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When is integral closure generated by one element

  1. Feb 5, 2015 #1
    This is not a homework problem, nor a textbook question. Please do not remove.
    Is there a concrete example of the following setup :
    [itex]R[/itex] is an integrally closed domain,
    [itex]a[/itex] is an integral element over [itex]R[/itex],
    [itex]S[/itex] is the integral closure of [itex]R[a][/itex] in its fraction field,
    [itex]S[/itex] is not of the form [itex]R{[}b{]}[/itex] for any element [itex]b[/itex] in [itex]S[/itex].

    For example, the integral closure of [itex]{\mathbb Z}(\sqrt 5)[/itex] is the set of elements of the form [itex](m + \sqrt 5 n)/2[/itex], where [itex]m^2 - 5n^2[/itex] is a multiple of 4. So, [itex]S[/itex] is generated by [itex]\sqrt 5/2[/itex] over [itex]\mathbb Z[/itex]: This does not fulfill the desired conditions.
  2. jcsd
  3. Feb 5, 2015 #2


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    i don't understand. in your example apparently R is not integrally closed. in fact under your conditions, that R is integrally closed and a is integral over R, this implies that R = R[a], doesn't it?

    Oh sorry, you meant integral over it but in some finite field extension of its fraction field. so your R was Z and your a was sqrt(5). got it. and maybe you meant Z[sqrt(5)] instead of Z(sqrt(5))?
    Last edited: Feb 5, 2015
  4. Feb 5, 2015 #3
    Yes, I meant [itex]{\mathbb Z}{[}\sqrt 5{]}.[/itex]
    In my example, [itex]R = {\mathbb Z},\ a = \sqrt 5[/itex] and [itex]S[/itex] is the integral closure of [itex]{\mathbb Z}[\sqrt 5][/itex] in [itex]{\mathbb Z}(\sqrt 5)[/itex], which has the form given above.
  5. Feb 5, 2015 #4


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    the answer is there does exist such a counter example Z[a] to the integral closure equaling Z[ b]. all your examples are numbers fields, and all number fields fall into your examples. You are asking for the structure of the ring of integers, which is known to have a finite basis as an abelian group. You want that basis to be the powers, up to some finite power, of a single integral element.
    It is well known since Richard Dedekind that not all rings of integers in number fields have so called "power bases". here is a reference: (or just search on "non monogenic fields".)

    Last edited: Feb 5, 2015
  6. Feb 6, 2015 #5
    Very useful. I was completely unaware of this topic. Thank you so many mathwonk.
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