About the primitive elements of $\mathbb{Q}(\sqrt{3},i\sqrt{5})$

Question

(I have already prove that $\mathbb{Q}(\sqrt{3},i\sqrt{5})=\mathbb{Q}(\sqrt{3}+i\sqrt{5})$ in case is useful); now I am asked to prove that if $v\in \mathbb{Q}(\sqrt{3},i\sqrt{5})$ has the property that every image of $v$ by elements of $G$ (Galois group of $\mathbb{Q}(\sqrt{3},i\sqrt{5})/\mathbb{Q}$) is different from each other, $v$ is a primitive element of the extension (i.e. $\mathbb{Q}(v)=\mathbb{Q}(\sqrt{3},i\sqrt{5})$). Here, I write my thoughts about it.

So, we can proceed by first calculating the Galois group of the extension, that has (this is just calculating) $4$ elements, that I represent in this table (taking on account that in this case each $\mathbb{Q}$-automorphism of the extension is determined by the image of $\sqrt{3}$ and $i\sqrt{5}$, and that each of this images can have values among the roots of $x^2-3$ and $x^2+5$, respectively):

Elements of $G$	Image of $\sqrt{3}$	Image of $i\sqrt{5}$
$\sigma_{1}$	$\sqrt{3}$	$i\sqrt{5}$
$\sigma_{2}$	$\sqrt{3}$	$-i\sqrt{5}$
$\sigma_{3}$	$-\sqrt{3}$	$i\sqrt{5}$
$\sigma_{4}$	$-\sqrt{3}$	$-i\sqrt{5}$

So, taking on account, that being $v\in \mathbb{Q}(\sqrt{3},i\sqrt{5})$,implies that $v=a+b\sqrt{3}+ci\sqrt{5}+di\sqrt{3}\sqrt{5}$, with $a,b,c,d\in\mathbb{Q}$, we can assure that, due to the fact that elements of $G$ are $\mathbb{Q}$-automorphisms, $\forall \sigma \in G$: $$ \sigma(v)=a+b\sigma(\sqrt{3})+c\sigma(i\sqrt{5})+d\sigma(\sqrt{3})\sigma(i\sqrt{5}) $$

So, if every image of $v$ by elements of $G$ is different from each other, this four expressions are different to each other:

• $a+b\sqrt{3}+ci\sqrt{5}+di\sqrt{3}\sqrt{5}$
• $a+b\sqrt{3}-ci\sqrt{5}-di\sqrt{3}\sqrt{5}$
• $a-b\sqrt{3}+ci\sqrt{5}-di\sqrt{3}\sqrt{5}$
• $a-b\sqrt{3}-ci\sqrt{5}+di\sqrt{3}\sqrt{5}$

But, from here, I don't know how to proceed...

Then I have thought another way... Supposing that $v$ is not a primitive element and trying to see that there is at least a pair among those expressions shown above that are the same. But I am not able to get anything clear from here neither... I am lost.

I would appreciate some hint/guidance! Thanks in advanced!

Answer 1

It's quite simple by Galois theory. Let $$K=\mathbb Q(\sqrt 3, i\sqrt 5), L=\mathbb Q(v)$$ $$H=\{g\in G: g|_L=\text{id}_L\}=\{g\in G: g(v)=v\} = \{\text{id}_G\}$$ then by Galois corresponence, $$[L:\mathbb Q] = [G:H] = |G|=[K:\mathbb Q]$$ Hence $L=K$

Answer 2

Indeed it has nothing to do with your specific Galois extension:

Let $K/F$ be a Galois extension with Galois group $G$ and $v \in K$.

Then, the minimal polynomial of $v$ over $F$ is $\prod_{v' \in Gv} (x - v')$ (proved below), with degree $\#\!\operatorname{Orbit}_G(v)$.

Therefore, $[K : F(v)] = \#\!\operatorname{Stab}_G(v)$ and $[F(v) : F] = \#\!\operatorname{Orbit}_G(v)$.

So if $\#\!\operatorname{Stab}_G(v) = 1$ (i.e. every image of $v$ by elements of $G$ is different from each other), then $K = F(v)$.

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Clearly the polynomial $p(x) = \prod_{v' \in Gv} (x - v')$ is fixed by $G$, so $p \in F[x]$.

Let $q | p$ be the irreducible factor with $v$ as a root. Then, in $K[x]$, we have $(x - v) | q$, so $(x - \sigma(v)) | q$ for any $\sigma \in G$, so $p | q$.

Therefore, $p$ is irreducible in $F[x]$.