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!

  • 3
    $\begingroup$ Theorem 3.8 here might be of interest : if $f$ is surjective and has finite kernel, then so is $f^{\times}$. $\endgroup$ – Watson Dec 4 '18 at 19:30
  • $\begingroup$ "Here" in @Watson's comment: Bartel and Lenstra Jr. - Commensurability of automorphism groups. $\endgroup$ – LSpice Nov 28 '20 at 19:42
  • $\begingroup$ (In my comment above, we also need $f$ to have finite cokernel in the theorem ; see also lemma 3.4 ibid. which just assumes $R$ to be artinian, which was also given here). $\endgroup$ – Watson Feb 20 at 17:49

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 $\pm 2^{\mathbf Z}5^{\mathbf Z}$. 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).


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.


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.


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].


Addendum I. I am raising this addendum at the top because it is more relevant to the question than what I have written before.

We denote by $R^{\times}$ the unit group of a unital ring $R$.

Claim 1. Let $R$ be a commutative and unital ring. Then the following are equivalent:

  • The natural map $R^{\times} \rightarrow (R/I)^{\times}$ is surjective for every ideal $I$ of $R$.

  • The Bass stable rank of $R$ is $1$.

This claim was inspired by Theorem 5.1 of [5], a preprint which cannot be more on topic!

Let us define the Bass stable rank of an associative and unital ring $R$. We call $\mathbf{r} \in R^n$ a unimodular row of $R$ if the components of $\mathbf{r}$ generate $R$. We denote by $\operatorname{Um}_n(R)$ the set of unimodular rows of size $n$. We say that $n > 0$ lies in the stable range of $R$ if for every $(r_1, \dots, r_{n + 1}) \in \operatorname{Um}_{n + 1}(R)$, there is $r \in R$ such that $(a_1 + r_1a_{n + 1}, \dots, a_n + r_n a_{n + 1}) \in \operatorname{Um}_n(R)$. It is well-known and easily checked that the stable range has no gap, i.e., if $n$ lies in the stable range of $R$, then so does $m$ for every $m > n$. It is also well-known that $\operatorname{sr}(R/\mathcal{J}(R)) = \operatorname{sr}(R)$ where $\mathcal{J}(R)$ denotes the Jacobson radical of $R$, that is, the intersection of the maximal ideals of $R$.

The Bass stable rank $\operatorname{sr}(R)$ of $R$, or simply stable rank of $R$, is the smallest element in the stable range of $R$.

The proof of Claim $1$ is actually straighforward.

Proof of Claim $1$. Assume that $R^{\times} \rightarrow (R/I)^{\times}$ is surjective for every ideal $I$ of $R$. Let $(a_1, a_2) \in \operatorname{Um}_2(R)$. Then $a_1 + I$ is unit of $R/I$ with $I = Ra_2$. Therefore we can find $u \in R^{\times}$ such that $a_1 + I = u + I$, i.e., there is $r \in R$ verifying $a_1 + r a_2 = u$. As a result $\operatorname{sr}(R) = 1$. Let us assume now that $\operatorname{sr}(R) = 1$. Let $I$ be an ideal of $R$ and let $a + I \in (R/I)^{\times}$. Then we can find $b \in I$ such that $(a, b) \in \operatorname{Um}_2(R)$. Because $R$ is of stable rank $1$, there is $r \in R$ such that $u \Doteq a + rb$ is a unit of $R$.

Semilocal rings, and Artinian rings in particular, are known to have stable rank $1$. More generally, so are the rings $R$ such that $R/\mathcal{J}(R)$ has Krull dimension $0$ (Justin Chen's semifields), e.g., von Neumann regular rings. The reader should be bear in mind that $$\operatorname{sr}(R) \le \dim_{Krull}(R) + 1$$ by the Bass Stable Range Theorem and its generalisation by R. Heitmann [2].

The class of local-global rings encompass all the previous classes.

Definition. A commutative and unital ring $R$ is local-global if every (possibly multi-variate) polynomial $f$ with coefficients in $R$ whose values generate $R$ represents a unit of $R$, i.e., some value of $f$ is a unit.

The following is well-known and immediate.

Claim 2. If $R$ is local-global then $\operatorname{sr}(R) = 1$.

Proof. Given $(a, b) \in \operatorname{Um}_2(R)$, consider the polynomial $f(X) = a + bX$.

Addendum II: I just discovered that the connection between surjections on unit groups and the stable rank 1 condition has been known for a long time. This is Lemma 6.2 of "Stable range in commutative rings" (1967) by D. Estes and J. Ohm. See also Corollaries 6.3 and 6.4 for results on the kernel of the induced group homomorphism. (The paper is wonderfully written.)

Former answer.

I would like to contribute in outlining two natural generalisations of the original question for which the Jacobson radical also comes into play. Eventually, I would like to quote two research results for which OP’s question is essential.

Considering that $R^{\times} = GL_1(R)$, it is natural to ask

When does the surjection $R \twoheadrightarrow R/I$ induce a surjective map $GL_n(R) \rightarrow GL_n(R/I)$?.

Here $I$ is an ideal of $R$ and the second map is the reduction of matrix coefficients modulo $I$. If $I$ is an ideal contained in the Jacobson radical $\text{rad}(R)$ of $R$, then $GL_n(R) \rightarrow GL_n(R/I)$ is surjective [5, Exercise I.1.12.iv].

Considering that $R$ is an $R$-module and that every element in $R^{\times}$ generates $R$, it is natural to ask

For $M$ an $R$-module, when do $R$-generating sets of $M/IM$ lift to $R$-generating sets of $M$?.

This holds if $I \subseteq \text{rad}(R)$ and $M/IM$ is finitely generated over $R$ by Nakayama’s lemma, see also the related concept of superfluous submodule.

I introduce now a result published in 2003 which directly relates to OP's question. Let $R$ be a Dedekind domain which is universal for $GE_2$. The latter property means that in the subgroup $GE_2(R) \subseteq GL_2(R)$ generated by the transvections and the diagonal matrices, the latter generators are only subject to the « obvious » or universal relations (equivalently $K_2(2, R)$ is generated by symbols as a normal subgroup of $St_2(2, R)$ [1, 4]). Then we have:

If $p$ is a prime element of $R$ such that the natural $R \twoheadrightarrow R/(p)$ induces a surjective homomorphism between unit groups, then the localization $RS^{-1}$ of $R$ with $S = \{1, p, p^2, \dots\}$ is also universal for $GE_2$.

This was proved in [3] and the result was shown to be sharp in [4].

I can’t resist mentioning a humble result of mine, which I would like to believe is entertaining.

A finitely generated group $G$ of rank $n$ is said to satisfy the generalised Andrews-Curtis conjecture if every $n$-tuple of elements which normally generate $G$ can be transitioned to an $n$-tuple of generators of $G$ by means of finitely many $AC$-moves, i.e., applying transvections or replacing one component by a conjugate a finite number of times. (If $G$ is the free group on $n$ generators, this is just the Andrews-Curtis conjecture, still unsettled.) Then the result reads as:

The solvable Baumslag-Solitar group $\langle a, b \,\vert\, aba^{-1} = b^n \rangle$ with $n \ge 1$ satisfies the generalised Andrews-Curtis conjecture if and only if the natural map $R^{\times} \rightarrow (R/(n - 1)R)^{\times}$ is surjective where $R = \mathbf{Z}[1/n]$.

[1] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.
[2] R. Heitmann, "Generating non-Noetherian modules efficiently", 1984.
[3] E. Abe and J. Morita, "Tits systems with affine Weyl groups in Chevalley groups over Dedekind domains", 1988.
[4] H. Yu and S. Chen, "Subrings in quadratic fields which are not $GE_2$", 2003.
[5] C. Weibel, "The K-book", 2013.
[6] J. Chen, "Surjections of unit groups and semi-inverses", 2017.


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