Corollary on splitting field of polynomial

Question

Need help in understanding a point in a proof: Let $F$ be a commutative field and $p\in F[X]$, $\deg (p) \geq 2$, an irreducible polynomial, then the ring $F[X]/pF[X]$ (quotient) is a field.

$pF[X]$ is an ideal of $F[X]$ - check
$F[X]/pF[X]$ is a commutative ring - check

Let's denote an element in $F[X]/pF[X]$ as $\bar{f}$
Let $\bar{0}\neq \bar{f}\in F[X]/pF[X]$. $f,p$ have no common divisors and as such there exist polynomials $x,y\in F[X]$ such that: $fx+py = 1$ due to which:
$$\bar{1} = \overline{fx+py} = \overline{fx}+\overline{py} = \overline{fx} $$

This is where I lose track. Why is $\bar{p}=\bar{0}$?

Isn't a quotient given as $F[X]/pF[X] :=\{\bar{x}=x+pF[X]\ : x\in F[X] \}$? So why does this make $\bar{p} =\bar{0}$

Answer 1

$\overline{p}$ means the image of $p$ under the canonical quotient map $F[x] \to F[x]/(p)$. The quotient collapses $p$ so the image is zero. Also, as you note, elements are of the form $x + p F[x]$. In order to write $p$ in that form, note that we just take $x = 0$.

Answer 2

$$\bar p=pF[x]=0+pF(x)=\bar 0$$

But an other way to check that $F[X]/(p)$ is a field is to show that $(p)$ is maximal. Since $F$ is a field, $F[X]$ is a PID, and thus all ideal are principal. Let $(g)\supset (p)$ an ideal that contain $(p)$. Therefore $g\mid p$ and thus $g=up$ since $p$ is irreducible, where $u$ is a unit. Therefore $g\in (p)$ and thus $(p)=(g)$. Therefore $(p)$ is maximal and thus $F[X]/(p)$ is a field.

Answer 3

When you "quotient out" by a polynomial, you are really quotienting out by the ideal generated by that polynomial. When it comes time to check which equivalence classes things are in you can think to yourself what do I have to add to $p$ to make it an element of the ideal generated by p. Well it is already in the ideal generated by p and so we don't have to add anything, that is $\overline{p} = 0$.