Please Check my proof: $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2 +\sqrt3)$

Question

$\mathbb{Q}(\sqrt2 + \sqrt3) \subset \mathbb{Q}(\sqrt2,\sqrt3)$ is obvious. Now for the converse. Since $p(x) = x^4 - 10x^2 + 1$ has $\sqrt2 + \sqrt3$ as a root, is monic and irreducible in $\mathbb{Q}$, $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}] = \text{deg}(p(x)) = 4$. However, $[\mathbb{Q}(\sqrt2,\sqrt3) : \mathbb{Q}] = 4$ and since $\mathbb{Q}(\sqrt2 + \sqrt3)$ is a subspace of $\mathbb{Q}(\sqrt2, \sqrt3)$ from the first implication, we must have equality as they have equal dimensions.

Is this correct? Thank you for your time.

Answer 1

You ought to give a reason for $p$ being irreducible. Also, your $p(x)$ has a typo in it; it should be $x^4-10x^{\color{red}2} + 1$).

You can prove it more directly. Let's set $t = \sqrt2+\sqrt3$ because it's easier to write. Then $t^3 = 11\sqrt2+9\sqrt3$. This gives us $$ \sqrt2 = \frac12(t^3-9t)\in\Bbb Q(t)\\ \sqrt3 = \frac12(11t-t^3)\in \Bbb Q(t) $$ which proves that $\Bbb Q(\sqrt2, \sqrt3)\subseteq \Bbb Q(t)$

Answer 2

I'll offer a slight variant on Arthur's argument. Let's prove $\sqrt{a},\,\sqrt{b}\in\mathbb{Q}(\sqrt{a}+\sqrt {b})$ for any $a,\,b\in\mathbb{N}$ that aren't perfect squares with $a<b$, e.g. $a=2,\,b=3$. Define $t:=\sqrt{a}+\sqrt{b}$ so $$\sqrt{b}-\sqrt{a}=\frac{b-a}{t},\,\sqrt{a}=\frac{1}{2}(t-\frac{b-a}{t}),\,\sqrt{b}=\frac{1}{2}(t+\frac{b-a}{t}).$$