# Prove that $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a}, \sqrt{b})$ without Galois theory

Let $a,b \in \mathbb{Z}$ be integers such that $\sqrt{a} \notin \mathbb{Z}$ and $\sqrt{b} \notin \mathbb{Z}$ (the number $a$ is allowed to be negative). I need to prove that

$$\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a}, \sqrt{b})$$

but I cannot use any Galois theory to do so, as we have not gotten to it yet in my course.

I was given a hint: "let $G$ and $H$ denote the left-hand side and the right-hand side, respectively; analyze possibilities for the dimension $[H:G]$ and use thee fact that $G(\sqrt{a}) = G(\sqrt{b}) = H$".

However, instead, I approached it similarly to Paramanand Singh's answer to this question. But, I am not sure such an approach is correct in this situation.

Anyway, this is what I did, according to Paramanand's method:

It is obvious that $\mathbb{Q}(\sqrt{a} + \sqrt{b}) \subseteq \mathbb{Q}(\sqrt{a}, \sqrt{b})$, since $\sqrt{a} + \sqrt{b}$ is a linear combination of $\sqrt{a}$ and $\sqrt{b}$

Now, for the other side, let $c = \sqrt{a}+\sqrt{b}$.

WTS: $e=\sqrt{a}$, $f = \sqrt{b}$ are rational functions of $c$.

$$(c-e)^{3} = b \, \to \, c^{3}-3c^{2}e + 3ac - ae = b \, \to \, 3c^{2}e + ae = -b + c^{3}+3ac \, \to \, e = \frac{c^{3} + 3ac - b}{3c^{2} + a}\, \to \, \sqrt{a} = \frac{c^{3} + 3ac -b }{3c^{2}+a}$$

So that $\sqrt{a}$ is a rational function of $c$.

Also, since $c = e + f = \sqrt{a} + \sqrt{b}$, we have that $\sqrt{b} = c - \sqrt{a}$, so that $\sqrt{b}$ is also a rational function of $c$.

Therefore $\mathbb{Q}(\sqrt{a}, \sqrt{b}) \subseteq \mathbb{Q}(\sqrt{a} + \sqrt{b})$

However, something about this is not sitting right with me. Is this, in fact, the correct way to prove it? Or should I do it the way my professor suggested? I would also like some help doing it that way.

I guess for that way, we could still start out by stating that "It is obvious that $\mathbb{Q}(\sqrt{a} + \sqrt{b}) \subseteq \mathbb{Q}(\sqrt{a}, \sqrt{b})$, since $\sqrt{a} + \sqrt{b}$ is a linear combination of $\sqrt{a}$ and $\sqrt{b}$" (if this is, in fact, true). Then, $\mathbb{Q}(\sqrt{a}, \sqrt)$ has degree $6$, right?

So, I would need to show that $\mathbb{Q}(\sqrt{a} + \sqrt{b})$ also has degree $6$, but I still don't see what this has to do with the hint...

If I let $G = \mathbb{Q}(\sqrt{a} + \sqrt{b})$ and $H = \mathbb{Q}(\sqrt{a}, \sqrt{b})$, then of course $G(\sqrt{a}) = \mathbb{Q}(\sqrt{a} + \sqrt{b})(\sqrt{a}) = \mathbb{Q}(\sqrt{a} + \sqrt{b})(\sqrt{b})$, but I still don't see how this helps!

Could someone please explain this to me? I would be most appreciative! Thank you!

Hint about your instructor's hint: The hint suggests that you look at $[H:G]$ by considering $[G[\sqrt{a}]:G]$ and $[G[\sqrt{b}]:G]$ which both equal $[H:G]$. Now since $\sqrt{a}\not\in\Bbb{Z}$ and $\sqrt{b}\not\in\Bbb{Z}$, $x^2-a$ and $x^3-b$ are irreducible over $\Bbb{Z}$. Looking at each of those polynomials over $G$ should give you a possibility for $[H:G]$ and considering them together will give the answer.
On the one hand considering $H=G[\sqrt{a}]$, we see $[H:G]|2$, but $H=G[\sqrt{b}]$ tells us $[H:G]|3$. Together these imply $[H:G]=1$
• can you please explain how we know that $[H:G]$ divides $2$ when $H = G(\sqrt{a})$ and that $[H:G]$ divides $3$ when $H = G(\sqrt{b})$? I assume it comes from the result that if $F, K, M$ are fields such that $F \subseteq K \subseteq M$, then $[M:F]=[M:K][K:F]$, but I can't figure out the inclusions here. – ALannister Aug 18 '17 at 3:14
• The polynomials $x^2-a$ and $x^3-b$ are irreducible over $\Bbb{Z}$. Over the field you called $G$ in your post $x^2-a$ must split or be irreducible (since it's a quadratic), so $[H:G]$ must be $1$ or $2$. The polynomial $x^3-b$ could have one root, split, or be irreducible over $G$. In the first two cases, we'd be done because $[H:G]$ would then equal $1$. Otherwise $[H:G]$ must equal $3$ in which case we come back to $[H:G]$ divides $2$ which is impossible if $[H:G]=3$ – sharding4 Aug 18 '17 at 3:26