If $j:K→K'$ is a field isomorphism, what is the meaning of $j(p)$, being $p$ a polynomial over $K$. I believe the title explains my question, though to clarify, I encountered a text in which the author defines $j:K→K'$ as a field isomorphism and $p$ as a polynomial that splits over $K$.
The polynomial $p$ is defined as a function, in other words it's equation contains the indeterminate $t$, while $j$ is simply defined as a mapping between $K$ and $K'$, so what could the author be referring to when talking about $j(p)$?
 A: 
The polynomial $p$ is defined as a function

No, that's wrong. It's a common misunderstanding: polynomials are not functions. The formal definition is that a polynomial over a (commutative) ring $R$ is an infinite sequence $(a_0, a_1, \ldots)$ where each $a_i\in R$ and there is $M$ such that $a_m=0$ for all $m>M$ (side note: we can omit the second condition to obtain formal power series). Smallest such $M$ is called the degree of the polynomial. The addition and multiplication are given by:
$$(a_0, a_1, \ldots)+(b_0,b_1,\ldots)=(a_0+a_1, b_0+b_1, \ldots)$$
$$(a_0, a_1, \ldots)\cdot (b_0, b_1, \ldots)=(c_0, c_1, \ldots)$$
$$c_i=\sum_{k=0}^{i}a_{k}\cdot b_{i-k}$$
That's all. With that we define a special $X$ (or $t$ in your case) polynomial by $X:=(0,1,0,\ldots)$ if $R$ has $1$. We also identify $r\in R$ with a polynomial $(r, 0,\ldots)$. Then every polynomial can be written in a form
$$a_0+a_1\cdot X+a_2\cdot X^2+\cdots+a_n\cdot X^n$$
for $a_i\in R$. The set of all polynomials is denoted by $R[X]$. With the addition and multiplication defined as above its a ring (even algebra over $R$).
Now if $f:R\to S$ is a ring homomorphism and $p=(a_0,a_1,\ldots)$ is a polynomial then $f$ can be extended to a ring homomorphism
$$F:R[X]\to S[X]$$
$$F((a_0,a_1,\ldots)) = (f(a_0), f(a_1), \ldots)$$
or with usual notation
$$F\big(\sum a_i\cdot X^i\big)=\sum f(a_i)\cdot X^i$$
And this is most likely what the author is referring to.
Side note: If $p=(a_0,a_1,\ldots)$ is a polynomial then it induces a polynomial function $\overline{p}:R\to R$ given by
$$\overline{p}(x)=a_0+a_1\cdot x + a_2\cdot x^2 + \cdots $$
(note that the sum is finite since $a_i=0$ eventually). Do not confuse it with addition and scalar multiplication in $R[X]$, here $x$ is simply an element of $R$ and we add and multiply in $R$. Denote by $R\{X\}$ the set of all polynomial functions (which is a ring under pointwise addition and multiplication).
It can be shown that a mapping
$$\Phi:R[X]\to R\{X\}$$
$$\Phi(p)=\overline{p}$$
is a ring homomorphism. It is surjective by definition.
You may think that I'm just doing some formal irrelevant distinction between polynomials and polynomial functions but that's not the case. Consider any finite ring $R=\{a_0,a_1,\ldots, a_n\}$. Define two polynomials:
$$P_1=0$$
$$P_2=(X-a_0)(X-a_1)\cdots(X-a_n)$$
These are two different polynomials because they have different coefficients ($P_2$ is of degree $n$). However the induced polynomial functions are the same because $\overline{P_2}(x)=0$ for any $x\in R$. Thus $\Phi$ is not injective over a finite ring.
It can be shown that $\Phi$ is an isomorphism if and only if the underlying ring $R$ is infinite.
