Exercise 4.1.8

Proof.

(a)

$x^2-3$ is irreducible over $\mathbb{Q}$. We show that it remains irreducible over $\mathbb{Q}[\sqrt{2}]$.

Suppose on the contrary that $f$ is reducible over $F$: $f=x^2-3=\mathit{uv},\mathit{uv}\in \mathbb{Q}[\sqrt{2}][x]$, where $u,v$ are nonconstant polynomials. Then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)\ge 1,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)\ge 1$, and as $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v)=1$,

$$u=\mathit{ax}+b,a,b\in \mathbb{Q}[\sqrt{2}],a\ne 0.$$

Then $\alpha =-b∕a\in \mathbb{Q}[\sqrt{2}]$ is a root of $u$, thus is a root of $f=x^2-3$. Since ${\sqrt{2}}^{2n}=2^n$ and ${\sqrt{2}}^{2n+1}=2^n\sqrt{2}$, every element of $\mathbb{Q}[\sqrt{2}]$ is of the form $c+d\sqrt{2},c,d\in \mathbb{Q}$.

We should have $\alpha =c+d\sqrt{2}=\pm \sqrt{3}.$ Then

$${\alpha }^2=c^2+2d^2+2\mathit{cd}\sqrt{2}=3.$$

If $\mathit{cd}\ne 0$, $\sqrt{2}=(c^2+2d^2-3)∕(2\mathit{cd})\in \mathbb{Q}$, in contradiction with the irrationality of $\sqrt{2}$. Thus $c=0$ or $d=0$.

$d=0$ gives $\sqrt{3}=\pm c\in \mathbb{Q}$: this is in contradiction with the irrationality of $\sqrt{3}$.

$c=0$ implies $\sqrt{\frac{3}{2}}=\pm d\in \mathbb{Q}$. But then $\sqrt{\frac{3}{2}}=\frac{p}{q},(p,q)\in \mathbb{Z}\times {\mathbb{N}}^{\text{∗}},p\wedge q=1$.

$3q^2=2p^2$, $q^2\mid 2p^2$ and $q^2\wedge p^2=1$. By Gauss Lemma, $q^2\mid 2,q\in {\mathbb{N}}^{\text{∗}}$, hence $q=1,3=2p^2$, thus $3$ is even: this is absurd.

Conclusion: $x^2-3$ is irreducible $\mathbb{Q}[\sqrt{2}]$.

(b)

$$\begin{array}{llll}\hfill f& =[(x-\sqrt{2}-\sqrt{3})(x+\sqrt{2}-\sqrt{3})][(x-\sqrt{2}+\sqrt{3})(x+\sqrt{2}+\sqrt{3})]& \hfill & \\ \hfill & =[{(x-\sqrt{3})}^2-2][{(x+\sqrt{3})}^2-2]& \hfill & \\ \hfill & =(x^2-2\sqrt{3}x+1)(x^2-2\sqrt{3}x+1)& \hfill & \\ \hfill & ={(x^2+1)}^2-{(2\sqrt{3}x)}^2& \hfill & \\ \hfill & =x^4-10x^2+1& \hfill & \end{array}$$

The equality $f=x^4-10x^2+1=(x^2-2\sqrt{3}x+1)(x^2-2\sqrt{3}x+1)$ show that $f$ is not irreducible over $\mathbb{Q}[\sqrt{3}]$.

Factorisation with Sage:

     K = NumberField(x^2-3, ’a’);L.<X> = PolynomialRing(K)
     p = X^4-10*X^2+1
     factor(p)

$$(X^2-2\mathit{aX}+1).(X^2+2\mathit{aX}+1).$$