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I have two questions:

a) How can I find the integral basis of the integral closure of $\Bbb Z$ in $\Bbb Q(\sqrt{3})$.
b) How can I show that an integral extension is not finite, for example how to show that $\Bbb Q(\sqrt{2},\sqrt[3]{2},\dots,\sqrt[p]{2},\dots)$ is not a finite extension of $\Bbb Q$.

Maybe these questions are easy but I have no idea how to deal with them, thank you for your help.

i.a.m
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2 Answers2

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Hints:

(1) The elements $\,1\;,\;\sqrt 3\,$ are integral and linearly independent (both over $\,\Bbb Q\,\,\,and\,\,\,\Bbb Z\,$)

(2) For any $\,n\ge 2\;,\;\;x^n-2\,$ is irreducible over $\,\Bbb Q\,$ ...

DonAntonio
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  • For (1), if I am not misunderstanding you, you still need to show that the integral closure is exactly $\mathbb{Z}[\sqrt{3}]$. –  Mar 26 '13 at 02:22
  • I think you are misunderstanding the beginning of my post, @Sanchez: these are hints and the OP must still complete stuff and work a little, – DonAntonio Mar 26 '13 at 02:23
  • @Don, Thank you for your Hints, but I'm really lost, can you say a word or two about how to continue from your hints. Thank you. – i.a.m Mar 26 '13 at 02:25
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    @DonAntonio, sure I know it's a hint, but I am suggesting that your hint may lead to the (easier) side of showing $\mathbb{Z}[\sqrt{3}]$ lying in the integral closure, while the more substantial part is to show that it is exactly the integral closure. –  Mar 26 '13 at 02:31
  • If you read Sanchez's comment then you have the continuation: what's the integral closure of the integers in the number field $,\Bbb Q(\sqrt 3),$ ? It's going to be tough if you try to deal with integral basis and stuff without first figuring out what the integral closure is. For (2) it's almsot straightforward: apply Eisenstein's Criterium to prove what I wrote there and then deduce that for any $,n\in\Bbb N,$ your extension contains a subextension of degree $,n,$ and thus cannot be a finite one... – DonAntonio Mar 26 '13 at 02:31
  • @Sanchez, there's a pretty standard way to find out what the integral closure of the integers in a quadratic extension $,\Bbb Q(\sqrt d),,,,,d\in\Bbb Z$ square-free, is (according as whether $,d=1,,or,,3,\pmod 4 ,$ and stuff). I think this must be known at least at the same time, if not before, dealing with this kind of questions. – DonAntonio Mar 26 '13 at 02:33
  • @DonAntonio, I would rather think that OP didn't even know about that fact given the way he asks about (1). Different people would have different perspectives on this I suppose. –  Mar 26 '13 at 02:35
  • @DonAntonio, I don't know how to find the clouser, can you tell me how to start? – i.a.m Mar 26 '13 at 02:37
  • @i.a.m, you can read it in trillions of places (well, I may have exaggerated and perhaps it's only billions). Check the following after googling 10 second:http://www.google.co.il/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=1&ved=0CC0QFjAA&url=http%3A%2F%2Fwww.math.uiuc.edu%2F~r-ash%2FAlgebra%2FChapter7.pdf&ei=2wlRUZWIBc2Shgf0roHYBA&usg=AFQjCNFgCvvgsFyg_N4Z5wk22jsy0mbikA&sig2=OGVyWeyD_sQE4ewfH9IF5A&bvm=bv.44342787,bs.1,d.Yms , in 7.2 and forward, page 5. In particular, 7.2.2. – DonAntonio Mar 26 '13 at 02:38
  • @DonAntonio, Ok, well do, thank you very much for your help. – i.a.m Mar 26 '13 at 02:40
  • @Sanchez, Thank you for you comments. – i.a.m Mar 26 '13 at 02:43
  • My pleasure, @i.a.m – DonAntonio Mar 26 '13 at 02:50
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Alternative hint for a). Since $\mathbb{Z}$ is integrally closed, $\alpha\in\mathbb{Q}(\sqrt{3})$ is in the integral closure of $\mathbb{Z}$ if and only if its polynomial is in $\mathbb{Z}[x]$. The minimal polynomial of $a+b\sqrt{3}$ has coefficients depending on $a$ and $b$--this should give you the necessary conditions.

Alex Youcis
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