Exercise 3.2.4 (Not principal ideals)

Proof.

(i)

Let $$I=⟨x,y⟩=\{f(x,y)\in \mathbb{Q}[x,y]:f\text{ has constant term }0\}.$$

(Here we write $x=t_1,y=t_2$.)

We show first that $⟨x,y⟩\ne \mathbb{Q}[x,y]$. If not, $1\in ⟨x,y⟩$, so

$$1=\mathit{xu}+\mathit{yv},u,v\in \mathbb{Q}[x,y].$$

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

$$⟨x,y⟩\ne \mathbb{Q}[x,y].$$

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

$$x=\mathit{pq},y=\mathit{pr},q,r\in \mathbb{Q}[x,y].$$

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

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

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

$$\begin{array}{llll}\hfill x& =\lambda (\mathit{\alpha x}+\mathit{\beta y}+\gamma ),& \hfill & \\ \hfill y& =\mu (\mathit{\alpha x}+\mathit{\beta y}+\gamma ).& \hfill & \\ \hfill & & \hfill \end{array}$$ This implies $\mathit{\lambda \beta }=0$ and $\mathit{\mu \alpha }=0$.

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

We have proved that $⟨x,y⟩$ is not a principal ideal, and thus $\mathbb{Q}[x,y]$ is not a principal ideal domain.

(ii)

Let $$J=⟨2,t⟩=\{f(t)\in \mathbb{Z}[t]:\text{ the constant term of }f\text{ is even }\}.$$

We first show that $⟨2,t⟩\ne \mathbb{Z}[t]$. If not, $1\in ⟨2,t⟩$, so $1=2u(t)+\mathit{tv}(t)$ for some polynomials $u(t),v(t)\in \mathbb{Z}[t]$. The evaluation with $t=2$ gives $1=2u(2)+2v(2)$, and $2\mid 1$: this is a contradiction. Therefore $⟨2,t⟩\ne \mathbb{Z}[t]$.

If $⟨2,t⟩$ was a principal ideal, generated by $p\in \mathbb{Z}[t]$, then $⟨2,t⟩=⟨p⟩$, and

$$2=u(t)p(t),y=v(t)p(t),u,v\in \mathbb{Z}[t].$$

The first equality shows that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=0$, thus $p=k\in \mathbb{Z}$, and

$$2=\mathit{ku}(t),t=\mathit{kv}(t).$$

The evaluation with $t=1$ gives $1=\mathit{kv}(1)$, therefore $k\mid 1$, and $p(t)=k=\pm 1$. But then $⟨2,t⟩=⟨p⟩=⟨\pm 1⟩=\mathbb{Z}[t]$. Previously, we proved that this is impossible. Therefore $⟨2,t⟩$ is not a principal ideal ideal, and $\mathbb{Z}[t]$ is not a principal ideal domain.