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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!

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    $\begingroup$ 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
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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).

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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.

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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.

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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].

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I would like to contribute in outlining two natural generalizations 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 [4, 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 $I \subset \text{rad}(R)$ 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 $GE_2(R) \subset GL_2(R)$ 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, 3]). 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 [2] and the result was shown to be sharp in [3].

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 generalized 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 generalized Andrews-Curtis conjecture if and only if the natural map $R \twoheadrightarrow R/(n - 1)R$ where $R = \mathbf{Z}[1/n]$ induces a surjective homomorphism between unit groups.


[1] "On the structure of the $GL_2$ of a ring", P. M. Cohn, 1966.
[2] "Tits systems with affine Weyl groups in Chevalley groups over Dedekind domains", E. Abe and J. Morita, 1988.
[3] "Subrings in quadratic fields which are not $GE_2$", H. Yu and S. Chen, 2003.
[4] "The K-book", C. Weibel, 2013.

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