## Comparison of group and ring theory

Yet again, I’m unable to sleep.

Is there any notion (which we could make reasonably precise) of an isomorphism between the roles played by normal subgroups in group theory and ideals in ring theory? Given any subset $S$ of a group (or ring, vector space, module, algebra, or other enrichment of group structure) $A$, we want to be able to talk about the set of equivalence classes $A/S$ modulo the equivalence relation $\sim$ defined by saying that $x \sim y$ if and only if $x-y \in S$. What must $S$ be like if we want $\sim$ to be an equivalence relation? Well, reflexivity occurs if and only if the identity element (that is, $0$) is in $S$; symmetry is related to whether $S$ is closed under negation; transitivity is related to whether $S$ is closed under addition.

The conditions above are what is required for this thing to even make sense, that is, for $\sim$ to even be an equivalence relation and therefore allow us to talk about equivalence classes. However, what we’ve constructed is merely a set at this point, and this is not so satisfactory. We want to endow this thing $A/S$ with the same kind of algebraic structure as $A$ (note: here I do not mean we want to turn $A/S$ into something isomorphic to $A$, but rather we want to associate to the set $A/S$ an object in the same category as $A$), in some natural way. In order to do this, we have to define algebraic operations on cosets. Considerations of well-definedness lead us directly to the notions of “normal subgroup”, “ideal”, “subspace”, and so on. These are essentially “subobjects that are well-behaved enough to be quotiented out”.

The reason one should care about the details of these “special” subobjects is clear: they are intricately related to structure-preserving transformations between algebraic structures, namely homomorphisms. Note that since vector spaces are essentially nothing more than abelian groups equipped with field actions (I haven’t really heard the latter term used, but it’s the best way I can describe what scalar multiplication really is), linear algebra is sculpted in a profound way by abelian group theory (i.e. every subspace is in particular a subgroup, which is automatically normal due to the abelianness). Because of the ubiquitous first isomorphism theorem (which is a rather simple observation with important consequences), an understanding of subobjects and quotient objects leads directly to an understanding of all morphisms between two algebraic structures. For example, we know that the kernel of any group homomorphism is a normal subgroup; the kernel of any ring homomorphism is an ideal; the kernel of a linear mapping is a subspace, and so on.

To elaborate on my previous remark, let $G$ and $H$ be groups with $\varphi : G \to H$ a homomorphism. Observe that FIT says that if $\pi : G \to G/\ker \varphi$ denotes the canonical projection, then there exists $\tilde{\varphi} : G/\ker \varphi \to H$ such that $\tilde{\varphi} \circ \pi = \varphi$. The map $\pi$ is surjective for trivial reasons, while certainly $\tilde{\varphi}$ is injective since we already “crushed” the kernel of $\varphi$ to a single element in the quotient group. Since any map is surjective onto its own image, we see that $\tilde{\varphi}$ gives us the isomorphism $\mathrm{im}(\varphi) \cong G/\ker \varphi$.

So what’s the general strategy if we want to hunt down all the homomorphisms between two algebraic structures $A_1$ and $A_2$? Well, you know that the possible kernels are just the “well-behaved subobjects” (normal subgroups, ideals, etc. as is suitable), so you try to list them out. Once you’ve pinpointed all of these, you can use the decomposition provided to you by FIT to reduce your problem to something easier, namely, you can work out what the quotient objects $A_1/N$ (where $N$ is one of the aforementioned subobjects) look like. Once you’ve done this, simply consider how many ways you can embed these quotients into $A_2$. A good start would probably be to ask how many subobjects of $A_2$ might be isomorphic to each quotient. However, one must be careful to note that this question will not fully answer the question of how many embeddings there are, although it will give a lower bound. Suppose $A_1/N \cong A$ for some subobject $A$ of $A_2$. The issue here is subtle, and one might fall into the trap of thinking that there is only one possible choice of embedding $\tilde{\varphi} : A_1/N \to A_2$ with $\mathrm{im}(\tilde{\varphi}) = A$, however, such a uniqueness is equivalent to the statement that $A$ admits no nontrivial automorphisms, which is definitely not true in general! This is why people often speak of things being “unique up to a unique isomorphism” rather than “unique up to isomorphism”; just because two objects are isomorphic does not imply that there is only one isomorphism between them.

Anyways, I hope to go into more detail on isomorphism theorems and other topics that I’ve been thinking about. Hopefully after enough time passes, I will have channeled my thoughts in a sufficiently coherent manner to cough up an insightful lecture or two. On the other hand, I might just end up talking about algebraic geometry instead… 🙄

(why the hell is there always construction outside my apartment at 4:00 am?)