An Integral Domain, Closed in its Field of FractionsDate: 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 Z[2^(1/3)]. 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) Finally, 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 http://mathforum.org/dr.math/ 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. Regards, Rob 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, Rob 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 http://www.springer.com/mathematics/numbers/book/978-3-540-55640-4 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/