What is the minimal polynomial of $\sqrt{3+4i}+\sqrt{3-4i}$ over $\mathbb{Q}$?

Question

My attempt:

I know that $P=\mathbb{Q(\sqrt{3+4i}+\sqrt{3-4i})} \subseteq \mathbb{Q(\sqrt{3+4i},\sqrt{3-4i})}=K$. But I don't know whether $K \subseteq P$. Assuming $K=P$. I can see that K is Galois over $\mathbb{Q}$(as the minimal polynomial $x^4-6x^2+25$ which splits into distinct roots in K) and that would be the minimal polynomial of $\sqrt{3+4i}+\sqrt{3-4i}$.

Thanks for any help!

Answer 1

As the minimal polynomial can be solved, depending on which roots you denote $\;\sqrt{3+4i}$ and $\sqrt{3-4i}$, you can see their sum has degree either $1$ or $2$ on $\mathbf Q$. Hence it can't generate an extension of degree $4$.

On the other hand, it is easy to see that $\;K=\mathbf Q(\sqrt{3+4i})=\mathbf Q(\sqrt{3-4i})$, since $$\sqrt{3+4i}\,\sqrt{3-4i}=\pm 5.$$

Answer 2

Note that $(2+i)^2=3+4i$ and $(2-i)^2=3-4i$. Depending on the choice of roots, the field has degree $1$ or $2$ over $\mathbb{Q}$.