# Is the extension $\mathbb{Q}(\sqrt{5},\sqrt{7})$ simple? If so/not, why?

Exercise sounds: Is the extension $$\mathbb{Q}(\sqrt{5},\sqrt{7})$$ simple? If so/not, why?I have the solution (on picture). Is it correct? Why do we prove in this way, why we must show that square, multiplication belong $$K$$? I saw the definition of the simple extension, but I don't understand why we prove in this way, why it isn't simple.

• The material you quote (image) in fact is a proof that $\mathbb Q(\sqrt 5, \sqrt 7)$ is a simple field extension of the rational numbers. Is that perhaps the the source of your confusion? – hardmath Mar 17 at 23:31
• Oh, I also thought that it is simple, but answers say no, so it make misled me, thanks. @hardmath – Bambeil Mar 17 at 23:35
• It might be a typo, or the problem posed in your textbook may have been phrased in a different way than you did above. It's always a good idea to include the source of problems you want help with. – hardmath Mar 17 at 23:37
• In fact, in characteristic 0 any finite field extension is simple. – Oleg Eroshkin Mar 18 at 0:17

It's simple because it is generated by adjoining a single element. This is shown by the equality $$\def\Q{{\Bbb Q}} \Q(\sqrt5,\sqrt7)=\Q(\sqrt5+\sqrt7)\ .$$
Now by definition $$\Q(\sqrt5,\sqrt7)$$ is a field containing $$\Q$$ and $$\sqrt5$$ and $$\sqrt7$$. It therefore contains $$\sqrt5+\sqrt7$$. By definition, $$\Q(\sqrt5+\sqrt7)$$ is the smallest field containing $$\Q$$ and $$\sqrt5+\sqrt7$$, so RHS is a subset of LHS. Your solution didn't give this working, perhaps because it is regarded as being too easy to bother.
Conversely, $$\Q(\sqrt5+\sqrt7)$$ contains $$\Q$$ and $$\sqrt5+\sqrt7$$. From your solution it therefore contains $$\sqrt5$$ and $$\sqrt7$$. Since LHS is the smallest field containing $$\Q$$ and $$\sqrt5$$ and $$\sqrt7$$ we have LHS a subset of RHS.
There is no specific reason to show that the square is in $$K$$, but it's a convenient way to do it.