In abstract algebra, how important is it to draw a distinction between polynomials and polynomial functions?

Question

In introductory analysis, it is common to define a single-variable polynomial as a function $p:\Bbb{R}\mapsto\Bbb{R}$ (or $\Bbb{C}\mapsto\Bbb{C})$ such that $$ p(x)=\sum_{i=0}^{n}a_ix^i \text{ for all $x$.} $$ This means that $x^2$ is technically not a polynomial, but rather the value of the polynomial $t\mapsto t^2$ at the point $x$. However, according to Wikipedia, in abstract algebra one often distinguishes between polynomials and polynomial functions. According to this definition, $x^2$ is literally a polynomial, whereas $t\mapsto t^2$ is the corresponding polynomial function.

My question is:

In abstract algebra, what is the purpose of defining polynomials as formal algebraic expressions, rather than as functions? How important is it to draw a distinction between polynomials and polynomial functions?

Answer 1

I will give you an example of why this is important. However, first, i will define what I mean by “polynomial” and what I mean by “polynomial function”. For this matter, let us restrict to polynomials over fields, it will be enough.

1. A polynomial $p$ over the field $K$ is a sequence of elements of $K$ that are all $0$ except for a finite number. We can write this sequence as follows: $$ p = (a_0,a_1,...,a_n,0,0,...),\ a_n\neq 0$$ Notice that, since we required that only a finite amount of the elements are nonzero, there will be a $last$ nonzero element, namely $a_n$, and in this case, we say that $n$ is the degree of the polynomial. We usually identify the latter sequence with the following formal expression: $$p = a_nt^n + \cdots + a_1t + a_0$$ The operations over polynomials (sum and product) are defined as it is usually known, and the reason for this will become clear later. With this sum and product, the set of all polynomials over $K$ (denoted by $K[t]$) is itself a ring.

The last thing there is to know about polynomials is the so-called evaluation homomorphism. Given $b\in K$, the evaluation homomorphism at $b$ is a function $\operatorname{ev}_b:K[t]\to K$ that maps every polynomial $p$ to $a_nb^n + \cdots + a_1b + a_0\in K$, being $a_i$ the coefficients of the polynomial. It is a ring homomorphism because of the way we defined sum and product of polynomials. The evaluation homomorphism corresponds, as it name says, to evaluating the polynomial at $b$.

2. A polynomial function over a field $K$ is a function $f_p:K\to K$ than can be written as $f_p(b) = \operatorname{ev}_b p$, for some polynomial $p\in K[t]$, that is, it is the result of evaluating a polynomial at every element of the field.

Now, because the evaluation function is a homomorphism, operating with formal polynomials and operating with polynomial functions may seem like the same (a polynomial defines a unique function and vice versa), but (here comes the important part) this is only true if $K$ is an infinite field (such as $\mathbb{C}$), i.e., in infinite fields the following property holds: two polynomials over $K$ define the same function iff they are equal as polynomials (i.e., their coefficients are all equal). However, in finite fields (say $K=\mathbb{F}_p$) we can always construct a nonzero polynomial which function is identically zero. Indeed, if we denote the (distinct) elements of the field as $\{\alpha_1,...,\alpha_p\}$, then the polynomial $$p = (t-\alpha_1)\cdots(t-\alpha_p)$$ is nonzero (has degree p), and yet all the elements of the field are roots of p, i.e., $p(b) = 0$ for all $b\in K$, so its corresponding polynomial function is the zero function.