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Welcome to my first MathOverflow posting!

This is a question about rings, all of them assumed to be both unital and associative. Let $f\colon R\to S$ be a map between rings such that $f(xy)=f(x)f(y)$ and $f(1+x)=1+f(x)$ for any $x,y\in R$. (This easily implies that $f(0)=0$ and $f(1)=1$.) Is $f$ additive, that is, does $f(x+y)=f(x)+f(y)$ for all $x,y\in R$?

In my paper Is addition definable from multiplication and successor? (can also be downloaded from the HAL preprint server), I verified that the answer to that question is positive in many cases, including the following:

  1. $R$ is finite;
  2. $R$ is commutative (or even a ring of $2\times2$ matrices over a commutative ring);
  3. $R$ is a full matrix ring, of order $>1$, over any ring;
  4. $f$ is the power function $x\mapsto x^n$ over $R$;
  5. $f$ is the determinant function on a ring of $3\times3$ matrices over a commutative ring...

and many other cases.

The answer to the question above is positive iff ring-theoretical addition has a positive existential (resp., positive primitive) definition from multiplication and successor. I have been unable to settle this problem, in any direction. My (modest) computer attempts suggest that there should be a counterexample lurking somewhere, perhaps even in the existing ring theory literature.

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    $\begingroup$ The question mathoverflow.net/q/426302/88133 has comments and answers that seem relevant. $\endgroup$
    – Zach Teitler
    Commented Sep 4 at 13:06
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    $\begingroup$ Have you tried the tropical ring? $\endgroup$
    – Andrej Bauer
    Commented Sep 4 at 14:02
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    $\begingroup$ About the tropical (semi) ring: in Section 11.1 of my paper I briefly discuss the example 0<a<b<1, with addition defined as maximum and multiplication zero unless of the factors is 1. Interchanging a and b yields a counterexample. However, there is no such example for finite cancellative semirings! $\endgroup$
    – Fred Wehrung
    Commented Sep 4 at 14:21
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    $\begingroup$ I'm a little confused by (4) and (5), which usually don't satisfy your hypotheses. Do you mean that, if they satisfy your hypotheses, then they satisfy the conclusion? $\endgroup$
    – LSpice
    Commented Sep 4 at 14:31
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    $\begingroup$ @James No I don't have such an axiomatization! In my paper I call "brachynomials" the terms of that language --- that is, finite compositions of successor and multiplication. I don't know whether there are brachynomials $p$ and $q$ in three variables, representing distinct polynomials, such that $p(x,y,x+y)=q(x,y,x+y)$ everywhere. $\endgroup$
    – Fred Wehrung
    Commented Sep 4 at 16:26

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