Cardinality of polynomial ring over a finite field 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!
 A: 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. 
A: By Lagrange interpolation, we can see that any function $\mathbb F\to\mathbb F$ can be represented as a polynomial function on $\mathbb F$, so the number of distinct polynomial functions is $|\mathbb F|^{|\mathbb F|}$. However, the ring of polynomials is not equal to the ring of polynomial functions. If two polynomials have different coefficients, then they are not the same polynomial, so $\mathbb F[x]$ is infinite. This is important because higher degree polynomials can be irreducible and we can adjoin new elements to the field that are roots of these polynomials, and the field (and hence the ring of polynomial functions) becomes larger.
Note that a field of characteristic $p>0$, if finite, can have $p^m$ elements for any $m>0$. These fields are obtained by adjoining roots of irreducible polynomials. There are also fields of characteristic $p$ that are not finite.
