When is integral closure generated by one element

In summary, there is a counterexample to the setup where R is an integrally closed domain, a is an integral element over R, S is the integral closure of R[a] in its fraction field, and S is not of the form R{[}b{]} for any element b in S. This is demonstrated by the integral closure of {\mathbb Z}(\sqrt 5) being the set of elements of the form (m + \sqrt 5 n)/2, where m^2 - 5n^2 is a multiple of 4. However, not all rings of integers in number fields have power bases, as shown by Richard Dedekind.
  • #1
coquelicot
299
67
Hello,
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.
 
Physics news on Phys.org
  • #2
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:
  • #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.
 
  • #4
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".)

http://wstein.org/129-05/final_papers/Yan_Zhang.pdf
 
Last edited:
  • Like
Likes coquelicot
  • #5
Very useful. I was completely unaware of this topic. Thank you so many mathwonk.
 

1. What is integral closure?

Integral closure is a concept in algebraic geometry and commutative algebra that involves closing a ring under integral extensions. In simpler terms, it is the process of adding new elements to a ring in order to make it "complete" or "well-behaved".

2. How is integral closure generated?

The integral closure of a ring R is generated by adding all elements that satisfy a certain polynomial equation over R. This means that for each element x in the integral closure, there exists a polynomial f(x) with coefficients in R such that f(x) = 0.

3. What is the significance of integral closure being generated by one element?

If integral closure is generated by only one element, it means that adding just one element is enough to make the ring complete. This is a desirable property as it simplifies calculations and makes the ring more manageable.

4. How do you determine if integral closure is generated by one element?

To determine if integral closure is generated by one element, one can use the "minimal polynomial" test. This involves finding the minimal polynomial of an element in the integral closure and checking if all other elements can be expressed as a power of this element. If so, then integral closure is generated by one element.

5. Can integral closure be generated by more than one element?

Yes, integral closure can be generated by more than one element. In fact, in most cases, it is generated by multiple elements. However, as mentioned earlier, having integral closure generated by one element is a desirable property.

Similar threads

  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
930
  • Linear and Abstract Algebra
Replies
7
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
6
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
1K
Back
Top