# When does a ring surjection imply a surjection of the group of units?

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?

• Theorem 3.8 here might be of interest : if $f$ is surjective and has finite kernel, then so is $f^{\times}$. – Watson Dec 4 '18 at 19:30

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 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 $$f(X) = a + bX$$.

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.