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

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!

• By $\sqrt{3+4i}$ do you mean $2+i$, or $-2-i$? And what do you mean by $\sqrt{3-4i}$? – Lord Shark the Unknown Jan 11 '19 at 18:36
• @LordSharktheUnknown This is a formal field, so I think it doesn't matter what they are as complex numbers? The definition is, start from $\mathbb{Q}(x,y,z)$, mod out by $x^2 = -1$, $y^2 = 3 + 4x$, and $z^2 = 3 - 4x$. Then look at the subfield generated by $y + z$. – 6005 Jan 11 '19 at 19:18
• Hmm, actually maybe it does matter. When adjoining $z = \sqrt{3 - 4i}$, we know that it's a solution to the equation $y^2 z^2 = 25$, i.e. $yz = 5$ or $yz = -5$. So $z$ is either $5/y$ or $-5/y$, and it isn't really a new element at all -- but which is it? – 6005 Jan 11 '19 at 19:25
• And as you pointed out it's the same problem with $y$, it can either be $2 + i$ or $2 - i$. – 6005 Jan 11 '19 at 19:27
• Cool, $\mathbb{3+4i}$ looks fancy. – Dietrich Burde Jan 11 '19 at 19:39

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 $$\sqrt{3+4i}\,\sqrt{3-4i}=\pm 5.$$
• Oh! yes. (albeit for me, it's a $\mathbf Q$). Thanks for pointing it! – Bernard Jan 11 '19 at 19:37
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 $$\mathbb{Q}$$.