes (not counting 2 and 5, which are no longer prime).

Answer 2

I don't know how satisfactory this will be, but at least its a first stab at an answer, and might highlight some of the issues.

There is one "obvious" condition which ensures $f^\times$ is surjective: if the kernel of $f$ is contained in the Jacobson radical of $R$, then $f^\times$ is surjective. We can think of $S$ as being $R/I$ for some ideal $I$, so that maximal ideals of $R/I$ correspond to maximal ideals of $R$ containing $I$. Since units are precisely elements that miss all maximal ideals, if every maximal ideal of $R$ contains $I$ then every unit in $R/I$ can be lifted to a unit in $R$ (in fact, every lift to an element of $R$ is a unit in this case).

For $I$ not contained in the Jacobson radical, $R$ will have maximal ideals not containing $I$, and the question of whether every unit in $R/I$ lifts to an element of $R$ missing every maximal ideal in $R$ seems subtle.

There are probably other, better, weaker conditions which will imply surjectivity, however.

It is also useful to keep in mind the following example: the map $k[x] \to k[x]/(x^2)$ is surjective and does not induce a surjection on units.

Answer 3

A sufficient condition is that $R$ is artinian (for example, finite). [Reduce to the local case and apply Jack's argument; or this proof which avoids the maximal ideal description of units].