Exercise 2.1.1

Proof. We show first that $⟨x,y⟩\ne F[x,y]$. If not, $1\in ⟨x,y⟩$, so

If we evaluate this identity at $x=0,y=0$, we obtain $1=0$, which is a contradiction, thus

If $⟨x,y⟩$ was a principal ideal, generated by $p\in F[x,y]$, then $⟨x,y⟩=⟨p⟩$, and

$\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(q)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x)=1$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)\le 1$, and $p\ne 0$.

If $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=0$, then $p=\lambda \in F^{\text{∗}}$, and $⟨x,y⟩=⟨\lambda ⟩=F[x,y]$, but we have proved that this is impossible.

Thus $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=1$, so $p=\mathit{\alpha x}+\mathit{\beta y}+\gamma ,\alpha ,\beta ,\gamma \in F$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(q)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(r)=0$, so $q=\lambda \in F^{\text{∗}},r=\mu \in F^{\text{∗}}$:

As $\lambda \ne 0,\mu \ne 0$, $\alpha =\beta =0$, whitch is in contradiction with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=1$.

We have proved that $⟨x,y⟩$ is not a principal ideal, and thus $F[x,y]$ is not a principal ideal domain. □