Find a between field $\mathbb{Q}\subseteq K\subseteq L$ with $[K:\mathbb{Q}]=2$.

Question

My exercise is

(1) to justify whether the field $L=\mathbb{Q}(\sqrt{2},i)$ is a Galois extension of $\mathbb{Q}$ and

(2) to find an element $z$ with $L=\mathbb{Q}(z)$ and a between field $\mathbb{Q}\subseteq K \subseteq L$ with $|K:\mathbb{Q}|=2$ and $K\neq \mathbb{Q}(\sqrt{2})$ and $K\neq\mathbb{Q}(i)$.

The field $L$ is a splitting field of the polynomial $f=(X^2-2)(X^2+1)\in\mathbb{Q}[X]$, and $L$ is a Galois extension over $\mathbb{Q}$.

For (2), I've found an element $z=\sqrt{2}i$ which equals $\mathbb{Q}(\sqrt{2},i)=\mathbb{Q}(\sqrt{2}i)$. How can I find $K$?

Answer 1

You $z$ is not OK, as $z^{2} \in \mathbb{Q}$. Try $w = \sqrt{2} + i$ instead.

But then your $z$ is OK to give you $K = \mathbb{Q}(z)$ - please check that indeed $K\neq \mathbb{Q}(\sqrt{2})$ and $K\neq\mathbb{Q}(i)$

Answer 2

To prove $z=\sqrt 2+i$ is a primitive element of $\mathbf Q(\sqrt2,i)$, you have to check

a) it satisfies a quartic equation,

b) this quartic equation is irreducible in $\mathbf Q$.

You'll easily find $z$ is a root of $p(x)=x^4-2x^2+9$. We can't apply Eisenstein criterion here, so we have to do it by hand. Indeed, with the rational roots theorem it's easy to check it has no root. Further, it can't be factored as a product of two quadratic polynomials. For if it did, we might suppose it factors as the product of two monic polynomials with integer coefficients, $x^2+ax+b$ and $x^2+a'x+b'$.

Write down the equations obtained by identification: $$\begin{cases}\begin{aligned}a+a'&=0, &b+b'+aa'&=-2, \\a(b-b')&=0, &bb'&=9.\end{aligned}\end{cases}$$ Then

• either $a\neq 0$,which implies $\;a'=-a$ and $b'=b$, hence $\;b^2=9$, $\;b=\pm3$ and finally $\;2b-a^2=-2$, so $\;a^2=2(b+1)= 8\;$ or $\;-4$. Both are impossible.
• or $a=a'=0$. We obtain $\;b+b'=-2,\enspace bb'=9$. $b$ and $b'$ are the roots of the quadratic equation $\;t^2+2t+9=(t+1)^2+8$. However this equation has no real roots.