Show $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$

Solution 1:

Michael Rozenberg has given a fine direct (i.e. two inclusions) proof that $\Bbb Q(\sqrt{5}+\sqrt{7}) = \Bbb Q(\sqrt{5},\sqrt{7})$, without needing a or b.

Your exercise as meant as an alternative proof. First find a minimal polynomial for $\alpha = \sqrt{5}+\sqrt{7}$. Note that $$\alpha^2 = 5+7 + 2\sqrt{35}$$ so

$$\alpha^2 - 12 = \sqrt{140}$$ and squaring elimates the final square root and

$$(\alpha^2 - 12)^2 =140$$ which simplifies to

$$\alpha^4 - 24\alpha^2 + 4 = 0$$

and so $p(x)=x^4 - 24x^2 + 4$ has $\sqrt{5} + \sqrt{7}$ as a root. If $p(x)$ is irreducible (Eisenstein does not apply as $p=2$ is the only candidate and fails) we know it is a minimal polynomial for $\alpha$. We'll leave it for now, we have $p(x)$ that has $\alpha$ as a zero.

The degree of $$[\Bbb Q(\sqrt{5},\sqrt{7}): \Bbb Q] = [\Bbb Q(\sqrt{5},\sqrt{7}): \Bbb Q(\sqrt{7})] \cdot [\Bbb Q(\sqrt{7}): \Bbb Q] = 2\times 2 = 4$$

by the standard degree formula. And $\alpha \in \Bbb Q(\sqrt{5},\sqrt{7})$ trivially and so the degree of $\alpha$ divides the degree of the extension it's in, i.e. $4$. So the minimal polynomial $m(x)$ of $\alpha$ (which always exists) has degree $4$ and by standard facts $m(x) | p(x)$. So $m(x)$ has degree dividing $4$ and $p$ has degree $4$ and both are monic, so it follows that $p(x)=m(x)$ and indeed $p(x)$ must be the minimal polynomial.

c. then follows as $\Bbb Q(\alpha)$ is an algebraic extension of $\Bbb Q$ of degree $4$ (because of $p$) inside the field extension $\Bbb Q(\sqrt{5},\sqrt{7})$ of degree $4$. Basic linear algebra..

Solution 2:

$$\sqrt7-\sqrt{5}=\frac{2}{\sqrt7+\sqrt5}\in\mathbb Q(\sqrt7+\sqrt5).$$ Thus, $$\sqrt7=\frac{\sqrt7+\sqrt5+\sqrt7-\sqrt5}{2}\in\mathbb Q(\sqrt7+\sqrt5)$$ and $$\sqrt5\in\mathbb Q(\sqrt7+\sqrt5).$$ Also, it's obvious that $$\mathbb Q(\sqrt7+\sqrt5)\subset\mathbb Q(\sqrt7,\sqrt5)$$