TITLE: Cardinality of polynomial ring over a finite field
QUESTION [2 upvotes]: My questions is relative to the cardinality.
Consider $\mathbb{F}$ a field of characteristic $p$, so it has the elements $\{0,1,\dots,p-1\}$.
Now consider the polynomials with variable $x$ and coefficientes over $\mathbb{F}$, that is, $\mathbb{F}[x]$.
Now, how many elements has $\mathbb{F}[x]$? My notes say that it is finite. Moreover, it's of the form $p^m$ (I don't know what means $m$), so $x^{p^m}=x$ (also I don't know why).
I suppose that you get infinite elements of the form $$a_nd^n+\cdots+a_2d^2+a_1d+a_0$$ with $n \geq 0, a_i \in \mathbb{F}$.
Sorry for the questions. I´ve tried, but I didn't the answer in this web!
Thanks!

REPLY [4 votes]: There is an important distinction between polynomials and functions that can be represented by polynomials.  The confusion comes in thinking about elements of $\mathbb{F}[x]$ as functions, which while reasonable, isn't the right way to approach it for the sake of abstract algebra.  A polynomial over $\mathbb{F}$ is, roughly, a formal expression of the ilk $f(x) = a_0 + a_1x + \cdots + a_nx^n$ with the expected way to add and multiply such expressions.  They are not technically functions. In this way, your ring $\mathbb{F}[x]$ is certainly infinite, as the expressions 
$$
1, x, x^2, x^3, \ldots
$$
are all distinct, no matter the base field.
A polynomial $f \in \mathbb{F}[x]$ can be regarded as a function $\mathbb{F} \to \mathbb{F}$ by evaluation in the obvious way, but this is a different creature. When regarded as a way to represent a function, you are right that there are only finitely many such ways to do this, since there are only finitely many self-maps on a finite set. 
The example that set it all straight for me was the polynomial $f(x) = x^2+x$ in $\mathbb{Z}_2[x]$.  Treating it as a function, it is clear that $f$ is identically $0$ (plug in $0$ or $1$ and always get back $0$).  However, $f$ is clearly not the additive identity for the ring of polynomials, so these are definitely different concepts.
Finally, be careful: a field having characteristic $p$ does NOT make it finite.  The field $\mathbb{F}_p(x)$ of rational functions with coefficients in $\mathbb{F}_p$ has characteristic $p$ but is infinite.
{"set_name": "stack_exchange", "score": 2, "question_id": 3658177}