Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently I ran across a weird example of a category in Jacobson's Basic Algebra II.

The category has, as objects, the class of rings. As morphisms, it uses all ring homomorphisms and antihomomorphisms of these rings.

Has anyone seen a use for this category?

I have the sense that it isn't well behaved, and so it might only be useful as a counterexample.

For example, it seems like products don't work. I didn't verify any details, but if you suppose there are three noncommutative rings $R$ and $S$ and $T$ for which there is a homomorphism of $R$ into $T$ and an anti homomorphism of $S$ into $T$, it seems like a product morphism from "$R\oplus S$" to $T$ is unlikely to exist in general.

Of course, I may just be blinded by familiarity with nice categories, so maybe there is a way around it...

Added I may in fact mean the coproduct and not the product. I never remember which is the messy one, for rings. Anyhow, the idea is that if you use the normal Cartesian product with coordinatewise ring product, it doesn't seem possible for a single product/coproduct morphism to combine a homomorphism with an antihomomorphism.

share|improve this question
I'm confused by your "product morphism": that would rather be a coproduct, no? The coproduct of rings is already nasty in the ordinary category of rings. –  t.b. Sep 11 '12 at 16:40
I think he just means composition of maps, not "product" in the sense of category theory, @t.b. –  Thomas Andrews Sep 11 '12 at 16:51
@ThomasAndrews No... it's pretty clear there's no issue with compositon of morphisms... –  rschwieb Sep 11 '12 at 17:04
@t.b. I think you're right... I always forget which one is the messy one. –  rschwieb Sep 11 '12 at 17:05
It doesn't help that I somehow left out a critical negating word... –  rschwieb Sep 11 '12 at 17:12

2 Answers 2

up vote 3 down vote accepted

This is useful is you would like a Hopf algebra to be a group object in the category of algebras. If $A$ is a Hopf algebra, then the antipode map $S: A \to A$, is an antiendomorphism of $A$. You can see this from the example of group algebras: if $G$ is a group, and $k[G]$ is the group algebra, then $k[G]$ has a Hopf algebra structure where $S: g \mapsto g^{-1}$ and we have $(gh)^{-1} = h^{-1} g^{-1}$.

I don't know that it's worth defining a category just to fix this issue. I have seen people spend far more time bothered by this issue than it is worth when learning Hopf algebras, so maybe this would have helped them.

share|improve this answer
Oo, good lead... thank you. –  rschwieb Sep 11 '12 at 18:19
I assume that you mean a group object in the nonstandard category of algebras where also anti-homomorphisms are allowed? But this probably has no products, and Hopf algebras are far from being group objects, since the multiplication is not defined on a product. Cocommutative Hopf algebras are group objects in the category of cocommutative coalgebras (here $\otimes=\times$). –  Martin Brandenburg Jun 6 at 20:48
@MartinBrandenburg You are right. I can save my answer some by pointing out that I was right that antipode is a map of rings in this peculiar category, but you are right that $\otimes$ is not product here. –  David Speyer Jun 7 at 0:09
That doesn't mean that the said category is useful. It only says that there are antihomomorphisms, which is clear anyway. –  Martin Brandenburg Jun 7 at 7:18

I also think that this category is kind of pathological, because it probably lacks limits and colimits. Perhaps a little bit more natural is the category with rings as objects and where the morphisms are either homomorphisms or antihomomorphisms, in the sense that we take the disjoint union of these sets (thus, a homomorphism of, say, commutative rings is distinguished from the corresponding antihomomorphism). Let me explain how this category can be constructed from a general method.

Let $\mathcal{C}$ be a category. Let $D : \mathcal{C} \to \mathcal{C}$ be a functor with $D \circ D = \mathrm{id}$. For $\mathcal{C}=\mathsf{Mon},\,\mathsf{Grp},\,\mathsf{Ring}$ or $\mathsf{Cat}$ we take $D$ to be the "opposite-object" functor. Define a new category $\mathcal{C}_D$ as follows: The objects are the objects of $\mathcal{C}$. For objects $x,y$ we define $$\hom_{\mathcal{C}_D}(x,y) = \hom_{\mathcal{C}}(x,y) \sqcup \hom_{\mathcal{C}}(x,D(y)).$$ The identity of an object in $\mathcal{C}_D$ is the identity in $\mathcal{C}$. The composition of $x \to y$ with $y \to z$ in $\mathcal{C}_D$ is the one in $\mathcal{C}$. The composition of $x \to y$ with $y \to D(z)$ is the composition $x \to D(z)$ in $\mathcal{C}$, regarded as an element in the second part of $\hom_{\mathcal{C}_D}(x,z)$. The composition of $x \to D(y)$ with $y \to z$ is the composition $x \to D(y) \to D(z)$ in $\mathcal{C}$, again regarded as an element in the second part of $\hom_{\mathcal{C}_D}(x,z)$. Finally, the composition of $x \to D(y)$ with $y \to D(z)$ is the composition $x \to D(y) \to D(D(z)) = z$, now regarded as an element in the first part of $\hom_{\mathcal{C}_D}(x,z)$. One checks that $\mathcal{C}_D$ is, indeed, a category.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.