Sketch 3

Strategies for $\text{Hom}$ Computations

Huh...

Right now it feels like I kind of suck at computing $\text{Hom}$ between two sets. We've had a few problems like this on our problem sets for algebra, and my go to methods have always been what I think is naive and not super fun: just looking at what elements could map where and figuring out what restrictions there might be on the maps. In this sketch I want to explore the usage of exact sequences, tensor products, and category theory for discovering $\text{Hom}$ sets (or modules or etc. etc. in enriched structures).

Date Started: February 29, 2024
Date Finished: Unfinished

My Ideas

$\mathsf{Grp}$

I'll start with the category of groups. I'll start with some simple examples, which I have even previously computed in previous sketches. Suppose we would like to compute $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$. Many people would find this one to be quite simple: any homomorphism out of $\mathbb{Z}$ is completely determined by its value at one, since $\mathbb{Z}$ is generated by 1. In other words, if $\phi \in \mathrm{Hom}(\mathbb{Z},\mathbb{Z})$, then $\phi (k)=k\phi (1)$ (here I'm implicitely utilizing the $\mathbb{Z}$-module structure on $\mathbb{Z}$ since it is an abelian group). Notice that in general, if we have fixed the value of $\phi$ mapping out of a group $G$ on a subset $S\subseteq G$ in a consistent way, the values of $\phi$ are determined on the smallest subgroup of $G$ containing $S$. This comes down to the classic idea on how I approach finding hom sets:

Claim. If $G$ is generated by $\{g_{\alpha }\mid \alpha \in S\}$ for some indexing set $S$, then any morphism out of $G$ is determined by its values on the generating set.

After we know this, we can look at the image. In the case of $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$, we know that we simply have to determine the value of any homomorphism on $1$, and we can see that $1$ can take any value in $\mathbb{Z}$. Thus, $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$. Similarly, we can see that $\mathrm{Hom}(\mathbb{Z},\mathbb{Q})\cong \mathbb{Q}$ and $\mathrm{Hom}(\mathbb{Z},\mathbb{R})\cong \mathbb{R}$. So far so good. In fact, we can see that for any group $G$, $\boxed{{\displaystyle \mathrm{Hom}(\mathbb{Z},G)\cong G}}$ since for every element $g\in G$, we have a homomorphism mapping $1$ to $G$, so that $k$ maps to $g^k$. This is quite nice.

But things aren't as nice when we consider mapping into $\mathbb{Z}$. Consider $\phi \in \mathrm{Hom}(\mathbb{Q},\mathbb{Z})$. Again, notice that the values of $\phi$ are determined on $\mathbb{Z}\subseteq \mathbb{Q}$ simply by choosing $\phi (1)$, namely $\phi (a)=a\phi (1)$. However, this becomes a problem because we need to map $\phi (\frac{1}{n})$ to a value such that $n\phi (\frac{1}{n})=\phi (1)\in \mathbb{Z}$. On the other hand, if we choose $N\in \mathbb{Z}$ such that $N>\phi (1)$, there clearly does not exist an integer such that $N\phi (\frac{1}{N})=\phi (1)$, that is, unless $\phi (1)=0$, in which case we have that $\phi (\frac{1}{n})=0$ for all $n$, so $\phi$ is the zero map. As such, there is only one possible map, so $\boxed{{\displaystyle \mathrm{Hom}(\mathbb{Q},\mathbb{Z})=0}}$. Notice that in this case there existed a larger subgroup of $\mathbb{Q}$ (the one generated by $\frac{1}{N}$) which was more restrictive than the one we initially picked ($\mathbb{Z}$). From this there's one thing in particular that I would like to notice: $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$ is a superset of $\mathrm{Hom}(\mathbb{Q},\mathbb{Z})$ since $\mathbb{Z}$ is a subset of $\mathbb{Q}$; any homomorhism from $\mathbb{Q}$ to $\mathbb{Z}$ must also be a homomorphism when restricted to $\mathbb{Z}$.

Claim. There is no unique maximal element in the lattice of subgroups of $\mathbb{Q}$.

Proof Sketch.
Suppose for contradiction that there was a unique maximal element, $H\le \mathbb{Q}$. If $\frac{1}{p^k}\in H$ for all $p$ prime and $k\in \mathbb{Z}$, then $H=\mathbb{Q}$. Thus there must be some prime $p$ and $j\in \mathbb{Z}$ such that $\frac{1}{p^j}$ is not in $\mathbb{Q}$, so simply take the smallest subgroup contianing $H$ and $\frac{1}{p^j}$. This subgroup still doesn't contain $\frac{1}{p^{j+1}}$ so it is not all of $\mathbb{Q}$, but it is larger than $H$, a contradiction. $◻$

To go back to another idea, maybe the maybe this idea that $\mathrm{Hom}$ reverses inclusions in one component will end up motivating categorical limits or functoriality, I guess we'll have to see.

Notice that the same issue happens with $\mathbb{Z}[\frac{1}{p}]$ for any $p$, we have that $\boxed{{\displaystyle \mathrm{Hom}(\mathbb{Z}[\frac{1}{p}],\mathbb{Z})=0}}$. We might instead consider $\mathrm{Hom}(\mathbb{Z}[\frac{1}{n}],\mathbb{Z}[\frac{1}{m}])$ for $n$ and $m$ integers. This becomes a much more tedious problem. We can see that $\varphi (a)=a\varphi (1)$ again on the integers, and $n^k\varphi (\frac{1}{n^k})=\varphi (1)$, so we must have that $\varphi (1)$ is always "divisible" by $n^k$ for all $k$, or that $n$ is invertible in $\mathbb{Z}[\frac{1}{m}]$. $n$ being invertible in $\mathbb{Z}[\frac{1}{m}]$ would imply that $n^k$ divides some power of $m$, so the primes dividing $n$ also divide $m$. It is interesting to note that this condition is precisely that $V(n)\subseteq V(m)$ in $\text{Spec}(\mathbb{Z})$. In this case, regardless of where we map $\varphi (1)$, things work, so that $\boxed{{\displaystyle \mathrm{Hom}(\mathbb{Z}[\frac{1}{n}],\mathbb{Z}[\frac{1}{m}])=\mathbb{Z}[\frac{1}{m}]}}$.

$\mathsf{CRing}$

$\mathsf{Mod}(A)$