minimal polynomial for $\sqrt{-3}+\sqrt{2}$ over $\mathbb{Q}$

Question

Almost have the answer. Let $a = \sqrt{-3}+\sqrt{2} \implies (\sqrt{-3}+\sqrt{2})(\sqrt{2}-\sqrt{-3}) = 5 \therefore $ a is a root of $x^4+2x^2+25$.

$\sqrt{2} = \frac{1}{2}(a+5/a) \in \mathbb{Q}(a), \sqrt{-3} = a-\sqrt{2}\in \mathbb{Q}(a)\implies \mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{-3})\subset \mathbb{Q}(a)$ and thus $\mathbb{Q}(\sqrt{2},\sqrt{-3})\subset \mathbb{Q}(a)$

$[\mathbb{Q}(a):\mathbb{Q}] \geq [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] \geq 4$.

Thus the degree of the minimal polynomial for $a$ over $\mathbb{Q} \geq 4$. So I'm not sure how to go from there to showing degree of the minimal polynomial for $a$ over $\mathbb{Q} = 4$? Because I think $x^4+2x^2+25$ should be the minimal polynomial. Thanks.

Answer 1

Note that $\sqrt{-3}\notin\mathbb{Q}(\sqrt{2})$, so $[\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})]=2$. Hence $$ [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=4 $$ Now note that $\mathbb{Q}(a)\subseteq\mathbb{Q}(\sqrt{2},\sqrt{-3})$, which together with what you proved yields equality.

Anyway, when you have proved that the degree is at least $4$ and $a$ is the root of a degree $4$ polynomial, you are finished.

Answer 2

If $a$ is a root of a 4th degree polynomial, then the minimal polynomial for $a$ divides a 4th degree

polynomial, so the degree of the minimal polynomial (or the extension field) is at most $4$.

On the other hand, you have shown that the degree of the extension field

(or the minimal polynomial) for $a=\sqrt{-3}+\sqrt2$ over $\mathbb Q$ is at least $4$.

Therefore, the degree of the minimal polynomial for $a$ is exactly $4,$

and $x^4+2x^2+25$ is itself indeed the minimal polynomial for $a$.