Universal properties in ring theory

Question

I am reading Hartley&Hawkes: Rings, Modules and Linear Algebra and I am having a little trouble making sense of universal properties. The two I am referring to are of direct sums and of polynomial rings. The statements are given below in questions 9 and 11 of Chapter 3:

(Universal property of direct sums.) Let $R_1, R_2$ be rings, let $R=R_1\oplus R_2$ (internal or external) and let $\pi _i: R\rightarrow R_i$ be the co-ordinate projections. Show that, given any ring $S$ and homomorphisms $\eta _i: S\rightarrow R_i$, there exists a unique homomorphism $\eta:S\rightarrow$ which makes the diagrams commute.Generalize.

(Universal property of polynomial rings.) Let $R$ be a commutative ring with 1. Show that given a commutative ring $S$, a homomorphism $\phi: R \rightarrow S$ and an element $a\in S$, there exists a unique homomorphism $\psi:R[x] \rightarrow S$ such that (i)$\psi (r)=\phi (r)$ for all $r\in R$, and (ii)$\psi (x)=a$. What happens if $R$ does not have a 1?

I think I am missing the big picture here.

First, what is meant by the universality of these properties?(I didn't take a course on Category Theory and I see that the notion of a universal property is defined in that context)

Second, later on when giving examples on module over a ring they gave an important example of turning a $k$-vector space $V$ into a $k[x]$-module by fixing some endomorphism $\alpha \in EndV$ and letting each polynomial $f(x)$ act on $v \in V$ as $fv=f(\alpha)(v)$. Beside proving this using directly the axioms of a module(which is clear), they also proved it using universal property of polynomial rings(referring to the exercise above) :

the map $f\rightarrow f(\alpha)$ is a ring homomorphism of $k[x]$ into $EndV$ and therefore makes V into a $k[x]$-module...

Can anybody clarify how?

Answer 1

There's probably one too many questions in your post, so I will address the primary one; your question about $k[x]$ modules might make a good post itself, if properly contextualized.

The general "meaning" of a universal property (if there is one) is that this object and these homomorphisms are very very special, amongst all such objects and homomorphisms. They sit there and they smile happily from high atop the universe of all such objects and homomorphisms, down upon all the others. And they have no competitor: if there happens to be a second object and homomorphisms sitting there and smiling happily from high atop the universe of all such objects and homomorphisms, when each of the two objects smiles upon each other, they recognize a fellow soul, because the two objects are isomorphic (which, in the mathematics, is tantamount to saying that the two are the same).

Let's take the direct sum for example.

Given two rings $R_1$, $R_2$, this object $R_1 \oplus R_2$, and these two homomorphisms $\pi_1 : R_1 \oplus R_2 \to R_1$ and $\pi_2 : R_1 \oplus R_2 \to R_2$, are universal amongst all such objects and homomorphisms, in the following precise mathematical sense. If you pick any other such object and homomorphisms, i.e. any other ring $S$ and two homomorphisms $\eta_1 : S \to R_1$ and $\eta_2 : S \to R_2$, then the following very strong conclusion holds: there exists a unique homomorphism $\eta : S \to R_1 \oplus R_2$ making all the diagrams commute.

And the object $R_1 \oplus R_2$ and its homomorphisms $\pi_1$ and $\pi_2$ have no competitor, in the following sense. Suppose that you have another ring $S'$ and a pair of homomorphisms $\pi'_1 : S' \to R_1$ and $\pi'_2 : S' \to R_2$ which satisfy the same universal property (namely, for any such ring $S$ and any pair of homomorphisms $\pi_1 : S \to R_1$ and $\pi_2 : S \to R_2$ there exists a unique homomorphism $\eta' : S \to S'$ which makes the diagrams commute). Then $S'$ is isomorphic to $R_1 \oplus R_2$.

If you haven't seen the proof of isomorphism of $S'$ and $R_1 \oplus R_2$, it's very instructive to work out, or even just to read. In outline:

• One lets $\eta : S' \mapsto R_1 \oplus R_2$ be the unique homomorphism arising from universality of $R_1 \oplus R_2$, $\pi_1$ and $\pi_2$;
• Next one lets $\eta' : R_1 \oplus R_2 \to S'$ be the unique homomorphism arising from universality of $S'$, $\pi'_1$ and $\pi'_2$;
• And now one proves that $\eta$ and $\eta'$ are inverse isomorphisms: their composition, either way, is the identity.

Each of the various little bits and pieces of the definition of universal properties is used in making this proof work.

Now, with this metaphor in mind, you might want to work through the universal property of polynomial rings in the same manner:

Answer 2

Considering what universal properties are in these concrete examples, it's exactly what's written in the images you provided. What the author did is write a proposition and named it "universal property of...". In general, universal properties specify objects up to a unique isomorphism, which means there is really just one object that has the desired property, up to a (unique) relabelling. That means that we can describe objects by what they are meant to do and not care about some particular (set-theoretic) implementation of it (except when you are proving that such an object exists in the first place).

In this example, you can define $k[X]$ just by stating the universal property, compared to the set-theoretic definition using finite sequences in $k$ and explicitly defining ring operations on them. The set-theoretic definition is important in a way that it makes sure that $k[X]$ does exist, but after that, you pretty much never use is again. The same thing is with tensor product, if you are familiar with it. There is a set-theoretic definition of it but once you know a tensor product can be explicitly constructed, you never actually explicitly use the construction ever again, just the universal property.

You can check out the Wikipedia page on Universal property, but there is no need to dwell too much on the formalism, the more examples you encounter, the more it will be clear why it's such an important concept in all of mathematics.

To address your second question, first note that one can define a $k$-vector space by taking an abelian group $V$ and a ring homomorphism $\pi\colon k\to \operatorname{End}(V)$ that maps $1_k$ to the identity homomorphism on $V$. The relationship with the standard definition is the following: you already have the structure of abelian group on $V$ and the only thing you are left to define is scalar multiplication which can be done by defining $a\cdot v := \pi(a)(v)$. Note that $\pi$ being a ring homomorphism automatically takes care of all the axioms for scalar multiplication. Obviously, you can do the converse as well, if you already have a vector space $V$, you can define $\pi(a)(v) := a\cdot v$.

By the universal property of polynomial rings you specified, we can now pick $\alpha\in\operatorname{End}(V)$ and it follows that there is a unique ring homomorphism $\pi_\alpha\colon k[X]\to \operatorname{End}(V)$ such that $\pi_\alpha(a) = \pi(a),\, \forall a\in k,$ and $\pi_\alpha(X) = \alpha$ (such homomorphism is called evaluation homomorphism because you take some $\alpha$ and evaluate it on some polyomial $p(X)$ by substituting every instance of $X$ by $\alpha$ and do the ring operations in the codomain). This $\pi_\alpha$ defines $k[X]$-module structure on $V$ in the same way as we already discussed, for a polynomial $p\in k[X]$ define action $p\cdot v := \pi_\alpha(p)(v) = p(\alpha)v$. Now, all of this might sound fancy, but you already know how to do this, for example if we take $\alpha\in \operatorname{End}(V)$ and it's characteristic polynomial $\chi$, Hamilton-Cayley theorem tells you that $\chi(\alpha) = 0$.

However, if you've been observant there is a technical issue: $\operatorname{End}(V)$ is not a commutative ring in general. We can avoid this issue by noting that $\operatorname{End}(V)$ can be replaced with its commutative subring $k[\alpha] = \{a_n\alpha^n + \ldots + a_1\alpha + a_0 I_V \mid n\in\mathbb N_0,\ a_i\in k,\, i = 0,\ldots,n\}.$