Subtraction-free identities that hold for rings but not for semirings?

Question

Here is a concrete, if seemingly unmotivated, aspect of the question I am interested in:

Question 1. Let $a$ and $b$ be two elements of a (noncommutative) semiring $R$ such that $1+a^3$ and $1+b^3$ and $\left(1+b\right)\left(1+a\right)$ are invertible. Does it follow that $1+a$ and $1+b$ are invertible as well?

The answer to this question is

• "yes" if $ab=ba$ (because in this case, $1+a$ is a left and right divisor of the invertible element $\left(1+b\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$).

• "yes" if $R$ is a ring (because in this case, $1+a$ is a left and right divisor of the invertible element $1+a^3 = \left(1+a\right)\left(1-a+a^2\right) = \left(1-a+a^2\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$).

• "yes" if $1+a$ is right-cancellable (because in this case, we can cancel $1+a$ from $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(\left(1+b\right)\left(1+a\right)\right) = 1+a$ to obtain $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = 1$, which shows that $1+a$ is invertible), and likewise if $1+b$ is left-cancellable.

I am struggling to find semirings that are sufficiently perverse to satisfy none of these cases and yet have $\left(1+b\right)\left(1+a\right)$ invertible. (It is easy to find cases where $1+a^3$ is invertible but $1+a$ is not; e.g., take $a = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ in the matrix semiring $\mathbb{N}^{2\times 2}$.)

The real question I'm trying to answer is the following (some hopefully reasonably clear handwaving included):

Question 2. Assume we are given an identity that involves only positive integers, addition, multiplication and taking reciprocals. For example, the identity can be $\left(a^{-1} + b^{-1}\right)^{-1} = a \left(a+b\right)^{-1} b$ or the positive Woodbury identity $\left(a+ucv\right)^{-1} + a^{-1}u \left(c^{-1} + va^{-1}u\right)^{-1} va^{-1} = a^{-1}$. Assume that this identity always holds when the variables are specialized to arbitrary elements of an arbitrary ring, assuming that all reciprocals appearing in it are well-defined. Is it then true that this identity also holds when the variables are specialized to arbitrary elements of an arbitrary semiring, assuming that all reciprocals appearing in it are well-defined?

There is a natural case for "yes": After all, the same claim holds for commutative semirings, because in this case, it is possible to get rid of all reciprocals in the identity by bringing all fractions to a common denominator and then cross-multiplying with these denominators. However, this strategy doesn't work for noncommutative semirings (and even simple-looking equalities of the form $ab^{-1} = cd^{-1}$ cannot be brought to a reciprocal-free form, if I am not mistaken). Question 1 is the instance of Question 2 for the identity \begin{align} \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \end{align} (where, of course, the only purpose of the $\left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1}$ factors is to require the invertibility of $1+a^3$ and $1+b^3$). Indeed, if $\alpha$ and $\beta$ are two elements of a monoid such that $\beta\alpha$ is invertible, then we have the chain of equivalences \begin{align} \left(\alpha\text{ is invertible} \right) \iff \left(\beta\text{ is invertible} \right) \iff \left( \alpha \left(\beta\alpha\right)^{-1} \beta = 1 \right) \end{align} (easy exercise).

Answer 1

Tim Campion's idea works, though his example needs a little fixing. As in Tim's answer, we will find a rig with two elements $X$ and $Y$ such that $X+Y=1$ but $XY \neq YX$.

Let $(M,+,0)$ be any commutative monoid. Let $R$ be the set of endomorphisms of $M$ obeying $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(0)=0$. Then $R$ is a rig, with $(\alpha+\beta)(x) = \alpha(x) + \beta(x)$ and $(\alpha \beta)(x) = \alpha(\beta(x))$.

Let $M$ be $\{ 0,1,2 \}$ with $x+y \mathrel{:=} \max(x,y)$. Define \begin{gather*} \alpha(0) = 0,\ \alpha(1) = 0,\ \alpha(2) = 2 \\ \beta(0) = 0,\ \beta(1) = 1,\ \beta(2) = 1. \end{gather*} Then $\alpha+\beta=\mathrm{Id}$ but $\alpha \beta \neq \beta \alpha$.

Answer 2

The answer to your first question is yes (which was very surprising to me, to be honest). I have no idea whether the second question also has a positive answer. (By the way, don't let the work below fool you. This took me an entire week of serious computations to discover the main idea.)

We will assume $(1+a^3)u=1$ and $d(1+a)=1$. We find that $$ d+au = d(1+(1+a)au) = d((1+a^3)u+(a+a^2)u)=d(1+a)(1+a^2)u=(1+a^2)u. $$ Thus, we compute $$ (1+a)d = d[1+(1+a)ad] = d[1+ad+a^2d] = d[u+a^3u+a^2d+ad] $$ $$ =d[a^2(d+au) + ad + u] = d[a^2(1+a^2)u+ad+u] = d[a(d+au)+a^4u+u] $$ $$ =d[a(1+a^2)u+(1+a^4)u] = d(1+a)(1+a^3)u=1. $$

---

Edited to add: A similar idea works for higher odd powers. The fifth power case is sufficient to give the main idea.

Assume $(1+a^5)u=1=d(1+a)$. We find $$ d+(a+a^3)u = d[(1+a^5)u + (1+a)(a+a^3)u] = d(1+a)(1+a^2+a^4)u=(1+a^2+a^4)u. $$ Then we compute $$ (1+a)d=d^3[(1+a)^2+(1+a)^3ad] = d^3[(1+a)^2(1+a^5)u + (a+3a^2+3a^3+a^4)d] = d^3[u+2au+a^2u+2a^6u + (a+3a^2+3a^3)d + a^4[(a+a^3)u+d]] = \cdots $$ and you keep reducing monomials with $d$ to monomials involving only $a$ and $u$.

Answer 3

The answer to the second question is no in general.

For instance, in an associative ring, the elements $x(x+y)^{-1}$ and $y(x+y)^{-1}$ necessarily commute — in other words, if $a + b = 1$, then $a$ and $b$ commute (since we have $b = 1-a$, so $ab = a(1-a) = a - a^2 = (1-a)a = ba$).

This need not be so in an an associative semiring. For instance, consider the join-semilattice $S$ of subsets of $\{x,y\}$ (so $S$ is a 4-element idempotent commutative monoid), and let $R$ be the semiring of $(\emptyset, \cup)$-preserving endomorphisms of $S$. Let $X \in R$ fix $x$ and carry $y$ to $\emptyset$; let $Y \in R$ fix $y$ and carry $x$ to $\emptyset$. Then $X + Y = 1$ but $X$ and $Y$ do not commute see David Speyer’s answer for a counterexample.

(One way of looking this is that we're asking whether every localization of a finitely-generated free noncommutative semiring $\mathbb N\{x_1,\dotsc, x_n\}(f(x_1,\dotsc,x_n)^{-1})$ injects into its group completion (i.e. is additively cancellative) — the above example shows that the the answer is no when we look at $\mathbb N\{x,y\}(x+y)^{-1}$.)

(By the way — I was initially trying to prove that the answer to (2) was yes. The localization $\mathbb N\{x,y\}(x+y)^{-1}$ was the first one I tried after monomial localizations. Since (2) already fails there, I suspect that the phenomenon is widespread, making the positive answer to (1) all the more surprising!)