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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.

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    $\begingroup$ Have you tried any fields of characteristic $p\not=0$? $\endgroup$ – gobucksmath Apr 22 '17 at 14:00
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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 $\sqrt 5 + \sqrt 7$.

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    $\begingroup$ I think all you need to do is square $\alpha-\sqrt{5}$ (where $\alpha=\sqrt{5}+\sqrt{7}$) to get an expression for $\sqrt{5}$ in terms of $\alpha$. $\endgroup$ – ancientmathematician Apr 22 '17 at 14:04
  • $\begingroup$ @ancientmathematician Yes, that is a nice way to do it. Anyway, it's not too hard to verify that $\alpha^3 = 26 \sqrt 5 + 22 \sqrt 7$, and from this point on, it's a matter of solving simultaneous linear equations. $\endgroup$ – Kenny Wong Apr 22 '17 at 14:06
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    $\begingroup$ Oh, thank you. The question in the book says "Is $\mathbb{Q}(\sqrt{5},\sqrt{7})$ simple? If not, why not?". Supposing that the example given wasn't a simple extension, what would be the method to go about proving it given that the only tools given so far are the definition of a simple extension and the minimum polynomial? $\endgroup$ – Jake Apr 22 '17 at 14:09
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    $\begingroup$ I don't know an elementary answer to this I'm afraid. Perhaps you could ask this as a separate question? However, all examples of the form $\mathbb Q(\alpha_1, \alpha_2, \dots, \alpha_n) : \mathbb Q$, where $\alpha_i$ are solutions of rational polynomial equations, are simple. This is a consequence of something called the primitive element theorem, which I believe appears later in the book. $\endgroup$ – Kenny Wong Apr 22 '17 at 14:12
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    $\begingroup$ I think that given that at this stage you have no theorems to help you'd just have to show that no element could generate an extension of such a degree. For example, let $K:=\mathbb{F}_2(s^2,t^2)$ where $s,t$ are independent transcendentals, and consider the extension $K[s,t]$ of degree 4. You can prove easily that all elements satisfy a quadratic polynomial at worst). $\endgroup$ – ancientmathematician Apr 22 '17 at 14:24
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Every finite separable extension is simple extension. Thus, you can not prove that this is not simple.

The statement that I have quoted is the statement for what is called Primitive element theorem.

If you want to construct some extensions which are not simple think about extensions that are not separable.

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  • $\begingroup$ Thank you for the answer, I suppose I'll have to wait until the chapter on separability. $\endgroup$ – Jake Apr 22 '17 at 14:27
  • $\begingroup$ @Jake yes. You have to wait until chapter on separability or consider looking at some infinite extensions. As you do not get to see many examples of infinite extensions, you have to wait till you read about separability. $\endgroup$ – user87543 Apr 22 '17 at 14:38
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Observe, for example, that

$$\sqrt7-\sqrt5=\frac2{\sqrt7+\sqrt5}\in\Bbb Q(\sqrt5+\sqrt7)\;$$, and thus

$$\sqrt7=\frac{(\sqrt7+\sqrt5)+(\sqrt7-\sqrt5)}2\in\Bbb Q(\sqrt5+\sqrt7)$$

and similarly for $\;\sqrt 5\;$ , and thus $\;\Bbb Q(\sqrt5,\,\sqrt7)=\Bbb Q(\sqrt5+\sqrt7)\;$

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  • $\begingroup$ Thank you for the explanation. $\endgroup$ – Jake Apr 22 '17 at 14:26

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