4.2 Algebraic Integer Domains

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Let _ 2 C be an algebraic number. If _ satisfies a monic polynomial with

integral coefficients, then _ is called an algebraic integer . More generally,

suppose that R _ S are integral domains and that _ 2 S. Say that _ is

integral over R if _ satisfies a monic polynomial in R[x]. Thus, the algebraic

integers are precisely the complex numbers which are integral over Z.

Lemma 4.2.1 Let R _ S be integral domains, and let _ 2 S. Then _ is

integral over R if and only if R[_] is a finitely generated R-module.

Note that the proof of the above actually reveals the following.:

Lemma 4.2.2 Let R _ S be integral domains. If S is a finitely generated

R-module, then every element of S is integral over R.

Lemma 4.2.3 Let R _ S _ T be integral domains. If T is a finitely generated

S-module, and if S is a finitely generated R-module, then T is a finitely

generated R-module.

As an immediate consequence we have the following proposition.

Proposition 4.2.4 Let _, _ be algebraic integers. Then __ and _ + _ are

also algebraic integers.

One could consider the ring Zalg _ C of all algebraic integers. However,

this ring doesn’t have very interesting factorization properties. For example,

Zalg has no primes. Indeed, if a 2 Zalg, then

p

a 2 Zalg. Rather than

considering all algebraic integers, it is more appropriate to consider the

following subrings of Zalg.

92 CHAPTER 4. DEDEKIND DOMAINS

Definition. Let Q _ E _ C, where [E : Q] < 1. Set

OE = {algebraic integers _|_ 2 E} = E \ Zalg.

Call the ring OE an algebraic integer domain.

Definition. Let R be an arbitrary integral domain. Say that R is integrally

closed if, whevever _ 2 F(R) and _ is integral over R, then _ 2 R.

Here F(R) is the field of fractions of the integral domain R.

The following is a sufficient, but not a necessary condition for an integral

domain to be integrally closed.

Lemma 4.2.5 If R is a u.f.d., then R is integrally closed.

Proposition 4.2.6 Let E be a field with [E : Q] < 1, and set R = OE.

(a) If _ 2 E, then n_ 2 R, for some n 2 Z.

(b) F(R) = E.

(c) R is integrally closed.

(d) R \ Q = Z.

An important class of algebraic integer domains are the quadratic integer

domains , defined as the domains of the form OE, where [E : Q] = 2. We’ll

simplify the notation slightly, as follows. First note that E = Q[

p

m], where

m is a square-free integer. Thus, denote Qm = OQ[

p

m].

Proposition 4.2.7

Qm =

(

{a + b

p

m| a, b 2 Z} if m 6_ 1(mod 4)

{a+b

p

m

2 | a, b 2 Z, a _ b(mod 2)} if m _ 1(mod 4)

Notice that the above proposition, together with Lemma 5, readily identifies

many integral domains which cannot possibly be u.f.d.’s. Indeed, if m

is square-free and satisfies m _ 1(mod 4), then the ring

R0 = {a + b

p

m|a, b 2 Z}

is properly contained in R = Qm, and yet it is clear that F(R0) = F(R).

Thus R0 is not integrally closed and hence cannot be a u.f.d.

4.2. ALGEBRAIC INTEGER DOMAINS 93

Perhaps unfortunately, not all quadratic integer domains are u.f.d.’s.

The simplest example is the ring R = Q−5, which by the above proposition

is simply the ring

Z[

p

−5] = {a + b

p

−5|a, b 2 Z}.

We already observed in Chapter 3 that R is not a u.f.d.

Unsolved Problem: Are there finitely or infinitely many real quadratic

integer domains which are also u.f.d’s?

In the next section we’ll see that an algebraic integer domain is a p.i.d

if and only if it is a u.f.d.

Exercises

1. Let F be a field and let x be indeterminate over F. Prove that the

ring R = F[x2, x3] is not integrally closed, hence is not a u.f.d.. (C.f.

Exercise 3 of Section 3.2.)

2. Prove the above assertion that if a is an algebraic integer, so is

p

a.

3. Let [E : Q] < 1, and set G = Gal(E/Q). If a 2 OE, and if _ 2 G,

then _ (a) 2 OE.

4. Show that Q−6 is not a u.f.d..

5. Let _ 2 C, and assume that f(_) = 0, for some monic polynomial

f(x) 2 Z[x]. Prove that _ is an algebraic integer.

6. Here’s another proof of the fact that if _, _ are algebraic integers,

so are _ + _ and __. Let f(x), g(x) be the minimal polynomials of

_, _, respectively. Let _1 = _, _2, . . . , _r be the roots of f(x), and let

_1 = _, _2, . . . , _s be the roots of g(x). If we set

h(x) =

Ys

j=i

f(x − _j),

then argue that h(x) is a monic polynomial Z[x]. Then show that

h(_ + _) = 0. Apply Exercise 5, above. Give a similar argument to

show that __ is also an algebraic integer.

94 CHAPTER 4. DEDEKIND DOMAINS

7. Show that if m > 0, and is square-free, then U(Qm) is infinite.

8. __ Let _ 2 C be a primitive n-th root of unity, and let E = Q[_]. Show

that OE = Z[_]. (You probably won’t get this one but give it a little

thought. You should at least see how Proposition 4.2.7 above makes

this statement true for n = 3.)

9. As we have seen, the ring Q−5 = Z[

p

−5] is not a u.f.d. Many odd

things happen in this ring. For instance, find an example of an irreducible

element _ 2 Z[

p

−5] and an element a 2 Z[

p

−5] such that

_ doesn’t divide a, but _ divides a2. (Hint: look at factorizations of

9 = 32.)

10. The following result is well-known to virtually every college student.

Let f(x) 2 Z[x], and let a

b be a rational root of f(x). If the fraction

a

b is in lowest terms, then a divides the constant term of f(x) and b

divides the leading coefficient of f(x). If we ask the same question in

the context of the ring Z[

p

−5], then the answer is negative. Indeed if

we consider the polynomial f(x) = 3x2 − 2

p

−5x − 3 2 Z[

p

−5], then

the roots are 2+

p

−5

3 and −2+

p

−5

3 . Since both 3 and ±2+

p

−5 are nonassociated

irreducible elements, then the fractions can be considered to

be in lowest terms. Yet neither of the numerators divide the constant

term of f(x).

11. We continue on the theme set in Exercise 10, above. Let R be an

integral domain with field of fractions F(R). Assume the following

condition on the domain R: Let f(x) = anxn + · · · + a0 2 R[x],

with a0, an 6= 0, and assume that a

b 2 F(R) is a fraction in lowest

terms (i.e., no common non-unit factors) satisfying the polynomial

f(x). Then a divides a0 and b divides an. Now prove that for such

a ring every irreducible element is actually prime. (Hint: Let _ 2 R

be an irreducible element and assume that _|uv, but that _ doesn’t

divide either u or v. Let uv = r_, r 2 R, and consider the polynomial

ux2 − (_ + r)x + v 2 R[x].)

12. Let K be a field such that K is the field of fractions of both R1,R2 _ K.

Must it be true that K is the field of fractions of R1 \ R2? (Hint: A

counter-example can be found in the field K = F(x).)

13. Let Q _ K be a finite algebraic extension. If K is the field of fractions

of R1,R2 _ K, prove that K is also the field of fractions of R1 \ R2.

4.2. ALGEBRAIC INTEGER DOMAINS 95

14. Again, let Q _ K be a finite algebraic extension. This time, let

{R_| _ 2 A} consist of the subrings of K having K as field of fractions.

Show that K is not the field of fractions of \_2AR_. (In fact,

\_2AR_ = Z.)

96 CHAPTER 4. DEDEKIND DOMAINS

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