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The algebra text book says, a polynomial in one variable over $mathbb{R}$ is given by,

$f(x)= a_n x^n + a_{n-1} x^{n-1} + cdots + a_1x + a_0$

Where $x$ is an unknown quantity which commutes with real numbers, called "indeterminate".

So I have a few question,

1. Is polynomial always a function?if not then what is a polynomial in general?


2. And what's up with the "indeterminate" thingy? Is it wrong to simply call it a variable?


3. What exactly is the $x$ in the expression? A number? A matrix? Or...some other object? Why does it have to "commute" with real numbers?

No, a polynomial is not a function. However, for each polynomial $p(x)=a_0+a_1x+cdots+a_nx^n$ you may consider the polynomial function$$begin{array}{rccc}pcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&a_0+a_1x+cdots+a_nx^n.end{array}$$And distinct polynonials will be associated with dstinct functions. However, although this is true over the reals, it doesn't hold in general. For instance, if you are working over the field $mathbb{F}_2$, then the polynomial $x^2-x$ and th null polynomial are distinct polynmials, but the function$$begin{array}{ccc}mathbb{F}_2&longrightarrow&mathbb{F}_2\x&mapsto&x^2-xend{array}$$is the null function.

So, a polynomial (over the reals) is an expression of the type $a_0+a_1x+cdots+a_nx^n$, where $x$ is an entity about which all we assume is that it commutes with each real number. Usually, it is called an “indeterminate” since it is not a specific real number.

1. Is polynomial always a function?

No, never.

2. And what's up with the "indeterminate" thingy? Is it wrong to simply call it a variable?

Yes, it is wrong. Formally a polynomial is a sequence $(a_0, a_1, ldots)$ such that $a_i=0$ eventually. It's a sequence of coefficients, it's not a function. Each coefficient is taken from a fixed ring, e.g. reals, complex numbers or finite fields. The underlying ring is also called a ring of scalars. With that we can define a specific polynomial addition, polynomial multiplication and scalar multiplication.

3. What exactly is the $x$ in the expression? A number? A matrix? Or...some other object? Why does it have to "commute" with real numbers?

$x$ is simply a special polynomial of the form $(0,1,0,0,ldots)$. It is a theorem that every polynomial can be uniquely written as a finite sum $sum a_ix^i$ with $a_i$ scalars.

It is important to distinguish polynomials from polynomial functions. For example if $K$ is a finite field, i.e. $K={a_1,ldots, a_m}$ then take polynomial $f(X)=(X-a_1)cdots(X-a_m)$. For example if $K=mathbb{Z}_2={0,1}$ then that would produce polynomial $(X-1)X=X^2-X=X^2+X$ which formally is $(0,1,1,0,0,ldots)$.

Clearly $f(x)=0$ for all $xin K$ meaning $f$ is a constant $0$ as a function $f:Kto K$. But as a polynomial $f(X)$ is non-zero of positive degree $m$.

Also note that over infintie fields these two coincide. Indeed, there is a ring epimorphism $K[X]to K{X}$ with polynomials on the left side and polynomial functions on the right. The kernel is non-trivial if and only if $K$ is finite. And in that case it is equal to the principal ideal $langle (X-a_1)cdots(X-a_m)rangle$.

In particular there are always infinitely many polynomials. But there might be finitely many polynomial functions, depending on the underlying ring of scalars.