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An Integral Domain, Closed in its Field of Fractions

Date: 01/20/2012 at 17:25:56
From: Rob
Subject: Is Z[2^(1/3)] integrally closed in Q[2^(1/3)]?

Hi Dr. Math,

My question is, is the integral domain Z[2^(1/3)] integrally closed in its
field of fractions, Q[2^(1/3)]? Here Z denotes the integers, and Q denotes
the rationals.

It's clear to me that Z[2^(1/3)] is contained in the ring of integers of
Q[2^(1/3)], but not conversely. And I know every element of Q[2^(1/3)]
takes this form:

   a + b2^(1/3) + c4^(1/3)    with a, b, and c rational
But I don't see why a + b2^(1/3) + c4^(1/3) being integral over Z[2^(1/3)]
necessarily implies a, b, and c are actually integers; and I don't know
how to show that any integral element of Q[2^(1/3)] is actually in

Date: 01/23/2012 at 05:04:36
From: Doctor Jacques
Subject: Re: Is Z[2^(1/3)] integrally closed in Q[2^(1/3)]?

Hi Rob,

The short answer is yes. To show this, we need to use a theorem called
Dedekind's Criterion. Rather than give you a proof of that theorem now,
let me show you how to use it.

We have an algebraic integer w, with minimal polynomial T(x). Let O be the
maximal order of Q(w), i.e., the ring of all algebraic integers in Q(w).
Z[w] is a subring of O; let n = [O:Z[w]] be its index as a module. Z[w] is
integrally closed in Q(w) if Z[w] = O, or n = 1.

The first step is to reduce the possible values of n. We use the fact that
n^2 must divide the discriminant of Z[w], which is the discriminant of the
polynomial T(x). In this case, 

   T(x) = x^3 - 2

The discriminant is equal to 
   -108 = -2^2*3^3.
This means that n = 1, 2, 3 or 6.

We will restrict the possible values of n one prime at a time. If p is a
prime that does not divide n, we say that Z[w] is p-maximal. Obviously, if
Z[w] is p-maximal for all candidate primes p, we have n = 1.

In what follows, we will fix a prime p (a possible divisor of n), and work
with polynomials over Z and polynomials over Z_p (i.e., modulo p). We will
use uppercase letters for polynomials over Z and lowercase letters for the
corresponding polynomials over Z_p.

We must check the primes 2 and 3. Let us start with p = 2.

We have T(x) = x^3 - 2, and therefore t(x) = x^3. The first step is to
factor t(x) over Z_p. In this case, the factorization is obvious:
t(x) = (x)^3.

We define g(x) as the product of the _distinct_ irreducible factors of
t(x), giving g(x) = x. Note that, because p divides the discriminant of
t(x), the factorization will always contain repeated factors.

We define 

   h(x) = t(x)/g(x) = x^2.

The next step is to "reconstruct" T(x) over Z. We take any polynomials G
and H congruent to g and h modulo p. In this case, we can simply take 
G(x) = x and H(x) = x^2.

Because we used reduction modulo p, G(x)H(x) - T(x) will be divisible 
by p. We compute:

   F(x) = (G(x)H(x) - T(x))/p
        = (x^3 - (x^3 - 2))/2
        = 1

We reduce F(x) modulo p, giving f(x) = 1. We define d(x) as the greatest
common divisor (gcd) of the polynomials f, g, and h, computed over Z_p.

The theorem says that Z[w] is p-maximal if and only if d(x) = 1 (or a
constant). This is the case here, and we have just proved that Z[w] is
2-maximal: n can only be 1 or 3.

We repeat the process with p = 3. We have:

   t(x) = x^3 + 1 = (x + 1)^3 (over Z_3)
   g(x) = (x + 1)
   h(x) = (x + 1)^2

We reconstruct T(x) using G(x) = x + 1 and H(x) = (x+1)^2, and we compute:

   F(x) = ((x + 1)^3 - x^3 + 2)/3
        = (3x^2 + 3x + 3)/3
        = x^2 + x + 1

We reduce F(x) modulo 3, giving

   f(x) = x^2 + x + 1 = (x - 1)^2     (mod 3)

We compute the gcd:

   d(x) = gcd((x + 1), (x + 1)^2, (x - 1)^2) = 1

This shows that Z[w] is 3-maximal. As the only possible prime factors of n
were 2 and 3, we have proved that Z[w] = O is integrally closed.

To see what happens when this is not the case, let us take w as a root of 

   T(x) = x^3 - 10.

The discriminant is -2700, and the candidate primes are 2, 3, and 5. We
will look at the case p = 3.

We have:

   T(x) = x^3 - 10
   t(x) = x^3 - 1 = (x - 1)^3
   g(x) = (x - 1)
   h(x) = (x - 1)^2
   F(x) = ((x - 1)^3 - (x^3 - 10))/3
        = (-3x^2 + 3x + 9)/3
        = -x^2 + x + 3
   f(x) = -x^2 + x = -x(x - 1)      (mod 3)


   d(x) = gcd (g(x), h(x), f(x)) = x - 1

As this is not constant, Z[w] is not 3-maximal. Indeed, this element is
not in Z[w]:

   (1 + w + w^2)/3

But it is an algebraic integer: its minimal polynomial is 

   x^3 - x^2 - 3x - 3

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum 

Date: 01/24/2012 at 12:02:53
From: Rob
Subject: Thank you (Is Z[2^(1/3)] integrally closed in Q[2^(1/3)]?)

Dear Doctor Jacques,

Many thanks for your response. As a layman and self-learner, I greatly
appreciate the time and effort you put into helping me.


Date: 01/24/2012 at 18:53:44
From: Rob
Subject: Is Z[2^(1/3)] integrally closed in Q[2^(1/3)]?

Dear Doctor Jacques,

After further reflection, I wanted to ask if you have a reference where a
statement and proof of Dedekind's criterion may be found?

Thanks and regards,

Date: 01/25/2012 at 03:17:24
From: Doctor Jacques
Subject: Re: Is Z[2^(1/3)] integrally closed in Q[2^(1/3)]?

Hi Rob,

You can find it in chapter 6 of this book:

  A Course in Computational Algebraic Number Theory
  Henri Cohen 

There may also be a more elementary (but rather messy) way of proving
this, but the method is not practical for larger examples and cannot be
generalized easily.

As we have seen, the index of Z[w] must be a divisor of 6. We can write a
general element of Q(w) as:

   u = (a + bw + cw^2)/6

We want to show that, if u is an algebraic integer, then a, b, and c must
be multiples of 6, which implies that u is in Z[w].

To do that, note that multiplication by w is a linear transformation of
the vector space Q(w)/Q, and can therefore be represented as a matrix
relative to the basis {1, w, w^2}. This matrix is easily seen to be:

   W = [0 0 2]
       [1 0 0]
       [0 1 0]

The matrix representing multiplication by u is (aI + bW + cW^2)/6:

   U = (1/6)[a 2c 2b]
            [b  a 2c]
            [c  b  a]

The minimal polynomial of u is the minimal polynomial of that matrix. If u
is irrational (b and c are not both 0), then that polynomial is also the
characteristic polynomial f(x) = det(xI - U), i.e., the one that is used 
to compute eigenvalues. If I did not make any mistake, that polynomial
should be:

   f(x) = x^3 - ax^2/2  + (a^2 - 2bc)x/12
              - a^3/216 + abc/36 - b^3/108 - c^3/54

If u is an algebraic integer, the coefficients of that polynomial must be
integers. You can use this to write three congruences modulo 2 and 
modulo 3, and it should be possible to show that these congruences imply 
that a, b, and c are multiples of 6, and therefore the coefficients of 
any algebraic integer u in the field are integers.

- Doctor Jacques, The Math Forum 
Associated Topics:
College Modern Algebra

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