TITLE: Is R[x] a ring of functions or a ring of expressions?
QUESTION [5 upvotes]: I know that these two notions are isomorphic, but for a ring $R$, is $R[x]$ a ring of functions (e.g. $f:x\mapsto x^3+1$) or a ring of expressions (e.g. $f(x)=x^3+1$). If it were the former, the variable $x$ would be meaningless and $R[x]=R[y]$, but if it were the latter, $x$ would be important and $R[x] \cong R[y]$ (isomorphic), but $R[x]\neq R[y]$.

REPLY [2 votes]: The polynomial algebra over $R$ is what's known as a "free object in the category of (commutative) $R$-algebras." That means that $R[x]$ is the unique $R$-algebra such that for any other $R$-algebra $S$ and any element $a$ of $S$ there is a unique map from $R[x]$ to $S$ taking $x$ to $a$.
What this means is that $R[x]$ consists of functions. But these are not functions $R \to R$, they are what we might call "universal functions." That is, for any $R$-algebra $S$, an element $f \in R[x]$ gives a function $f : S \to S$. Two elements of $R[x]$ are equal if and only if they determine the same functions $S \to S$ for every $S$.
For example, let $R = \mathbf{Z}/3\mathbf{Z}$ and $f = x^3 - x \in R[x]$. Then the function $f : R \to R$ equals the zero function. On the other hand, if we take $S = R[i]$ (where $i^2 = -1$) then $f : S \to S$ does not equal the zero function because $i^3 - i \ne 0$. Thus $f \ne 0$ in $R[x]$ because there exists some $R$-algebra $S$ (namely $R[i]$) for which $f : S \to S$ is not the zero function.
Two polynomials, defined naïvely, are equal if and only if their respective coefficients are equal. To make this precise, we define $R[x]$ to be the set of sequences of coefficients with the appropriate addition and multiplication and show that this ring satisfies the "universal function" property of a free object. One key step of the proof is to consider $f : S \to S$ where $S = R[x]$.