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I have to determine the degree of $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$ over $\mathbb{Q}$ and show that $\sqrt{2}+\sqrt{3}$ is a primitive element ? Could someone please give me any hints on how to do that ?

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What's the degree of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}(\sqrt{3})$ ? What's the degree of $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$ ? This determines the degree you're looking for. Next, show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ to get the second part. – Bey Oct 30 '12 at 18:39
@Bey The last equality is just the definition of "primitive", which I knew already :( – user47580 Oct 30 '12 at 18:43
Well, obviously $\sqrt{2}+\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. Now show that $\sqrt{2}, \sqrt{3}\in\mathbb{Q}(\sqrt{2}+\sqrt{3})$. – Bey Oct 30 '12 at 18:46
@Bey Show that $\sqrt{2},\sqrt{3}\in \mathbb{Q}(\sqrt{2}+\sqrt{3})$ turned out to be more difficult that I thought. After playing some time I could show that roots like $\sqrt{6}$ or $\sqrt{12}$ are in there, but not for $\sqrt{2},\sqrt{3}$ (frustration :( ). What is the right polynomial ?? – user47580 Oct 30 '12 at 21:47
up vote 6 down vote accepted

Clearly $[\mathbb Q(\sqrt 2):\mathbb Q]\le 2$ becasue of the polynomial $X^2-2$ and $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt 2)]\le 2$ because of the poylnomial $X^2-3$. In fact, $\sqrt 2\notin \mathbb Q$ implies $[\mathbb Q(\sqrt 2):\mathbb Q]=2$. We also have $\sqrt 3\notin \mathbb Q(\sqrt 2)$ because $(a+b\sqrt 2)^2 = 3$ implies $(a^2+2b^2) + 2ab\sqrt 2 = 3$, hence $2ab = 0$ and $a^2+2b^2=3$; thus either $a=0$ and $b^2=\frac 32$, or $b=0$ and $a^2=3$. But both $\sqrt{\frac32}$ and $\sqrt 3$ are irrational. Therefore $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt 2)]=2$ and finally $$[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=4.$$

For the second part , note that $\mathbb Q(\sqrt 2+\sqrt 3)$ contains $(\sqrt 2+\sqrt 3)^2=2+2\sqrt 6+3$, hence also $\sqrt 6$ and $\sqrt6(\sqrt 2+\sqrt 3)=2\sqrt 3+3\sqrt 2$, and finally both $3(\sqrt2+\sqrt 3)-(2\sqrt 3+3\sqrt 2)=\sqrt 2$ and $(2\sqrt 3+3\sqrt 2)-2(\sqrt2+\sqrt 3)=\sqrt 3$.

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$$x=\sqrt 2+\sqrt 3\Longrightarrow x^2-2\sqrt 2\,x+2=3\Longrightarrow x^4-2x^2+1=8x^2\Longrightarrow$$

$$\Longrightarrow x^4-10x^2+1=0$$

Can you now prove the polynomial $\,t^4-10t^2+1\in\Bbb Q[t]\,$ is irreducible?

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