For a nonconstant $p$ in $F[x]$, the ideal $(p)$ is maximal and $F[x]/(p)$ is a field exactly when $p$ is irreducible

In a commutative ring the multiples of $p$ form an ideal containing $p$ and lie in every ideal containing $p$, so [L2] identifies $(p)$ with the set of multiples of $p$. Suppose $p$ is irreducible and $f+(p)$ is a nonzero residue class; then $p\nmid f$. If a common divisor $d$ of $p,f$ were a nonunit, a factorization $p=de$ and [L6] would make $e$ a unit, so $d$ would be associate to $p$ and $d\mid f$ would imply $p\mid f$, a contradiction. Thus every common divisor is a unit, and [L1] gives $A p+B f=1$, whence [L4] gives $(B+(p))(f+(p))=1+(p)$; every nonzero class is invertible, so the quotient is a field.

Conversely, suppose the quotient is a field and $p=ab$. By [L4], the two residue classes have product zero, so one is zero; say $a\in(p)$. The characterization established in step 1.1 gives $a=pc$, and hence $p=ab=pcb$. A direct leading-coefficient argument shows that $F[x]$ has no zero divisors, because $F$ is a field, so cancellation of the nonzero polynomial $p$ gives $cb=1$ and makes $b$ a unit. The other case similarly makes $a$ a unit, and [L6] makes $p$ irreducible.

Steps 1.1 and 2.1 prove that irreducibility is equivalent to quotient fieldness, and [L5] identifies quotient fieldness with maximality of $(p)$ in the sense of [L3].