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My general question is to find, if this is possible, two real numbers $a,b$ such that $K=\Bbb Q(a,b)$ is not a simple extension of $\Bbb Q$. $\newcommand{\Q}{\Bbb Q}$

Of course $a$ and $b$ can't be both algebraic, otherwise $K$ would be a separable ($\Q$ has characteristic $0$) and finite extension, which has to be simple. So I tried with $\Q(\sqrt 2, e)$ but any other example would be accepted.

The field $\Q(\sqrt 2, e)$ has transcendence degree $1$ over $\Q$, but I'm not sure if this imply that it is isomorphic to $\Q(a)$ for some transcendental number $a$ (the fact that two fields have the same transcendence degree over another field shouldn't imply that the fields are isomorphic).

I'm not sure about the relation between the algebraic independence of $a$ and $b$, and the fact that $\Q(a,b)/\Q$ is a simple extension. Notice that $\Q(\pi, e)$ is probably unknown to be a simple extension of $\Q$.

Thank you for your help!

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  • $\begingroup$ For every transcendental number $\alpha$, ${\mathbb Q}(\alpha) \cong {\mathbb Q}(X)$, so all simple extensions of ${\mathbb Q}$ of transcendance degree $1$ are isomorphic. $\endgroup$ Aug 17 '16 at 19:10
  • $\begingroup$ @Magdiragdag: yes, obviously, but is every extension of $\Bbb Q$ with transcendence degree $1$ over $\Bbb Q$ of the form $\Bbb Q(X)$? $\endgroup$
    – Watson
    Aug 17 '16 at 19:12
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    $\begingroup$ @Watson No, as ${\mathbb Q}(\sqrt 2, e)$ is now an obvious counterexample. It has transcendance degree 1 and is not isomorphic to ${\mathbb Q}(X)$, since it contains $\sqrt 2$ and ${\mathbb Q}(X)$ does not contain an element whose square is $2$. $\endgroup$ Aug 17 '16 at 19:14
  • $\begingroup$ @Interested people: here is another example of non-simple extension: math.stackexchange.com/questions/333392 $\endgroup$
    – Watson
    Aug 22 '16 at 13:05
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The extension $\mathbb{Q}(\sqrt{2},e)\supset\mathbb{Q}$ is not simple. If $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},e)$, then $\mathbb{Q}(u)$ is infinite-dimensional over $\mathbb{Q}$, so $u$ is transcendental. But then $\mathbb{Q}(u)$ is purely transcendental over $\mathbb{Q}$ while $\mathbb{Q}(\sqrt{2},e)$ is not.

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  • $\begingroup$ Ah yes… so easy! Well done! $\endgroup$
    – Watson
    Aug 17 '16 at 19:14
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You have to show that $$ \mathbb{Q}(X) \subsetneq \mathbb{Q}(e,\sqrt{2}) $$ for any $X \in \mathbb{Q}(e,\sqrt{2})$.

  • If $X$ is algebraic, then $[\mathbb{Q}(X) : \mathbb{Q}]$ is finite while $[\mathbb{Q}(e,\sqrt{2}): \mathbb{Q}]$ is infinite.

  • If $X$ is not algebraic, then it is transcendental. It suffices to show that $\mathbb{Q}(X)$ does not contain a square root of $2$. Since $\mathbb{Q}(X)$ is isomorphic to the fraction field of polynomials, you need to show that there do not exist polynomials $p(X)$, $q(X)$ with rational coefficients such that $$ \left( \frac{p(X)}{q(X)} \right)^2 = 2. $$ Can you take it from here?

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  • $\begingroup$ Thank you for your answer. If $a$ and $b$ are required to be algebraically independent over $\Bbb Q$, then 1) do you know if $\Bbb Q(a,b)/\Bbb Q$ can be a simple extension? 2) do you know if there is also an example where $a,b$ are algebraically independent over $\Bbb Q$ and $\Bbb Q(a,b)/\Bbb Q$ is not simple? $\endgroup$
    – Watson
    Aug 17 '16 at 20:20
  • $\begingroup$ (I can also a new question for this, if necessary ;-)) $\endgroup$
    – Watson
    Aug 17 '16 at 20:20
  • $\begingroup$ (This is just to know what the eventual algebraic independence of $e$ and $\pi$ could tell us about the extension $\Bbb Q(e,\pi)$, e.g. is it simple or not). $\endgroup$
    – Watson
    Aug 17 '16 at 20:22
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    $\begingroup$ @Watson The answer to (1) is no, because $\mathbb{Q}(a,b)$ has trasncendence degree 2. For (2), yes, you can construct many examples for instance with the Lindemann-Weierstrass theorem. So, $\mathbb{Q}(e, e^{\sqrt{2}})$ would work for example. It would be fine to ask a new quesiton, too, and link back to this one. $\endgroup$
    – 6005
    Aug 17 '16 at 20:26
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Call $K=\Bbb{Q}(\sqrt{2})$. Then $\Bbb{Q}(\sqrt{2} , e)$ is isomorphic to $K(x)$, the fraction field of $K[x]$. If this were a simple extension of $\Bbb{Q}$, it would be isomorphic to $\Bbb{Q}(x)$, the fraction field of $\Bbb{Q}[x]$. So $$\Bbb{Q}(x) \cong K(x)$$ But this contradicts the fact that $X^2-2 \in \Bbb{Q}(x)[X]$ has no root in $\Bbb{Q}(x)$.

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