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The category $\mathbf{Grp}$ of groups has a zero object, namely the trivial group $1$. Since $\mathbf{Grp}$ is furthermore complete, we have the notion of a kernel of a group homomorphism. The kernel of a homomorphism $G\xrightarrow{\ f\ }H$ is a monomorphism $\operatorname{Ker}f\xrightarrow{\ker f}G$, so $\operatorname{Ker}f$ can be interpreted as a subgroup of $G$ with underlying set a subset of $G$. Namely, $\operatorname{Ker}f=\{x\in G\mid fx=1\}$.

In the category $\mathbf{Ring}$ of rings, we don't have a zero object and thus there is no natural definition of a kernel. Anyway, we define the kernel of a ring homomorphism $R\xrightarrow{\ f\ }S$ as the set $\operatorname{Ker}f=\{x\in R\mid fx=0\}$, analogous to the case of (additive) groups. However, it is not clear to me, what kind of object this is, i.e. to which category it belongs.

It seems wrong to think of it as a set, because it does have extra structure, e.g. as an (abelian) group. So maybe one should define the kernel of a ring homomorphism as an abelian group (the kernel of the underlying group homomorphism). But also this seems a little arbitrary, because one can easily define Rings and kernels of such without ever coming across abelian groups. Also, this doesn't give rise to a one-to-one correspondence, since not every kernel of a homomorphism of underlying abelian groups gives rise to a kernel of a ring homomorphism.

So, is it possible to define a kernel of a ring homomorphism without "leaving" the category of rings?

The same problem comes with all ideals. Is an ideal a set? A group? An abelian group? A module? A non-unitary ring? If one indeed defines left-ideals of $R$ as submodules of $R$, viewed as a left-module over itself, a right-ideal as a left-ideal of $R^{\operatorname{op}}$ and a two-sided ideal as a left-and-right ideal, how does this intuitively give rise to the fact that

two-sided-ideal$\iff$kernel of underlying group homomorphism of some ring homomorphism?

I hope you can understand the trouble I have.

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  • $\begingroup$ Have a look into universal algebra :) $\endgroup$ – Shaun Jun 22 '14 at 15:09
  • $\begingroup$ What if you workrd with non unital rings? The zero ring is a zero object. $\endgroup$ – Olivier Bégassat Jun 22 '14 at 15:44
  • $\begingroup$ @OlivierBégassat: One could do so, but this does not yield a qualitative difference to abelian or non-abelian groups. I have forgetful functors from $\mathbf{Ring}$ to both $\mathbf{Ab}$ and $\mathbf{Rng}$. But why take a ring homomorphism, consider its image under the (non-full) forgetful functor into one of these categories and consider the kernel of this new object of some other category? This reasoning doesn't make sense to me. $\endgroup$ – Jakob Werner Jun 22 '14 at 16:05
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    $\begingroup$ The proper thing to look at in the non-additive context is the kernel pair of a morphism. But in any theory of algebraic structures with a Mal'cev operation, such as anything extending the theory of groups, the kernel pair contains no more information than the kernel. $\endgroup$ – Zhen Lin Jun 22 '14 at 16:52
  • $\begingroup$ @NikolajK: Rhineland-Palatinate, why? $\endgroup$ – Jakob Werner Jul 23 '14 at 10:54
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A left ideal of a ring $R$ is just a left $R$-submodule of $R$. Thus, the natural category in which left ideals live is the category of left modules. The latter has a zero object and kernels can be computed as usual. If $f : R \to S$ is a homomorphism of rings, you can view $S$ as an left $R$-module via $r s := f(r) s$, so that $f$ becomes a homomorphism of left $R$-modules and the kernel of this homomorphism (in the category of left $R$-modules) is what you usually call the kernel of the ring homomorphism of $f$. Similarly we can deal with right ideals. And for two-sided ideals we work with the category of $(R,R)$-bimodules and observe that every ring homomorphism $R \to S$ induces an $(R,R)$-bimodule structure on $S$. By the way, all these definitions also work when we replace $\mathsf{Ab}$ by any abelian tensor category. For example, (quasi-coherent) ideal sheaves also fit into this picture.

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  • $\begingroup$ I guess I should really start learning to work with bimodules. $\endgroup$ – Jakob Werner Jun 23 '14 at 8:10
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Yes. Instead of taking the kernel you can take the kernel pair. In general, the kernel pair of a morphism is an attempt to recover a universal equivalence relation compatible with that morphism. If $f : R \to S$ is a morphism, then the kernel pair of $f$ is the pullback of the diagram $R \xrightarrow{f} S \xleftarrow{f} R$. In the case of rings, if $f$ has kernel $I$ then this is the ring

$$\{ (r_1, r_2) \in R \times R : r_1 \equiv r_2 \bmod I \}.$$

This sort of object is called a congruence in universal algebra, and should be thought of as an equivalence relation internal to the category of rings. One way to state one of the isomorphism theorems is that $f$ is surjective iff it is the coequalizer of the two projections from its kernel pair to $R$, or equivalently iff it is an effective epimorphism.

For groups, abelian groups, and rings, the ability to take inverses (in the third case, for addition) turns out to imply that you can replace the study of kernel pairs with the study of kernels (in the third case, at the price of leaving the category, as you noticed). But when you can't do this you really need to look at kernel pairs; for example if you start studying monoids or semirings.

See this blog post for a non-categorical introduction to the idea of internal equivalence relations and these three posts for variations on the relationship between kernel and cokernel pairs and various flavors of monomorphism and epimorphism. For example, with no hypotheses on the category, a morphism is a monomorphism iff its kernel pair exists and is trivial and, dually, an epimorphism iff its cokernel pair (the kernel pair in the opposite category) exists and is trivial.

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    $\begingroup$ Isn't the equalizer of $ f $ with itself the identity on $\operatorname {dom} f $? $\endgroup$ – Jakob Werner Jun 22 '14 at 17:52
  • $\begingroup$ @Jakob: sorry, I misspoke. The kernel pair is a pullback, not an equalizer. $\endgroup$ – Qiaochu Yuan Jun 22 '14 at 17:54
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Viewing rings as the commutative algebra objects of a symmetric monoidal category seems to make the setting more amenable to abstraction (which seems to be the goal here). So do that: $\mathbf{CRing} = \mathbf{CAlg}(\mathbf{Ab})$, where $\mathbf{Ab}$ is the symmetric monoidal category of abelian groups.

Hovey defines a Smith ideal (in https://arxiv.org/abs/1401.2850) as a monoid (in the pushout product monoidal structure, see paper) in the arrow category $\mathrm{Arr}(\mathcal{C})$ for a monoidal category $(\mathcal{C},\otimes,\mathbb{1})$.

So a Smith ideal for $\mathbf{Ab}$ ends up being an ideal inclusion $(j : I \to R) \in \mathrm{Arr}(\mathcal{C})$

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