# Can $\mathbf{Rng}$ be made into a pre-additive category?

Question: Can $$\mathbf{Rng}$$ be made into a pre-additive category?

Rngs are just rings, without the requirement of an identity. Accordingly, we do not require rng homomorphisms to preserve the identity (should it exist). These objects (rngs) and morphisms (rng homomorphisms) form the category $$\mathbf{Rng}$$.

Wikipedia says that:

Despite the existence of zero morphisms, Rng is still not a pre-additive category.

For a category to be pre-additive, we require a zero object, and an abelian group structure on the hom-sets. The pointwise sum of two rng homomorphisms is generally not a rng homomorphism - so certainly, $$\mathbf{Rng}$$ is not a pre-additive category with the obvious choice of addition (of homomorphisms). However, how do we know that $$\mathbf{Rng}$$ cannot be made into a pre-additive category, possibly by choosing some other addition operation on the hom-sets? All we need is an abelian group structure, and there are no restrictions on the group operation.

Thanks a lot!

• Doesn't a pre-additive category have to have the composition respect the hom addition in a distributive way? I would think that would be the obstruction. So, there are some restrictions on the group operation. Feb 18 at 13:52
• That's right, there should be some restrictions on the choice of the group operation - but it's hard to point out exactly what they are. @Randall Feb 18 at 13:57
• You can't just "make the hom-sets" the trivial group. You can't choose the hom-sets, and they have more than $1$ element in general. Feb 18 at 14:08
• But that is very clearly not a group operation. Feb 18 at 14:22
• The product of two rngs $A$ and $B$ is $A \times B$. Is this also the coproduct (with the injections $a \mapsto (a,0)$ and $b \mapsto (0,b)$)? Feb 18 at 14:45

In a preadditive category, any finite product or coproduct is a biproduct (see here for a proof). This is not true in $$\mathbf{Rng}$$: the product of two rngs $$A$$ and $$B$$ is just the usual cartesian product $$A\times B$$, but this is typically not a coproduct under the inclusion maps $$i:A\to A\times B$$ and $$j:B\to A\times B$$ (which are the identity on one coordinate and zero on the other). In particular, these inclusions satisfy $$i(a)j(b)=0$$ for all $$a\in A,b\in B$$, so if they satisfied the universal property of the coproduct, the same would have to be true of any rng $$C$$ with homomorphisms from $$A$$ and $$B$$. This is clearly false in general--for instance, if $$A=B=C$$ is a rng with nonzero multiplication, you could take the identity homomorphisms $$f:A\to C$$ and $$g:B\to C$$ and these do not satisfy $$f(a)g(b)=0$$ for all $$a\in A,b\in B$$.