Recuperating usual polynomials from the universal property

Question

Let $R$ be a ring. There is an inclusion map $i : R \hookrightarrow R[X]$, where $R[X]$ is the polynomial ring. The universal property of the polynomial ring $R[X]$ states:

For any ring $S$, any ring homomorphism $f : R \rightarrow S$, and any $s \in S$, there is a unique ring homomorphism $\widehat{f} : R[X] \rightarrow S$ such that $\widehat{f} \circ i = f$ and $\widehat{f}(X) = s$.

This property characterizes the ring $R[X]$ up to isomorphism. But we're already familiar with polynomials being of the form $r_0 + r_1 X + ... + r_n X^n$, so we should be able to establish this from the universal property. In other words, consider the following property:

For any $p \in R[X]$, there exist unique $n \in \Bbb{N}$ and $r_0, ..., r_n \in R$ with $r_n \neq 0$ such that $p = r_0 + r_1 X + ... + r_n X^n$.

Question: How can we prove the above property directly from the universal property?

I noticed that you can evaluate $p$ at $0$ and get $r_0$. Then I thought of considering $p - r_0$; this evaluates to zero, so maybe we should be able to factor $p-r_0 = X \cdot p_1$, but I don't even know if we can do polynomial division just based on the universal property. Even if we can, and we try to proceed by induction, I'm not sure why it's guaranteed to stop. So, I'm not too sure how to proceed.

Answer 1

It is easy to prove that all polynomials are of the form $r_0 X^0 + \cdots + r_n X^n$ from the universal property. For let $S \subseteq R[X]$ be the set of all polynomials of this form. Then $S$ is a subring of $R[X]$, and the inclusion map $i : R \to R[X]$ factors through $S$; moreover, $X \in S$. Apply the universal property to get an $R$-algebra homomorphism $f : R[X] \to S$ sending $X$ to $X$. Then $f : R[X] \to R[X]$ is an $R$-algebra homomorphism sending $X$ to $X$, as is the identity map; by the universal property, $f$ is the identity map, so $S = R[X]$. That is, all elements of $R[X]$ can be written as $r_0 X^0 + \cdots + r_n X^m$. If you then want the additional requirement that $r_n \neq 0$ or $n = -1$, you simply take the largest $n$ such that $r_n \neq 0$ (or $n = -1$ if no such $n$ exists).

Uniqueness, however, is not so trivial. To prove uniqueness, we must explicitly define a ring $A$ to consist of formal expressions of the form $r_0 Y^0 + \cdots + r_n Y^n$, modulo the appropriate equivalence relation. We must then give $A$ the obvious $R$ algebra structure. We then consider the unique map $R[X] \to A$ which is an $R$-algebra homomorphism sending $X$ to $Y$; this map allows us to assert uniqueness. Unfortunately, this technique essentially relies on building $R[X]$ explicitly rather than using the universal property (as $A$ is $R[X]$ “in disguise”).

Unfortunately, your technique for evaluating the polynomials doesn’t always work. In particular, if $R$ is a finite nonzero ring, there are only finitely many functions $R \to R$ even though there are infinitely many polynomials in $R[X]$. Thus, a polynomial is not determined by evaluating it against every element of $R$. To be explicit, if $R = \{a_0, \ldots, a_n\}$, then let $P = (X - a_0)(X - a_1) \cdots (X - a_n)$. Then $P \neq 0$, but for all $x \in R$, $P(x) = 0$.

However, if $R$ is an infinite field, then polynomials are determined uniquely by their evaluations against elements of $R$.

Answer 2

You can get uniqueness without doing the construction, i.e. in a way where you still don't actually know $R[X]$ exists. The idea is to let $R[X]$ act on infinite sequences $(r_0, r_1, r_2, \ldots)$ where $R$ scales pointwise and $X$ is the shift operator $(r_0, r_1, r_2, \ldots) \mapsto (0, r_0, r_1, \ldots)$. This gives a unique map $R[X] \to \mathrm{End}_R(\oplus_{i=0}^\infty R)$ by the universal property. Moreover, $$(r_0 + r_1 X + \cdots + r_n X^n) \cdot (1, 0, 0, \ldots) = (r_0, r_1, r_2, \ldots, r_n, 0, 0, \ldots),$$ giving uniqueness. While I've phrased this argument in coordinates, that's not really necessary.