a) In $\mathbb{R}$, $\sqrt{2}$ and $\sqrt{3}$ are algebric over $\mathbb{Q}$. Find the polynomial of degree $4$ over $\mathbb{Q}$ satisfiable by $\sqrt{2}+\sqrt{3}$
b) Wich is the degree of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$?
c) With is the degree of $\sqrt{2}\cdot \sqrt{3}$ over $\mathbb{Q}$?
I've found (a) here: Find the minimal polynomial of $\sqrt2 + \sqrt3 $ over $\mathbb Q$
For $b$, what is the degree of an element over $\mathbb{Q}$ and how to justify it?
$\endgroup$ 01 Answer
$\begingroup$It is: $a=\sqrt{2}+\sqrt{2}$, then $a^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$. Hence $a^2-5=2\sqrt{6}$ and $(a^2-5)^2=24$, so
$a^4-10a^2+1=0$ and the polynomial is given by $X^4-10X^2+1\in\mathbb{Q}[X]$, which is irreducible.
Edit: So the degree of $\sqrt{2}+\sqrt{3}$ is $4$. It is the degree of the minimal polyomial.
c) $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ and it is $X^2-6$ the minimal polynomial.
$\endgroup$ 3
