# More classes of rings spanned by units?

I will formulate questions only after a rather protracted introduction.

Rings in this text are unital and nontrivial, algebras are associative, modules are unitary.

$\newcommand{\Uspan}{\mathfrak{S}} \newcommand{\End}{\mathrm{End}} \newcommand{\isomorph}{\cong} \newcommand{\Mtxalg}{\mathbb{M}} \newcommand{\id}{\mathrm{id}} \newcommand{\Finfield}{\mathbb{F}} \newcommand{\Jrad}{\mathfrak{J}} \newcommand{\set}[1]{\{#1\}}$Let $R$ be any ring. The set $\Uspan(R)$ of all sums of units of $R$ (including $0$, the sum of no units at all)
is a subring of $R$. We shall say that $R$ is spanned by units if $\Uspan(R)=R$. If a commutative ring $k$
is spanned by units and $A$ is a $k$-algebra, then $\Uspan(A)$ is a $k$-subalgebra of $A$.

(A) $~$If $R$ is any ring and $M$ is a free right $R$-module having a basis consisting of at least two vectors, then the ring $\End(M)$ is spanned by units.

Proof. $~$First suppose $M$ has a finite basis consisting of $n\geq 2$ vectors, so that $\End(M)\isomorph\Mtxalg_n(R)$. Let $x\in R$. If $1\leq i,j\leq n$ and $i\neq j$, then $aE_{ij}=(I+aE_{ij})-I$, where $I+aE_{ij}$ is invertible with the inverse $I-aE_{ij}$. Now let $1\leq i\leq n$, and pick some $j$, $1\leq j\leq n$, different from $i$; also denote by $I'$ the sum of all $E_{kk}$ with $1\leq k\leq n$ and $k\neq i,j$. We have $aE_{ii}=A-B$, where $A:=aE_{ii}+E_{ij}+E_{ji}+I'$, $B:=E_{ij}+E_{ji}+I'$, $A^{-1}=E_{ij}+E_{ji}-aE_{jj}+I'$, and $B^{-1}=B$. Now suppose that $M$ has an infinite basis. In this case $\End(M)\isomorph\Mtxalg_2(\End(M))$, and the result follows by the first part of the proof.$~$ Done.

The idea for the second part of the proof comes straight from vector spaces whose algebra of endomorphisms is generated by its idempotents.

There exist rings that are not spanned by units. For example, if $k$ is any commutative ring, then $k[X]$ is not spanned by units.

A subring of a ring spanned by units may not be spanned by units. For example, if $k$ is a field, then the field $k(X)$ is clearly spanned by units, while its subring $k[X]$ is not spanned by units.

It is trivial that a division ring is spanned by units.

A finite-dimensional algebra $A$ over an infinite field $F$ is spanned by units.

Proof. $~$An $u\in A$ is invertible iff $u_L\colon A\to A : x\mapsto ax$ is bijective iff $\det(u_L)\neq 0$. Let $a\in A$. The monic polynomial $\det((\lambda 1_A+a)_L)$ of degree $n=\dim_F A$ in $\lambda\in F$ has at most $n$ zeros, therefore $u:=\lambda 1_A+a$ is invertible for some $\lambda$, and we see that $a=u-\lambda 1_A$ is a unit or a sum of two units.$~$ Done.

Let $R_1$, $\ldots$, $R_s$ ($s\geq 1$) be rings, and suppose that $2\cdot1_{R_k}$ is invertible in $R_k$ for $k=1,\,\ldots,\,s$.
If each of the rings $R_1$, $\ldots$, $R_s$ is spanned by units, then so is the ring $R := R_1\times\cdots\times R_s$.

Proof. $~$It suffices to prove that for each $k=1,\,\ldots,\, s$ and every unit $x$ of $R_k$ the element $\overline{x}\in R$ with the $k$-th coordinate $x$ and all other coordinates $0$ is a sum of units of $R$. Denote by $e$ the element of $R$ with the $k$-th coordinate $0$ and all other coordinates $1$; then $\overline{x}=(2\overline{x}+e)-(\overline{x}+e)$ expresses $\overline{x}$ as a sum of two units of $R$.$~$ Done.

Let $F\isomorph\Finfield_{2^k}$ with $k\geq2$. If each of $F$-algebras $A_1$, $\ldots$, $A_s$ ($s\geq 1$) is spanned by units, then so is their product $A:=A_1\times\cdots\times A_s$.

Proof. $~$Let $x$ be a unit of $A_k$. Let $\overline{x}$ denote the element of $A$ with the $k$-th coordinate $x$ and all other coordinates $0$, and let $e$ be the element of $A$ with the $k$-th coordinate $0$ and all other coordinates $1$. Let $1$, $\alpha_1$, $\ldots$, $\alpha_m$, where $m=|F|-2=2^k-2>0$ is even, be the non-zero elements of the field $F$. Then $1=\alpha_1+\cdots+\alpha_m$ and $\overline{x}=(\alpha_1\overline{x}+e)+\cdots+(\alpha_m\overline{x}+e)$.$~$ Done.

Let $R$ be a ring and $J$ its Jacobson radical. If the quotient ring $R/J$ is spanned by units, the so is the ring $R$.

Proof. $~$Let $x\in R$. By assumption $x+J=y_1+\cdots+y_m+J$ for some $y_1,\ldots,y_m\in R$
such that $y_1+J$, $\ldots$, $y_m+J$ are units of $R/J$, hence $y_1$, $\ldots$, $y_m$ are units of $R$. Since $x=y_1+\cdots+y_m+w$ for some $w\in J$, we have $x=y_1+\cdots+y_m+(1+w)-1$ expressed
as a sum of units of $R$.$~$ Done.

Every finite-dimensional algebra $A$ over a field $F$ consisting of at least three elements is spanned by units.

Proof. $~$Apply Artin-Wedderburn theorem to $A/J$, where $J$ is the Jacobson radical of $R$, then use the results, gathered above, about rings spanned by units.$~$ Done.

The two-element field $\Finfield_2$ is indeed a trublesome exception. For any integer $n\geq 2$ the only unit of the $\Finfield_2$-algebra $\Finfield_2^n$ is its identity element $1_A$, and $\Uspan(A)=\set{0,1_A}\neq A.$

(B) $~$Every algebraic algebra $A$ over a field $F$ of at least three elements is spanned by units.

Proof. $~$For every $a\in A$ the finite-dimensional algebra $F[a]$ is spanned by units.$~$ Done.

Questions. $~$ Results (A) and (B) describe two classes of rings spanned by units. Are there any other (rich enough) classes of rings spanned by units, perhaps already known, perhaps newly discovered by you? Is there a structural characterization of rings spanned by units? (Very probably this is too much to hope for.) Can you give some interesting properties of rings spanned by units, or of some special kinds of such rings? Do you know, or see, any useful applications of the property of a ring being spanned by units? (I know of one such application. In fact I started exploring rings spanned by units because I needed this property to finish the proof of a certain hefty theorem.)

• See also here. – Bill Dubuque Sep 10 '20 at 23:15

The study of rings (additively) generated by units is quite old, going back at least to the 1950's. The terminologies that you should look for include S-ring and k-good ring.

In case you haven't seen it, I can recommend Srivastava's A survey of rings generated by units as a good starting point to find out what is known. It will give you many good references, such as Raphael's Rings which are generated by their units and Henriksen's Two classes of rings generated by their units.

There are a lot of interesting things to say, especially along the lines of rings whose elements are sums of $n$ units. A $k$-good ring is one in which every element is a sum of $k$ units. Henriksen showned that every ring is Morita equivalent to a $3$-good ring.

For example, every linear transformation of a vector space over a division ring is a sum of two invertible transformations, excepting the case of a $1$ dimensional space over $F_2$.

From Srivastava's article:

Let X be a completely regular Hausdorff space. Then every element in the ring of real-valued continuous functions on X is the sum of two units [18]. Every element in a real or complex Banach algebra is the sum of two units [22].

Since rings of linear transformations have this nice proprety, it's natural to ask about von Neumann regular rings. It turns out, though, that it's possible for an element of a VNR ring to not be expressible as a sum of units. But not all is lost: all unit-regular VNR rings in which $2$ is a unit are S-rings (in fact, they are $2$-good!), and some self-injective VNR rings are S-rings.

I think that there is not much to say about a structural characterization of such rings, though. As you've noted, the condition does not pass meaningfully through important constructions. If polynomial rings aren't ever S-rings, and every ring is Morita equivalent to an S-ring, we have not learned much about structure.

• Have downloaded the three papers you suggested. Looking forward to the fun of reading them. – chizhek Aug 7 '14 at 12:52
• @1947chizek Enjoy :) – rschwieb Aug 7 '14 at 12:53