I'm reading through Stewart's Galois theory textbook, and it's easy to find examples both in the textbook and on the internet regarding proofs that certain algebraic extensions are simple. For example, that $\mathbb{Q}(\sqrt2, \sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt{3})$
My question is, how do I go about proving that an extension is not simple? For example $\mathbb{Q}(\sqrt5,\sqrt7)$. So far I think I've shown that $\mathbb{Q}(\sqrt5,\sqrt7) \neq \mathbb{Q}(\sqrt5 + \sqrt7)$ and also that $\sqrt5, \sqrt7$ are linearly independent over $\mathbb{Q}$, therefore $\mathbb{Q}(\sqrt5,\sqrt7) \neq \mathbb{Q}(\sqrt5)$ and $\mathbb{Q}(\sqrt5,\sqrt7) \neq \mathbb{Q}(\sqrt7)$
I don't think this is sufficient; I need to show in general that there is no $\alpha \in \mathbb{C}$ such that $\mathbb{Q}(\sqrt5,\sqrt7) = \mathbb{Q}(\alpha)$, and I don't think I've exhausted all possibilities with the three cases above. I thought I should see how the minimum polynomial for $\alpha$ might look, but I'm not sure where to begin there.
Also, the question as to whether $\mathbb{Q}(\sqrt5,\sqrt7)$ is simple is given in Stewart right after the simple extension is defined, before any tools like the tower law, normality, seperability etc. are introduced. Therefore I think it's expected that the reader solves this using simple definitions.
Best Answer
I believe the example you're looking at is simple, and $\mathbb Q(\sqrt 5, \sqrt 7)$ does equal $\mathbb Q(\sqrt 5 + \sqrt 7)$.
Since this example comes so early in the book, and you're looking for an elementary way to see this, I suggest you expand $(\sqrt 5 + \sqrt 7)^2$ and $(\sqrt 5 + \sqrt 7)^3$. Then you could try to write $\sqrt 5$ and $\sqrt 7$ as polynomials in $\sqrt 5 + \sqrt 7$.