Additive group functor preserving limits and colimits

Question

I have a homework question that goes as follows:

Let $A: Rng \to Gp$ denote the "additive group" functor, i.e. Taking a ring $R=(R,+,\cdot, 0,1)$ to its underlying additive group and taking a ring homomorphism $f: R\to S$ to the same map $f$, views as a group homomorphism $(R,+,0)\to (S,+,0)$. Does $A$ preserve limits? Does $A$ preserve colimits? Give a proof or counter example.

I think it preserves limits. I use the reasoning that $A$ preserves products and equalizers, so therefore it preserves limits. My problem is that it seems intuitively clear to me why this functor would preserve products and equalizers, it almost seems "trivial" with this functor, but is it really? Can someone help me see where the non-triviality comes in?

And then my question with colimits is, do we just show that A preserves pushouts and coequalizers? My intuition would then lead me to believe that it would preserve these things, but if it does not can you give me some reasoning to help me see why it wouldn't preserve these things?

Thanks!!!

Answer 1

It might be easier to just use the explicit description of limits and co-limits. It is a general result in universal algebra in any category of generalized algebras (e.g. rings or groups) that the limit of a diagram $I$ is the subset of the product of the objects of $I$ satisfying the functional equations of the arrows of $I$. Similarly, the co-limit of $I$ is the quotient of the co-product of the objects of $I$, the equivalence relations are given by relations generated by the arrows of $I$.

Apply it to the case of rings and groups. The product (resp. coproduct) of rings is direct product (resp. tensor product), and the product (resp. co-product) of groups is the direct product (resp. free product). However, since the underlying group of a ring is abelian, for your case free product of the groups in the image of the additive functor is the direct sum. I think it is routine from this point onwards to study how do coherence relations demanded by the arrows of a diagram $I$ behave under this description. Actually, it is immediate that the additive functor preserves limits.

Furthermore, the discussion above also immediately shows the additive functor does not preserve colimits or even finite coproducts as $\mathbb Z\otimes \mathbb Z=\mathbb Z$ as rings while $\mathbb Z\oplus \mathbb Z\neq \mathbb Z$ as groups.

Answer 2

The functor $\mathcal{A}:\mathbf{Ring}\to\mathbf{Ab}$ has a left adjoint (exercise: what is it?), therefore it preserves limits. As Victor Zhang states, $\mathcal{A}$ doesn't preserve coproducts. Another counter-example in addition to the one given there is $\mathbb{Q}\otimes\mathbb{Z}_2 \cong 0$ as rings while $\mathbb{Q}\oplus\mathbb{Z}_2$ is certainly non-trivial as an commutative group.