Galois theory and Field extensions

Question

Let $\alpha$ be a complex number with $$\alpha^2 = \sqrt3 -\sqrt5.$$ Prove that the extension of the field Q with $\alpha$ over Q is NOT Galois.

Answer 1

Hint:

The minimal polynomial of $\alpha$ over $\Bbb Q$ is $$x^8-16x^4+4=(x^4-8)^2-60$$ For your extension to be Galois it needs to contain all the roots of the minimal polynomial.

Answer 2

In order for an extension to be galois it must be separable and normal. Separable is not going to be a problem over $\mathbb{Q}$. In order for it to be normal it must contain all of the roots of the irreducible polynomial. As noted above the minimal polynomial of this is $$x^8 - 16x^4 + 4 = 0.$$ Let $\beta$ be $i\alpha$ then this is clearly also a root of the above irreducible polynomial since the polynomial above is all in terms of $x^4$. It follows that $\beta^2$ = $-\alpha^2$ = $\sqrt{5} - \sqrt{3}$ and so it is easy to see that $\beta$ is real. If $\mathbb{Q(\alpha)}$ contains all of the roots of this polynomial then clearly $\mathbb{Q(\beta)}$ $\subset$ $\mathbb{Q(\alpha)}$. It is clear from the fact that the minimal polynomial is irreducible, and from viewing $\mathbb{Q(\alpha)}$ and $\mathbb{Q(\beta)}$ as the images of the evaluation map at the two different roots then they are both isomorphic to $$\frac{\mathbb{Q}[X]}{(x^8-16x^4+4)}$$ and thus both isomorphic to one another. Due to the containment and isomorphism we see that $\mathbb{Q}(\alpha)$ = $\mathbb{Q}(\beta)$. However we have $\mathbb{Q}(\beta)$ $\subset$ $\mathbb{R}$ but $\mathbb{Q}(\alpha)$ $\nsubseteq$ $\mathbb{R}$ so we have our contradiction.

Answer 3

Avoiding minimal polynomials, let us do Galois theory by introducing the biquadratic field $K=\mathbf Q(\sqrt 3, \sqrt 5)$. Its 3 quadratic subfields are $\mathbf Q(\sqrt 3), \mathbf Q(\sqrt 5), \mathbf Q(\sqrt {15})$, so $K$ necessarily coincides with $\mathbf Q(\sqrt 3 - \sqrt 5)$, and $L=\mathbf Q(\alpha)$ is a quadratic extension of $K$. It is classically known (and easy to show) that $L/\mathbf Q$ is normal iff, for any $h\in G$=$Gal(K/\mathbf Q)$, all prolongations $\tilde h$ of $h$ to $\mathbf Q$-embeddings of $L$ into $ \mathbf{\bar Q}$ stabilize $L$, i.e. $h(\sqrt 3 - \sqrt 5)$ differs multiplicatively from $(\sqrt 3 - \sqrt 5)$ by a square in $K^*$. But $G$ is generated by 2 automorphisms $\sigma, \tau$ defined by $\sigma (\sqrt 3)=\sqrt 3, \sigma (\sqrt 5)=-\sqrt 5$, and $\tau(\sqrt 3)=-\sqrt 3, \tau(\sqrt 5)=\sqrt 5$, so that the normality of $L/\mathbf Q$ is equivalent to $(\sqrt 3 - \sqrt 5)(\sqrt 3 + \sqrt 5)$ and $(\sqrt 3 - \sqrt 5)(-\sqrt 3 - \sqrt 5)$ being squares in $K^*$, or $-2$ and $2$ being squares in $K^*$. This cannot happen because of the known list of quadratic subfields of $K$.