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I'm looking for a few examples of particular field extensions $\mathbb{Q}(\alpha,\beta) / \mathbb{Q}$.

(i) Is it possible to find a non-simple one? So I would like to show that there exists a field $\mathbb{Q}(\alpha,\beta)$ that cannot be written as $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\gamma)$ for any $\gamma \in \mathbb{Q}(\alpha,\beta)$.
(ii) If (i) is not possible could I still find one such that $\mathbb{Q}(\alpha,\beta)\not = \mathbb{Q}(\alpha + \beta)$

I know about the Primitive element theorem , in particular that every finite separable extension is simple. The thing is that I don't know how to construct a non-finite or non-separable extension, if it even exists, so I guess (i) might not be possible.

For (ii), reading the Wikipedia page of the theorem, I saw that there exists only finitely many $\gamma := \alpha + c\beta$ with $c\in \mathbb{Q}$ that generate $\mathbb{Q}(\alpha , \beta)$, so there could actually be a case where $\alpha + \beta$ satisfies (ii).

Thank you for any answer you might have.

Arctic Char
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Measure me
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    one of the most tractable transcendental field extensions of $\mathbb{Q}$ is the field $\mathbb{Q}(x)$ of single-variable rational polynomials in $\mathbb{Q}$; this is the fraction field of the polynomial ring $\mathbb{Q}[x]$, and its elements are of the form $f/g$ for some $f,g\in\mathbb{Q}[x]$ with $g\neq 0$. (multiplication and addition are defined in the same way as we extend multiplication and addition on $\mathbb{Z}$ to multiplication and addition on $\mathbb{Q}$) – Atticus Stonestrom Jul 20 '21 at 00:51
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    now, we can do the analogous thing in two variables and obtain the field $\mathbb{Q}(x,y)$; this is the fraction field of the polynomial ring $\mathbb{Q}[x,y]$. exercise: try to show that $\mathbb{Q}(x,y)$ is not simple. (for a hint, try to show that, for any elements $a,b\in\mathbb{Q}(x)\setminus\mathbb{Q}$, there exists a non-zero polynomial $p(s,t)\in\mathbb{Q}[s,t]$ such that $p(a,b)=0$. is the same true of $\mathbb{Q}(x,y)$? – Atticus Stonestrom Jul 20 '21 at 00:52
  • That's that actually a nice example. Although I didn't mention it, I was referring to number fields. – Measure me Jul 20 '21 at 00:56
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    oh!! in that case, what you ask is indeed impossible; number fields are by definition finite extensions of $\mathbb{Q}$, and all finite field extensions of characteristic zero are separable. so the primitive element theorem tells us that every number field is a simple extension of $\mathbb{Q}$ :) – Atticus Stonestrom Jul 20 '21 at 01:00
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    Thank you!! That's very helpfull and the link too, clears up some stuff. Also I didnt think about it, but number fields actually have to be of finite degree. – Measure me Jul 20 '21 at 01:08
  • yep, exactly! :) happy it helped! – Atticus Stonestrom Jul 20 '21 at 01:10
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    Comments as answers :( – Martin Brandenburg Jul 20 '21 at 06:59
  • So, since (i) is impossible, we know that the primitive element theorem applies to guarantee that only finitly many $c\in \mathbb{Q}$ are such that $\alpha + c\beta$ generates $\mathbb{Q}(\alpha,\beta)$. How could one show that there exists an extention where $c = 1$ is not one of these few values? – Measure me Jul 20 '21 at 10:16
  • Making $\alpha+\beta$ (so $c=1$) not work is easy. Try $\alpha=\sqrt3+\sqrt2$, $\beta=\sqrt3-\sqrt2$. If you want neither $c=0$ nor $c=1$ to work I have to think harder :-) – Jyrki Lahtonen Jul 20 '21 at 10:47
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    Not too hard. With $\alpha=\sqrt3+\sqrt2$, $\beta=\sqrt5-\sqrt2$ none of $\alpha,\beta,\alpha+\beta$ generate $$\Bbb{Q}(\alpha,\beta)=\Bbb{Q}(\sqrt5,\sqrt3,\sqrt2)=\Bbb{Q}(\sqrt5+\sqrt3+\sqrt2).$$ – Jyrki Lahtonen Jul 20 '21 at 11:28
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    @AtticusStonestrom Could you please avoid using the comments to answer? This defeats the format of the site, and prevents us from creating a good repository of Q&As. (i.e. please consider turning those comments into an answer.) – Pedro Jul 21 '21 at 07:59
  • dear @PedroTamaroff apologies, you are of course right; I think I sometimes struggle to decide what the boundary is between something that's just a comment and something that really is an answer. Guenterino has now already completely answered this particular question below, but I will keep this in mind in future :) – Atticus Stonestrom Jul 21 '21 at 16:59

1 Answers1

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In general, non-finite extensions do exist. For example the extension $\mathbb{R}/\mathbb{Q}$ is going to be non-finite. But also the algebraic extension $\bar{\mathbb{Q}}/\mathbb{Q}$, where $\bar{\mathbb{Q}}$ is the algebraic closure, is non-finite. Both of these extensions are not primitive, but they are also not generated by adjoining two elements only.

To find such a field let's think about what you wrote here:

I know about the Primitive element theorem , in particular that every finite separable extension is simple. The thing is that I don't know how to construct a non-finite or non-separable extension, if it even exists, so I guess (i) might not be possible.

You already know that every finite, seperable extension is primitive. Now include your condition that we only want $2$ generators, i.e. $\mathbb{Q}(a,b)$. We know, that if $a,b$ are algebraic elements over $\mathbb{Q}$, then the extension is going to be finite and seperable (all algebraic extensions in characteristic $0$ are seperable). So, the only thing that we have left to do is adding non algebraic elements. For example, one could add variables, to obtain $\mathbb{Q}(x,y)$. Or, you could add $\pi$ and a variable: $\mathbb{Q}(x,\pi)$. Both of these extensions satisfy condition (i) and (ii) alike.

If you want to find fields where an algebraic extension satisfies (i), you need to stick to fields with characteristic $p>0$.

Edit: I just saw in the comments under your question that you are looking for number fields. The above discussion should make clear why this is not possible.

Guenterino
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