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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?

Paprika
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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.

Cornman
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  • How to show that is is irreducible? – Paprika Sep 10 '17 at 23:52
  • You have more possible ways. It is enough to observe the integers that divide 1. So 1 and -1. $(1)^4-10\cdot (1)^2+1=-8$ so no root. And $(-1)^4-10\cdot (-1)^2+1=-8$. Still no root. You also can use the "reduction criteria" for $\mathbb{F}_3[X]$ and show that is has no root in this field. Which stats that the polynomial is irreduible in $\mathbb{Q}[X]$. – Cornman Sep 10 '17 at 23:56
  • If a polynomial $\in \mathbb Q[X]$ has no roots in $\mathbb Q$, then it is not necessarily true that the polynomial is irreducible in $\mathbb Q[X]$, because $\mathbb Q$ is not an algebraically closed field. Knowing that this polynomial has no roots helps, though, in reducing the problem of proving irreducibility to checking that it doesn't factor as a product of two quadratics. – Cauchy Sep 11 '17 at 05:40