What are such "off-by-a-factor" ring maps called?

Suppose $$R,S$$ are (non-unital) rings. What is the term for a function $$f:R\rightarrow S$$ such that

• $$f(a+b)=f(a)+f(b)$$ for all $$a,b\in R$$, and

• there is some $$u\in S$$ such that for all $$a,b\in R$$ we have $$uf(ab)=f(a)f(b)$$?

I'm currently calling these "weak homomorphisms" (and "weak embeddings" in the injective case), but I suspect they have an actual name.

For example, let $$R=\mathbb{Q}$$ and let $$S$$ be the ring of polynomials with rational coefficients whose constant terms are integers. The map $$R\rightarrow S:q\mapsto qx$$ is of course not a homomorphism, and indeed there is no homomorphism from $$R$$ to $$S$$, but it does satisfy the weaker property above via $$u=x$$.

(I'm primarily running into this notion in the context of certain models of Robinson arithmetic, the idea being that "sufficiently generic" such models admit weak embeddings from lots of rings and this leads to some interesting structural properties, but I'm also interested in them in other contexts - including non-unital ones.)

• Note that $u=f(1)$ which is more explicit. Also this notion turns up in the context of multiplicative functions. It would be nice to have a name for the "off-by-a-factor" version. Jun 20, 2020 at 13:06
• @Somos Oops, that was a typo - (very belatedly) fixed! Oct 7, 2021 at 19:01
• In recent years the notion of approximate groups has been developed. This seems to be somewhat an analogue/generalisation. Oct 7, 2021 at 19:06
• @TheSimpliFire I read the Wikipedia article on approximate groups. I don't see the connection because they are not defined in terms of functional equations. Oct 7, 2021 at 19:10
• @NoahSchweber - are these functions $f$ assumed to be maps of abelian groups? Oct 7, 2021 at 19:49

I don't think I know of a term, but I'm going to show two situations where these objects behave very differently. The first, which seems closest to the kind of situation you want, makes this object 'essentially a homomorphism'. The second means that it is no restriction at all. It all depends on the choice of $$u$$.
I'm going to assume that $$R$$ and $$S$$ are commutative, although it is just so I don't need to keep track of sidedness. Let $$R^2$$ denote the set of products $$r_1r_2$$, for $$r_i\in R$$. If $$X\subset R$$, let $$\overline{f(X)}$$ denote the subring generated by $$f(X)$$. (Note that $$\overline{f(R)}$$ need not be equal to $$f(R)$$...) We may assume that $$S=\overline{f(R)}$$ without loss of generality.
The first situation is where $$u$$ is not a zero divisor in $$S$$, in particular where $$S$$ is an integral domain. In the localized ring $$S'=S[u^{-1}]$$, $$u$$ is a unit. Let $$f':R\to S'$$ be given by $$f'(r)=u^{-1}f(r)$$. Then $$f'$$ is a ring homomorphism.
To see this, notice that $$f'(r_1+r_2)=u^{-1}(f(r_1)+f(r_2))=u^{-1}f(r_1)+u^{-1}f(r_2))=f'(r_1)+f'(r_2).$$ Secondly, $$f'(r_1)f'(r_2)=u^{-2}f(r_1)f(r_2)=u^{-1}f(r_1r_2)=f'(r_1r_2).$$ (Commutativity is needed here, or at least that $$u$$ is central.)
The opposite position is when $$u=0$$. Then, for any abelian group $$S$$ and homomorphism from $$(R,+)$$ to $$S$$, it extends to a map $$R\to S'$$ where $$S'$$ is the ring with set $$S$$ and multiplication $$0$$.