Definition of polynomial solutions over polynomial field $F[x]$ I'm studying abstract algebra and ran into the problem of solving equations where solutions are polynomials.
The problem is as follows: Given B a member of a polynomial field $F[x]$, having coefficients in field $F$. Find all the possible polynomials X in $F[x]$ where $X = B$. That is finding the tuples of coefficients that satisfy $X = B$.
What is not clear to me is how one defines the equation $X = B$: $X = B$ for all $x$ in $F$ or for at least one $x$?
 A: Assuming Zev's interpretation is correct:
Remember that every element of $F[x]$ defines an element of $F^F$ (the set of all functions from $F$ to $F$) by "evaluation": given $a(x)\in F[x]$, 
$$a(x) = a_0 + a_1x + \cdots + a_nx^n,\qquad a_i\in F,$$
we define the function $a\colon F\to F$ by 
$$a(r) = a_0 + a_1r + \cdots + a_nr^n,$$
that is, $a$ is the function induced by mapping $x\to r$ and using the universal property of $F[x]$. 
The question is then: when do two polynomials define the same polynomial function ?
(Early on, we are used to thinking about polynomials and polynomial functions as one and the same; but formally they are not the same, and this question is precisely designed to explore the distinction). 
Theorem. If $F$ is infinite, then two polynomials in $F[x]$ define the same polynomial function $F\to F$ if and only if they are identical. If $F$ is finite, of order $p^n$, then $a(x)$ and $b(x)$ in $F[x]$ define the same polynomial function if and only if they leave the same remainder when divided by $x^{p^n}-x$.
Proof. Suppose that $a(x),b(x)\in F[x]$ are such that $a(r)=b(r)$ for all $r\in F$. Let $c(x) = b(x)-a(x)$. Then $c(r)=0$ for all $r\in F$. If $F$ is infinite, this implies that the degree of $c$ is $0$ (since a polynomial of degree $n$ has at most $n$ distinct roots). This proves the first part.
Assume now that $F$ is finite of order $p^n$. The multiplicative group of nonzero elements of $F$ is a cyclic group of order $p^{n}-1$, so by Lagrange's Theorem we know that $r^{p^n-1}=1$ for all $r\in F$, $r\neq 0$, and therefore, $r^{p^n}=r$ for all $r\in F$. In particular, the polynomial $x^{p^n}-x$ defines the constant function $0$. By the Factor Theorem,
$$x^{p^n}-x = \prod_{r\in F}(x-r).$$
Assume first that $a(x)$ and $b(x)$ define the same function $F\to F$. Proceeding as above, we see that $c(x)$ vanishes at every element of $F$. By the Factor Theorem, $x-r$ divides $c(x)$ for every $r\in F$, so $x^{p^n}-x | c(x)$. Therefore, $x^{p^n}-x$ divides $a(x)-b(x)$, which means that $a(x)$ and $b(x)$ have the same remainder when divided by $x^{p^n}-x$.
Conversely, if $a(x) \equiv b(x) \pmod{(x^{p^n}-x)}$, then we can write $b(x) = a(x) + k(x)(x^{p^n}-x)$. Evaluating at $r\in F$ we have
$$b(r) = a(r) + k(r)(r^{p^n} - r) = a(r)+k(r)(0) = a(r),$$
so $b(x)$ defines the same function $F\to F$ as $a(x)$, as claimed. $\Box$
