Proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$

Question

Here is the question I want to answer:

Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$

Here is what I managed to prove so far:

I managed to write $\sqrt{2}$ as a function in $\sqrt{2} + \sqrt{3}$ and $\sqrt{3}$ as a function in $\sqrt{2} + \sqrt{3}$ as can seen below:

$\textbf{Finding f(x).}$

Since $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{2.3}.$ Then we have that \begin{align*} (\sqrt{2} + \sqrt{3})^2 (\sqrt{2} + \sqrt{3}) &= (5 + 2 \sqrt{2.3}) (\sqrt{2} + \sqrt{3} )\\ &= 5 \sqrt{2} + 5 \sqrt{3} + 4 \sqrt{3} + 6 \sqrt{2} \\ &= 11 \sqrt{2} + 9 \sqrt{3} \\ &= 9 (\sqrt{2} + \sqrt{3}) + 2 \sqrt{2} \end{align*}

Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 9x + 2 \sqrt{2}.$$ And so we have that $$\sqrt{2} = \frac{x^3}{2} - \frac{9x}{2} = f(x).$$\

$\textbf{Finding g(x).}$ \begin{align*} (\sqrt{2} + \sqrt{3})^3 &= 11 \sqrt{2} + 9 \sqrt{3}\\ &= 11 \sqrt{2} + 9 \sqrt{3} + 2 \sqrt{3} - 2 \sqrt{3}\\ &= 11 (\sqrt{2} + \sqrt{3}) - 2 \sqrt{3} \end{align*}

Then if we put $x = \sqrt{2} + \sqrt{3},$ we will get that $$x^3 = 11x - 2 \sqrt{3}.$$ And so we have that $$\sqrt{3} = \frac{- x^3}{2} + \frac{11x}{2} = g(x).$$\

And I understand that this proves this inclusion $\mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2} + \sqrt{3}] $

My question is how can I prove the other inclusion $\mathbb Q(\sqrt2+\sqrt3)\subseteq \mathbb Q(\sqrt2,\sqrt3)$? could anyone help me in showing this please pointing out for me what exactly the definition of $\mathbb Q(\sqrt2,\sqrt3)$ for me.

Answer 1

You are asking about definition for $\Bbb{Q}(\sqrt{2},\sqrt{3})$.

Let me add some more.

Suppose $F$ is a field of a subfield $K$ and $\alpha$ is an element of $K$. Then the collection of all subfields of $K$ containing both $F$ and $\alpha$ is nonempty(why!). Since the intersection of subfields is again a subfield, it implies that there is a unique smallest subfiles containing both $F$ and $\alpha$. Similarly, you can replace $\alpha$ by a collection $\alpha,\beta,\dots$ of elements of $K$.

We conclude a definition from above,

Definition Let $K$ be an extension of the filed $F$ and let $a,b,c,\dots\in K$ be collection of elements in $K$. Then the smallest field containing both $F$ and the elements $a,b,c\dots$ denoted $F(a,b,c,...)$ is called the field generated by $a,b,c\dots$ over $F$.

Clearly $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\Bbb{Q},\sqrt{2}$ and $\sqrt{3}$. i.e., $\sqrt{2},\sqrt{3}\in \Bbb{Q}(\sqrt{2},\sqrt{3})$.

Note that $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is a field. So by the closure property w.r.t. addition, $\sqrt{2}+\sqrt{3}\in\Bbb{Q}(\sqrt{2},\sqrt{3})$

$\Rightarrow \Bbb{Q}(\sqrt{2}+\sqrt{3})\subseteq \Bbb{Q}(\sqrt{2},\sqrt{3})$