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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.enter image description here

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    $\begingroup$ 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? $\endgroup$ – hardmath Mar 17 at 23:31
  • $\begingroup$ Oh, I also thought that it is simple, but answers say no, so it make misled me, thanks. @hardmath $\endgroup$ – Bambeil Mar 17 at 23:35
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    $\begingroup$ 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. $\endgroup$ – hardmath Mar 17 at 23:37
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    $\begingroup$ In fact, in characteristic 0 any finite field extension is simple. $\endgroup$ – Oleg Eroshkin Mar 18 at 0:17
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The working is correct, but in fact it proves that the extension is simple.

It's simple because it is generated by adjoining a single element. This is shown by the equality $$\def\Q{{\Bbb Q}} \Q(\sqrt5,\sqrt7)=\Q(\sqrt5+\sqrt7)\ .$$

The main part of the solution is proving that equality. You have to show that LHS is a subset of RHS and conversely.

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.

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