MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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).

share|cite|improve this answer

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.

share|cite|improve this answer

There is a simple and reasonably general sufficient criterion for a ring surjection $f : R \to S$ to induce a surjection $f^\times : R^\times \to S^\times$ on unit groups (apologies for bumping an old post, but none of the other answers seemed to have this simple line of reasoning).

Proposition: Let $f : R \twoheadrightarrow S$. If $\ker f$ is contained in all but finitely many maximal ideals of $R$, then $f^\times$ is surjective.

Proof: Write $I := \ker f$, and $\text{mSpec}(R) \setminus V(I) = \{m_1,...,m_n\}$. Then $\{I, m_1,...,m_n\}$ are pairwise comaximal. Pick $v \in S^\times$, and write $v = f(u)$ for some $u \in R$ (notice $u \not \in m$, for any $m \in \text{mSpec}(R) \cap V(I)$). By Chinese Remainder, there exists $a \in R$ with $a \equiv 0 \pmod{I}$, $a \equiv 1-u \pmod{m_i}$ for $i = 1,...,n$. Then $u + a \in R^\times$, and $f(u+a) = f(u) = v$.

This immediately yields that if $R$ is semilocal (has only finitely many maximal ideals), then every surjection out of $R$ induces a surjection on units. This generalizes the case where $R$ is Artinian (or finite). The case that $I$ is contained in the Jacobson radical of $R$ can also be recovered via the reduction:

Proposition: Let $\overline{R} := R/\text{rad}(R)$ (where $\text{rad}(R)$ is the Jacobson radical). Then $f^\times : R^\times \to (R/I)^\times$ is surjective iff $\overline{f}^\times : \overline{R}^\times \to (\overline{R}/\overline{I})^\times$ is surjective.

This also yields the semilocal case, since then $\overline{R}$ is a finite product of fields. Concerning the limitations of the first proposition: although the condition that $I$ avoids only finitely many maximal ideals seems strong, it is in a sense sharp: e.g. $\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ for $p$ prime, $p > 3$ does not induce a surjection on units. One final remark that may be of interest:

Proposition: If $R = \bigoplus_{i=0}^\infty R_i$ is $\mathbb{N}$-graded and $I \subseteq R_+$ is a homogeneous prime concentrated in positive degree, then $R^\times \to (R/I)^\times$ is surjective.

share|cite|improve this answer
Very very nice! – Filippo Alberto Edoardo Jan 4 '14 at 11:32

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].

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.