# Universal constructions throughout mathematics

Here are the notes I wrote for the talk I am giving at the Canadian Undergraduate Mathematics Conference.

My talk is meant as a friendly introduction to category theory and to the notion of a universal arrow.

Edit: corrected a couple of mistakes in my original document. If you find any more, please let me know.

# Multiplicative sets

This problem is taken from Atiyah-Macdonald.

A multiplicatively closed subset $S$ of a ring $A$ is said to be saturated if

$xy \in S \Leftrightarrow x \in S \mbox{ and } y \in S$.

Prove that

(i) $S$ is saturated $\Leftrightarrow A-S$ is a union of prime ideals.

(ii) If $S$ is any multiplicatively closed subset of $A$, there is a unique smallest saturated multiplicatively closed subset $\overline{S}$ containing $S$, and that $\overline{S}$ is the complement in $A$ of the union of the prime ideals which do not meet $S$. ($\overline{S}$ is called the saturation of $S$.)

Solution:

(i) This is trivial if $0 \in S$, since then $S=A$ and $A$ is a union of prime ideals. So suppose otherwise. Let $a \in A-S$; we will show that $a$ is contained in a prime ideal contained in $A-S$. Let $\Sigma$ denote the collection of ideals containing $(a)$ and contained in $A-S$, partially ordered by inclusion. $\Sigma$ is nonempty since $(a)$ \in $\Sigma$; indeed, since $a \notin S$ and $S$ is saturated, $ca \notin S$ for every $c \in A$. Moreover, it is clear that every chain in $\Sigma$ has an upper bound — simply take the union of all ideals in the chain. So, by Zorn’s theorem, $\Sigma$ has a maximal element $P$. We will show that $P$ is prime. Suppose otherwise; then there exist $u, v \in A-P$ such that $uv \in P$. The ideals $(P, u)$ and $(P, v)$ strictly contain $P$, so they do not lie in $\Sigma$ by the maximality of $P$. So there exists $s \in S$ such that $s=m+ux$, with $m \in P$, and $s' \in S$ such that $s'=m'+ux'$, where $m' \in P$. Multiplying these two equations, we have $ss'=mm'+mux'+m'vx+uvxx'$. This lies in $P$ since $uv \in P$, and it lies in $S$ since $S$ is multiplicatively closed; this is a contradiction. Thus $P$ is a prime ideal.

(ii) Let $\overline{S}$ denote the intersection of all saturated, multiplicatively closed subsets of $A$ containing $S$. This is nonempty since for example $A$ is such a set. It is easy to see that $\overline{S}$ is saturated and multiplicatively closed, and clearly it is the smallest saturated, multiplicatively closed subset of $A$ containing $S$. Since $A\backslash \overline{S}$ is a union of prime ideals by (i), and since $A \backslash \overline{S} \subset A \backslash S$, we see that  $A \backslash \overline{S}$ is a union of prime ideals contained in $A \backslash S$, i.e. a union of prime ideals not intersecting $S$. Since $\overline{S}$ is minimal, $A\backslash \overline{S}$ must in fact be the union of all such prime ideals.