Exercise 3.1.5

Proof. Let $f\in F[x]$ be irreducible, and let $L=F[x]∕⟨f⟩$ the quotient ring.

(a)

Let $\bar{g}\in L,\bar{g}\ne \bar{0}$, that is to say $g+⟨f⟩\ne 0+⟨f⟩$, which is equivalent to $g\notin ⟨f⟩$, or $f\nmid g$ (in $F[x]$).

Let $u$ a common divisor of $f$ and $g$. Since $f$ is irreducible, either $u$ is a nonzero constant, or $f=\mathit{ku},k\in F^{\text{∗}}$, i.e., $u$ is associate to $f$. But in this last case, $u=k^{-1}f$ and $f$ divides $u$, which divides $g$, so $f\mid g$, in contradiction with the hypothesis.

So the only common divisors of $f,g$ are the nonzero constants, thus $f\wedge g=1$.

By Bézout theorem, there exist polynomials $A,B\in k[x]$ such that

$$1=\mathit{Af}+\mathit{Bg}.$$

(b)

As $\bar{f}=f+⟨f⟩=\bar{0}$, $\bar{1}=\bar{A}\bar{f}+\bar{B}\bar{g}=\bar{B}\bar{g}$, which we can write $$1+⟨f⟩=(B+⟨f⟩)(g+⟨f⟩).$$

So $B+⟨f⟩$ is the inverse of $g+⟨f⟩$ in $L=F[x]∕⟨f⟩$.