Find the $\text{irr}_{(\Bbb{Q(\sqrt{2}),\sqrt{2}+\sqrt{3}})}(x)$

Question

We want to find the minimal polynomial $\text{irr}_{(\Bbb{Q(\sqrt{2}),\sqrt{2}+\sqrt{3}})}(x)$ of $\sqrt{2}+\sqrt{3}\in\Bbb{R}$ over $\Bbb{Q(\sqrt{2})}$.

Answer: We can proove that $\Bbb{Q(\sqrt{2}+\sqrt{3})}=\Bbb{Q(\sqrt{2},\sqrt{3})}$.

We also have that the minimal polynomial of $\sqrt{3}$ over $\Bbb{Q}(\sqrt{2})$ is $x^2-3$, so $[\Bbb{Q}(\sqrt{2})(\sqrt{3}):\Bbb{Q}(\sqrt{2})]=[\Bbb{Q}(\sqrt{2},\sqrt{3}):\Bbb{Q}(\sqrt{2})]=2.$

But $\Bbb{Q(\sqrt{2}+\sqrt{3})}=\Bbb{Q}(\sqrt{2},\sqrt{3})=\Bbb{Q}(\sqrt{2}+\sqrt{3},\sqrt{2})=\Bbb{Q}(\sqrt{2}+\sqrt{3})(\sqrt{2})$, and from this

$$[\Bbb{Q}(\sqrt{2})(\sqrt{2}+\sqrt{3}):\Bbb{Q}(\sqrt{2})]=2=\deg \text{irr}_{(\Bbb{Q(\sqrt{2}),\sqrt{2}+\sqrt{3}})}(x) . $$

If we set $\beta:=\sqrt{2}+\sqrt{3} \implies \beta-\sqrt{2}=\sqrt{3} \implies \beta^2-2\sqrt{2}\beta-1=0.$ So, we take the irreducible and monic polynomial, which has $\beta$ as a root $$\text{irr}_{(\Bbb{Q(\sqrt{2}),\sqrt{2}+\sqrt{3}})}(x) =x^2-2\sqrt{2}x-1 \in \Bbb{Q}(\sqrt{2})[x]$$

Is this completely right? Thank you in advance.

Answer 1

My approach

Take $\sqrt2 + \sqrt3$ and its conjugate $\sqrt2-\sqrt3$.

Then, the minimal polynomial is $[x-(\sqrt2+\sqrt3)][x-(\sqrt2-\sqrt3)]$, which is $x^2-2\sqrt2x-1$, matching with your answer.

The proof that it is minimal is left to the reader as an exercise.

Answer 2

Another approach. Since $\sqrt3\notin\Bbb Q(\sqrt2\,)$, you get $\text{Irr}(\sqrt3,\Bbb Q(\sqrt2))=X^2-3$. Call this $f(X)$. Then clearly $f(X-\sqrt2\,)=\text{Irr}(\sqrt3+\sqrt2,\Bbb Q(\sqrt2))$. In other words, the minimal polynomial is $(X-\sqrt2)^2-3=X^2-2\sqrt2\,X+2-3$, and there you are.