Is polynomial in general the same as polynomial function? 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,


*

*Is polynomial always a function? If not then what is a polynomial in general?

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

*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?



  
*
  
*Is polynomial always a function?
  

No, never.


  
*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.


  
*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 $x\in K$ meaning $f$ is a constant $0$ as a function $f:K\to K$. 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.
