Showing $\mathbf Q(\sqrt2,\sqrt3)=\mathbf Q(\sqrt2+\sqrt3)$

Question

Showing $\mathbf Q(\sqrt2,\sqrt3)=\mathbf Q(\sqrt2+\sqrt3)$

BUT I want to show this using The Theorem of the Primitive Element,

So I have to verify that $c$ cannot be $1$ and I need the $\mathbf Q$-monomorphisms, what are these ?

Answer 1

Well, there's a mistake in my argument which is detailed in the comment section.

I think it is beneficial to leave this answer as is in case somebody comes upon this and can then learn from it too.

My original post:

Here's one way to do it:

Theorem

Let $F$ be an infinite field. If $F(\alpha_1,\alpha_2)$ is a simple and algebraic extension of $F$ then $\exists t \in F$ s.t $F(\alpha_1,\alpha_2) = F(\alpha_1 + t\alpha_2)$.

This can be generalized for any n BTW

Proof:

I'll assume you know (since we're using the primitive element theorem) that under these conditions there is only a finite number of sub-fields in $F$.

$F$ is infinite then $\exists t_1 \neq t_2 \in F$ s.t $F(\alpha_1 + t_1\alpha_2) = F(\alpha_1 + t_2\alpha_2)$.

Then $\alpha_2 = \frac{(\alpha_1 + t_1\alpha_2)-(\alpha_1 + t_2\alpha_2)}{t_1-t_2} \in F(\alpha_1 + t_1\alpha_2)$ then clearly $\alpha_1 \in F(\alpha_1 + t_1\alpha_2)$ as well.

This ends that theorem.

Now $\mathbb{Q}(\sqrt{2},\sqrt{3})$ applies to the above by the primitive root theorem.

Then $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+ t\sqrt{3})$

Now notice that $t \neq 0$ necessarily and that $t\sqrt{2} + t\sqrt{3} = a_0 + a_1\sqrt{2} + a_1t\sqrt{3} $

implying $ 0 = a_0 + (a_1-t)\sqrt{2} + (a_1t-t)\sqrt{3} $ and as $\{a_0,\sqrt{2},\sqrt{3}\}$ is a linearly independent set we have $a_1 = t = a_1t$ then $t=1$