Algebraic number theory (Math 784)

70 Pages·1996·0.425 MB·English

by  Filaseta M.

Math 784: algebraic NUMBER THEORY
(Instructor’s Notes)*
Algebraic Number Theory:
(cid:15) What is it? The goals of the subject include: (i) to use algebraic concepts to deduce
information about integers and other rational numbers and (ii) to investigate generaliza-
tions of the integers and rational numbers and develop theorems of a more general nature.
Although (ii) is certainly of interest, our main point of view for this course will be (i). The
focus of this course then will be on the use of algebra as a tool for obtaining information
about integers and rational numbers.
(cid:15) A simple example. Here, we obtain as a consequence of some simple algebra the
following:
Theorem 1. Let (cid:18) 2 Q with 0 (cid:20) (cid:18) (cid:20) 1=2. Then sin2((cid:25)(cid:18)) 2 Q if and only if (cid:18) 2
f0;1=6;1=4;1=3;1=2g.
It is easy to check that for (cid:18) 2 f0;1=6;1=4;1=3;1=2g, we have sin2((cid:25)(cid:18)) is rational; so we
are left with establishing that if sin2((cid:25)(cid:18)) 2 Q, then (cid:18) 2 f0;1=6;1=4;1=3;1=2g. Observe
that we can use Theorem 1 to immediately determine for what (cid:18) 2 Q the value of sin((cid:25)(cid:18))
is rational (see the upcoming homework assignment). Before getting to the proof of this
theorem, we give some background.
(cid:15) Some de(cid:12)nitions and preliminaries.
De(cid:12)nition. Let (cid:11) be a complex number. Then (cid:11) is algebraic if it is a root of some
f(x) 2 Z[x] with f(x) 6(cid:17) 0. Otherwise, (cid:11) is transcendental.
Examples and Comments:
(1) Rational numberps are algebraic.
(2) The number i = −1 is algebraic.
(3) The numbers (cid:25), e, and e(cid:25) are transcendental.
(4) The status of (cid:25)e is unknown.
(5) Almost all numbers are transcendental.
De(cid:12)nition. An algebraic number (cid:11) is an algebraic integer if it is a root of some monic
polynomial f(x) 2 Z[x] (i.e., a polynomial f(x) with integer coe(cid:14)cients and leading coef-
(cid:12)cient one).
Examples and Comments:
(1) Integers (sometimes called \rational integers") are algebraic integers.
(2) Rational numbers which are not rational integers are not algebraic integers. In other
words, we have
*These notes are from a course taught by Michael Filaseta in the Spring of 1997 and 1999 but based
on notes from previous semesters.
1
2
Theorem 2. If (cid:11) is a rational number which is also an algebraic integer, then (cid:11) 2 Z.
P
Proof. Suppose f(a=b) = 0 where f(x) = n a xj with a = 1 and where a and b are
j=0 j n
relatively prime integers with b > 0. It su(cid:14)ces to show b = 1. From f(a=b) = 0, it follows
that
an +an−1an−1b+(cid:1)(cid:1)(cid:1)+a1abn−1 +a0bn = 0:
It follows that an has b as a factor. Since gcd(a;b) = 1 and b > 0, we deduce that b = 1,
completing the proof. (cid:4)
(3) The number i is an algebraic integer.
(4) Transcendental numbers are not algebraic integers.
(5) If u 2 Q, then 2cos((cid:25)u) is an algebraic integer. This requires an explanation which
we supply next.
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Lemma. For each positive integer m, there is a g (x) = m b xj 2 Z[x] satisfying:
m j=0 j
(i) cos(m(cid:18)) = g (cos(cid:18))
m
(ii) b = 2m−1
m
(iii) 2k−1jb for k 2 f2;3;:::;mg.
k
Proof. We do induction on m. The cases m = 1 and m = 2 are easily checked. Suppose
the lemma holds for m (cid:20) n. Observe that
cos((n+1)(cid:18))+cos((n−1)(cid:18)) = 2cos(n(cid:18))cos(cid:18):
Then (i), (ii), and (iii) follow by considering gn+1(x) = 2xgn(x)−gn−1(x). (cid:4)
Write u = a=m with m a positive integer. By the lemma,
Xm
(cid:6)1 = cos(m(cid:25)u) = g (cos((cid:25)u)) = b (cos((cid:25)u))j;
m j
j=0
where b = 2m−1 and for some integers b0 we have b = 2j−1b0 for j 2 f2;3;:::;m−1g.
m j j j
Multiplying through by 2 and rearranging, we deduce that 2cos((cid:25)u) is a root of
xm +b0m−1xm−1 +(cid:1)(cid:1)(cid:1)+b02x2 +b1x+(2b0 (cid:7)2):
It follows that 2cos((cid:25)u) is an algebraic integer.
(cid:15) An application. Before proving Theorem 1, we determine for what (cid:18) 2 Q, the value
of cos((cid:25)(cid:18)) is rational. By the last example above, if (cid:18) 2 Q, then 2cos((cid:25)(cid:18)) is an algebraic
integer. If we also have cos((cid:25)(cid:18)) is rational, then Theorem 2 implies that 2cos((cid:25)(cid:18)) 2
Z. Since j2cos((cid:25)(cid:18))j (cid:20) 2, we deduce that 2cos((cid:25)(cid:18)) 2 f−2;−1;0;1;2g so that cos((cid:25)(cid:18)) 2
f−1;−1=2;0;1=2(cid:8);1g. It foll(cid:9)ows that both (cid:18) and cos((cid:25)(cid:18)) are rational if and only if (cid:18) 2
fk=2 : k 2 Zg[ k=3 : k 2 Z .
(cid:15) Completing the proof of Theorem 1. Let u = 2(cid:18), and suppose that sin2((cid:25)(cid:18)) 2 Q.
Then 2cos((cid:25)u) = 2−4sin2((cid:25)(cid:18)) is an algebraic integer which is also rational. By Theorem
3
2, we deduce 2 − 4sin2((cid:25)(cid:18)) 2 Z. It follows that 4sin2((cid:25)(cid:18)) is a non-negative rational
integer which is (cid:20) 4. We deduce that sin2((cid:25)(cid:18)) 2 f0;1=4;1=2;3=4;1g. Note that sin((cid:25)x) is
a positive increasing function for 0 (cid:20) x (cid:20) 1=2 so that there can be no more than 5 such
(cid:18). The result easily follows. (Note that the previous application involving cos((cid:25)(cid:18)) could
have been used to prove Theorem 1.)
Homework:
(cid:14)
(1) Prove that sin(1 ) is algebraic. (One approach is to begin by showing that the
coe(cid:14)cient of xj in g (x), as given in the lemma, is 0 if j 6(cid:17) m (mod 2). There are easier
m
approaches, but I won’t give hints for them.)
(2) (a) Using Theorem 1, determine the values of (cid:18) 2 Q\[0;2) for which sin2((cid:25)(cid:18)) 2 Q.
(b) Using Theorem 1, determine the values of (cid:18) 2 Q\[0;2) for which sin((cid:25)(cid:18)) 2 Q.
(3) Determine explicitly the set S satisfying: both (cid:18) 2 [0;1=2] and cos2((cid:25)(cid:18)) are rational
if and only if (cid:18) 2 S.
(cid:18) (cid:19)
1 1
(4) Let n denote a positive integer. Prove that cos−1 p is rational if and only if
(cid:25) n
n 2 f1;2;4g.
p
(5) Using Theorem 2, prove that if m is a positive integer for which m 2 Q, then m is
a square (i.e., m = k2 for some k 2 Z).
The Elementary Symmetric Functions:
(cid:15) The de(cid:12)nition. Let (cid:11)1;(cid:11)2;:::;(cid:11)n be n variables. Then
(cid:27)1 = (cid:11)1 +(cid:11)2 +(cid:1)(cid:1)(cid:1)+(cid:11)n
(cid:27)2 = (cid:11)1(cid:11)2 +(cid:11)1(cid:11)3 +(cid:11)2(cid:11)3 +(cid:1)(cid:1)(cid:1)+(cid:11)n−1(cid:11)n
(cid:27)3 = (cid:11)1(cid:11)2(cid:11)3 +(cid:11)1(cid:11)2(cid:11)4 +(cid:1)(cid:1)(cid:1)+(cid:11)n−2(cid:11)n−1(cid:11)n
. .
. .
. .
(cid:27)n = (cid:11)1(cid:11)2(cid:1)(cid:1)(cid:1)(cid:11)n
are the elementary symmetric functions in (cid:11)1;(cid:11)2;:::;(cid:11)n.
(cid:15) Why the terminology? The term \symmetric" refers to the fact that if we permute
the (cid:11)j in any manner, then the values of (cid:27)1;:::;(cid:27)n remain unchanged. More explicitly, a
function f((cid:11)1;:::;(cid:11)n) is symmetric in (cid:11)1;:::;(cid:11)n if for all (cid:30) 2 Sn (the symmetric group on
f1;2;:::;ng), we have f((cid:11)(cid:30)(1);:::;(cid:11)(cid:30)(n)) = f((cid:11)1;:::;(cid:11)n). The term \elementary" refers
to the following:
Theorem 3. Let R be a commutative ring with an identity. Then every symmetric
polynomial in (cid:11)1;:::;(cid:11)n with coe(cid:14)cients in R is expressible as a polynomial in (cid:27)1;:::;(cid:27)n
with coe(cid:14)cients in R.
Proof. For a symmetric h((cid:11)1;:::;(cid:11)n) 2 R[(cid:11)1;:::;(cid:11)n], we set T = Th to be the set of
n-tuples (‘1;:::;‘n) with the coe(cid:14)cient of (cid:11)1‘1 (cid:1)(cid:1)(cid:1)(cid:11)n‘n in h((cid:11)1;:::;(cid:11)n) non-zero. We de(cid:12)ne
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the size of h to be (k1;:::;kn) where (k1;:::;kn) is the element of T with k1 as large as
possible, k2 as large as possible given k1, etc. Since h((cid:11)1;:::;(cid:11)n) is symmetric, it follows
that (‘1;:::;‘n) 2 T if and only if each permutation of (‘1;:::;‘n) is in T. This implies
that k1 (cid:21) k2 (cid:21) (cid:1)(cid:1)(cid:1) (cid:21) kn. Observe that we can use the notion of size to form an ordering
on the elements of R[(cid:11)1;:::;(cid:11)n] in the sense that if h1 has size (k1;:::;kn) and h2 has size
(k10;:::;kn0 ), then h1 > h2 if there is an i 2 f0;1;:::;n−1g such that k1 = k10;:::;ki = ki0,
0
and ki+1 > ki+1. Note that the elements of R[(cid:11)1;:::;(cid:11)n] which have size (0;0;:::;0) are
precisely the constants (the elements of R).
Suppose now that (k1;:::;kn) represents the size of some symmetric g 2 R[(cid:11)1;:::;(cid:11)n]
with g 62 R. For non-negative integers d1;:::;dn, the size of h = (cid:27)1d1(cid:27)2d2 (cid:1)(cid:1)(cid:1)(cid:27)ndn is (d1 +
d2+(cid:1)(cid:1)(cid:1)+dn;d2+(cid:1)(cid:1)(cid:1)+dn;:::;dn−1+dn;dn). Taking d1 = k1−k2;d2 = k2−k3;:::;dn−1 =
kn−1 −kn, and dn = kn, we get the size of h is (k1;:::;kn). The coe(cid:14)cient of (cid:11)1k1 (cid:1)(cid:1)(cid:1)(cid:11)nkn
in h is 1. It follows that there is an a 2 R such that g −ah is of smaller size than g.
The above implies that for any symmetric element f 2 R[(cid:11)1;:::;(cid:11)n], there exist
a1;:::;am 2 R and h1;:::;hm 2 R[(cid:27)1;:::;(cid:27)n] such that f − a1h1 − (cid:1)(cid:1)(cid:1) − amhm has
size (0;0;:::;0). This implies the theorem. (cid:4)
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(cid:15) Elementary symmetric functions on roots of polynomials. Let f(x) = n a xj be
j=0 j
a non-zero polynomial in C[x] of degree n with not necessarily distinct roots (cid:11)1;:::;(cid:11)n.
Then it is easy to see that
Yn
f(x) = an (x−(cid:11)j) = anxn −an(cid:27)1xn−1 +an(cid:27)2xn−2 +(cid:1)(cid:1)(cid:1)+(−1)nan(cid:27)n;
j=1
where now we view the (cid:27)j as elementary symmetric functions in the numbers (cid:11)1;:::;(cid:11)n.
It follows that
((cid:3)) (cid:27)1 = −an−1; (cid:27)2 = an−2; :::; (cid:27)n = (−1)na0:
a a a
n n n
(cid:15) An example (almost Putnam Problem A-1 from 1976). Consider all lines which pass
through the graph of y = 2x4+7x3+3x−5 in 4 distinct points, say (x ;y ) for j = 1;2;3;4.
j j
We will show that the average of the x ’s is independent of the line and (cid:12)nd its value.
j
If y = mx+b intersects y = 2x4 +7x3 +3x−5 as indicated, then x1;:::;x4 must be
the four distinct roots of 2x4+7x3+(3−m)x−(b+5) = 0. From the previous section, we
deduce that the sum of the xj’s is (cid:27)1 = −7=2. The average of the xj’s is therefore −7=8.
Homework:
2
(1) Show that the average of the x ’s is independent of the line and (cid:12)nd its value.
j
(2) Prove or disprove that the average of the y ’s is independent of the line.
j
(3) In the proof of Theorem 3, we deduced that the size of g−ah is smaller than the size
of g. By continuing the process, we claimed that eventually we would obtain an element
of R[(cid:11)1;:::;(cid:11)n] of size (0;0;:::;0). Prove this as follows. Explain why the claim is true
if n = 1. Consider n (cid:21) 2. Let (k1;:::;kn) be the size of g with g 62 R, and let (k10;:::;kn0 )
5
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be the size of g−ah. Let b be an integer (cid:21) k1. Associate the integer nj=−01‘n−jbj with an
n-tuple (‘1;:::;‘n). Show that the integer associated with (k1;:::;kn) is greater than the
0 0
integer associated with (k ;:::;k ). Explain why (0;0;:::;0) is obtained by continuing
1 n
the process as claimed. (There are other approaches to establishing that (0;0;:::;0) will
be obtained, and you can feel free to establish this in a di(cid:11)erent manner.)
Algebraic Numbers and Algebraic Integers as Algebraic Structures:
(cid:15) The main theorems we deal with here are as follows.
Theorem 4. The algebraic numbers form a (cid:12)eld.
Theorem 5. The algebraic integers form a ring.
To prove these, we suppose that (cid:11) and (cid:12) are algebraic numbers or integers, and prove that
−(cid:11), (cid:11) + (cid:12), and (cid:11)(cid:12) are likewise. In the case that (cid:11) is a non-zero algebraic number, we
show that 1=(cid:11) is as well.
(cid:15) The case for −(cid:11). If f(x) is a polynomial with integer coe(cid:14)cients having (cid:11) as a root,
then we consider (cid:6)f(−x). If f(x) is monic, then one of these will be as well. Hence, if (cid:11)
is an algebraic number, then so is −(cid:11); and if (cid:11) is an algebraic integer, then so is −(cid:11).
(cid:15) The case for (cid:11) + (cid:12). Suppose (cid:11) is a root of f(x) 2 Z[x] and (cid:12) is a root of g(x) 2
Z[x]. Let (cid:11)1 = (cid:11);(cid:11)2;:::;(cid:11)n denote the complete set of roots of f(x) (counted to their
multiplicity so that the degree of f(x) is n) and let (cid:12)1 = (cid:12);(cid:12)2;:::;(cid:12)m denote the complete
set of roots of g(x). Consider the polynomial
Yn Ym (cid:0) (cid:1)
F(x) = x−((cid:11) +(cid:12) ) :
i j
i=1j=1
Taking R = Z[(cid:12)1;:::;(cid:12)m] in Theorem 3, we see that the coe(cid:14)cients of F(x) are symmet-
ric polynomials in (cid:11)1;:::;(cid:11)n. Thus, if (cid:27)1;:::;(cid:27)n correspond to the elementary sym-
metric functions in (cid:11)1;:::;(cid:11)n and A is some coe(cid:14)cient (of xk) in F(x), then A =
B((cid:27)1;:::;(cid:27)n;(cid:12)1;:::;(cid:12)m) for some polynomial B with integer coe(cid:14)cients. Now, the coef-
(cid:12)cients of F(x) are also symmetric in (cid:12)1;:::;(cid:12)m. Taking R = Z[(cid:27)1;:::;(cid:27)n] in Theorem
0 0
3 and (cid:27)1;:::;(cid:27)m to be the elementary symmetric functions in (cid:12)1;:::;(cid:12)m, we get that
0 0 0 0
A = B ((cid:27)1;:::;(cid:27)n;(cid:27)1;:::;(cid:27)m) for some polynomial B with integer coe(cid:14)cients. On the
other hand, ((cid:3)) implies that (cid:27)1;:::;(cid:27)n;(cid:27)10;:::;(cid:27)m0 are all rational so that A 2 Q. Thus,
F(x) 2 Q[x] and m0F(x) 2 Z[x] for some integer m0. Since (cid:11) + (cid:12) is a root of m0F(x),
we deduce that (cid:11)+(cid:12) is an algebraic number. If (cid:11) and (cid:12) are algebraic integers, then we
can take the leading coe(cid:14)cients of f(x) and g(x) to be 1 so that ((cid:3)) implies that each of
(cid:27)1;:::;(cid:27)n;(cid:27)10;:::;(cid:27)m0 is in Z so that F(x) 2 Z[x]. Since F(x) is monic, we obtain that in
this case (cid:11)+(cid:12) is an algebraic integer.
(cid:15) The case for (cid:11)(cid:12). The same idea as above works to show (cid:11)(cid:12) is an algebraic number
(or integer) by de(cid:12)ning
Yn Ym (cid:0) (cid:1)
F(x) = x−(cid:11) (cid:12) :
i j
i=1j=1
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(cid:15) The case for 1=(cid:11). Suppose (cid:11) 6= 0 and (cid:11) is a root of n a xj 2 Z[x]. Then it is
P j=0 j
easy to show that 1=(cid:11) is a root of nj=0an−jxj 2 Z[x]. Hence, 1=(cid:11) is an algebraic number.
Comments: The above completes the proofs of Theorems 4 and 5. Suppose (cid:11) is a non-
zero algebraic integer. We note that 1=(cid:11) is an algebraic integer if and only if (cid:11) is a root
of a monic polynomial in Z[x] with constant term (cid:6)1.
(cid:15) An additional result. Next, we prove the following:
Theorem 6. If (cid:11) is an algebraic number, then there is a positive rational integer d such
that d(cid:11) is an algebraic integer.
P
Proof. Suppose (cid:11) is a root of f(x) = n a xj 2 Z[x] with a 6= 0. By consider-
j=0 j n
Ping −f(x) if necessary, we may suppose an > 0. Since (cid:11) is a root of ann−1f(x) =
n a an−j−1(a x)j, it follows that a (cid:11) is a root of a monic polynomial. The result
j=0 j n n n
is obtained by taking d = a . (cid:4)
n
(cid:15) Comment. The above is simple enough that you should remember the argument
rather than the theorem. This has the advantage that if you know a polynomial f(x) that
(cid:11) is a root of, then you will know the value of d in the theorem.
Homework:
p p
(1) Find a polynomial in Z[x] of degree 4 which has 1+ 2+ 3 as a root. Simplify your
answer.
(cid:18)p (cid:19)
1 3 2
−1
(2) Prove that sin is irrational.
(cid:25) 3
(3) Let (cid:11) be non-zero. Prove that (cid:11) and 1=(cid:11) are both algebraic integers if and only if (cid:11) is
a root of a monic polynomial in Z[x] and (cid:11) is a root of a polynomial in Z[x] with constant
term 1.
Minimal Polynomials:
(cid:15) De(cid:12)nition. Let (cid:11) be an algebraic number. Then the minimal polynomial for (cid:11) (in
Q[x]) is the monic polynomial in Q[x] of minimal degree which has (cid:11) as a root. (Note the
(cid:12)rst homework assignment below.)
(cid:15) Goal for this section. We will establish:
Theorem 7. The minimal polynomial for an algebraic number (cid:11) is in Z[x] if and only if
(cid:11) is an algebraic integer.
(cid:15) A lemma of Gauss.
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De(cid:12)nition. Let f(x) = n a xj 2 Z[x] with f(x) 6(cid:17) 0. Then the content of f(x) is
j=0 j
gcd(an;an−1;:::;a1;a0). If the content of f(x) is 1, then f(x) is primitive.
Lemma. If u(x) and v(x) are primitive polynomials, then so is u(x)v(x).
Proof. It su(cid:14)ces to prove thPat the content of u(x)vP(x) is not divisible by each prime. Let
p be a prime. Write u(x) = n a xj and v(x) = m b xj. Let k and ‘ be non-negative
j=0 j j=0 j
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integers as small as possible such that p - a and p - b ; these exist since u(x) and v(x) are
k ‘
primitive. One checks that the coe(cid:14)cient of xk+‘ is not divisible by p. It follows that the
content of u(x)v(x) cannot be divisible by p, completing the proof. (cid:4)
Theorem 8 (Gauss’ Lemma). Let f(x) 2 Z[x]. Suppose that there exist u1(x) and
v1(x) in Q[x] such that f(x) = u1(x)v1(x). Then there exist u2(x) and v2(x) in Z[x] such
that f(x) = u2(x)v2(x) and degu2(x) = degu1(x) and degv2(x) = degv1(x).
Comment: The theorem implies that if f(x) 2 Z[x] has content 1, then a necessary and
su(cid:14)cient condition for f(x) to be irreducible over the rationals is for it be irreducible over
the integers. Also, we note that the proof will show more, namely that one can take u2(x)
and v2(x) to be rational numbers times u1(x) and v1(x), respectively.
Proof. Let d denote the content of f(x). Then there are positive rational integers a and b
and primitive polynomials u(x) and v(x) in Z[x] with degu(x) = degu1(x) and degv(x) =
degv1(x) satisfying u1(x)v1(x) = (a=b)u(x)v(x). Then there is a primitive g(x) 2 Z[x]
for which f(x) = dg(x) and bdg(x) = bf(x) = au(x)v(x). By the lemma, u(x)v(x) is
primitive. It follows that the content of au(x)v(x) is a. Since g(x) is primitive, the
content of bdg(x) is bd. Hence, a = bd. We set u2(x) = du(x) and v2(x) = v(x). Then
f(x) = u1(x)v1(x) = du(x)v(x) = u2(x)v2(x), and we deduce the theorem. (cid:4)
(cid:15) The proof of Theorem 7. It is clear that if the minimal polynomial for (cid:11) is in
Z[x], then (cid:11) is an algebraic integer. Now, consider an algebraic integer (cid:11), and let f(x) 2
Z[x] be monic with f((cid:11)) = 0. Let u1(x) be the minimal polynomial for (cid:11). We want
to prove that u1(x) 2 Z[x]. By the division algorithm for polynomials in Q[x], there
exist v1(x) and r(x) in Q[x] such that f(x) = u1(x)v1(x) + r(x) and either r(x) (cid:17) 0
or 0 (cid:20) degr(x) < degu1(x). Note that r((cid:11)) = f((cid:11)) − u1((cid:11))v1((cid:11)) = 0. Since u1(x)
is the monic polynomial of smallest degree having (cid:11) as a root, it follows that r(x) (cid:17) 0
(otherwise, there would be a k 2 Z for which (1=k)r(x) 2 Q[x] is monic, is of smaller
degree than degu1(x), and has (cid:11) as a root). Thus, f(x) = u1(x)v1(x) is a factorization
of f(x) in Q[x]. By Gauss’ Lemma and the comment after it, there exist u2(x) and v2(x)
in Z[x] with f(x) = u2(x)v2(x) and with u2(x) = mu1(x) for some non-zero rational
number m. By considering f(x) = (−u2(x))(−v2(x)) if necessary, we may suppose that
the leading coe(cid:14)cient of u2(x) is positive. Since f(x) is monic, we deduce that u2(x) is
monic. Comparing leading coe(cid:14)cients in u2(x) = mu1(x), we see that m = 1 so that
u1(x) = u2(x) 2 Z[x] as desired.
Algebraic Number Fields:
(cid:15) The de(cid:12)nition. If (cid:11) is an algebraic number, then Q((cid:11)) is de(cid:12)ned to be the smallest
(cid:12)eld containing both (cid:11) and the rationals.
(cid:15) Some simple observations. Let f(x) 2 Q[x] be the minimal polynomial for (cid:11). By
considering each integer j (cid:21) 0 successively and (cid:11)jf((cid:11)) = 0, one shows that (cid:11)n+j can be
expressed as a polynomial in (cid:11) with coe(cid:14)cients in Q and with degree (cid:20) n−1. It follows
that Q((cid:11)) is the set of all numbers of the form g((cid:11))=h((cid:11)) where g(x) and h(x) are in Z[x],
degg(x) (cid:20) n−1, degh(x) (cid:20) n−1, and h((cid:11)) 6= 0. By Theorem 4, every element of Q((cid:11))
is an algebraic number. For this reason, we refer to Q((cid:11)) as an algebraic number (cid:12)eld.
8
(cid:15) The ring of algebraic integers in Q((cid:11)).
Theorem 9. The algebraic integers contained in an algebraic number (cid:12)eld Q((cid:11)) form a
ring.
Proof. If (cid:11) and (cid:12) are in Q((cid:11)), then so are (cid:11)(cid:12) and (cid:11) −(cid:12) since Q((cid:11)) is a (cid:12)eld. If also (cid:11)
and (cid:12) are algebraic integers, then Theorem 5 implies (cid:11)(cid:12) and (cid:11)−(cid:12) are algebraic integers.
The result follows. (cid:4)
Homework:
(1) Prove that for every algebraic number (cid:11), the minimal polynomial for (cid:11) exists and is
unique.
(2) Prove that the minimal polynomial f(x) for an algebraic number (cid:11) is irreducible over
the rationals. In other words, prove that there do not exist g(x) and h(x) in Q[x] of degree
(cid:21) 1 satisfying f(x) = g(x)h(x).
(3) With the notation in the section above, let (cid:11)1;(cid:11)2;:::;(cid:11)n be the roots of f(x) with
(cid:11)1 = (cid:11). Show that w = h((cid:11)2)h((cid:11)3)(cid:1)(cid:1)(cid:1)h((cid:11)n) 2 Q[(cid:11)] (i.e., w can be expressed as a
polynomial in (cid:11) with rational coe(cid:14)cients). Also, show that w 6= 0. By considering
(g((cid:11))w)=(h((cid:11))w), show that every element of Q((cid:11)) can be written uniquely in the form
u((cid:11)) where u(x) 2 Q[x] and degu(x) (cid:20) n−1. (There are other ways to establish this; we
will in fact do this momentarily. The homework problem is to establish this result about
Q((cid:11)) by showing that one can \rationalize the denominator" of g((cid:11))=h((cid:11)).)
Quadratic Extensions:
p
(cid:15) De(cid:12)nition. Let m 2 Z with m not a square. Then Q( m) is a quadratic extension of
p
the rationals. Note that the minimal polynomial for m is x2−m (see the (cid:12)rst homework
exercises, problem (5)).
p
(cid:15) The elements of Q( m). We have discussed this in more generality already. If
p
(cid:12) 2 Q( m), then there are rational integers a, b, c, and d such that
p p p p
a+b m a+b m c−d m (ac−bdm)+(bc−ad) m
(cid:12) = p = p (cid:2) p = :
c+d m c+d m c−d m c2 −md2
p
Observe that the denominator is non-zero since m 62 Q. The above corresponds to what
p
took place in the last homework problem; we have shown that each element of Q( m) can
p
be expressed as a linear polynomial in m with coe(cid:14)cients in Q. Note that each element
p p
of Q( m) has a unique representation of the form a+b m with a and b rational.
(cid:15) When does Q((cid:11)1) = Q((cid:11)2)? One way to show two algebraic number (cid:12)elds are the
same (cid:12)eld is to show that (cid:11)1 2 Q((cid:11)2) (so that Q((cid:11)2) is a (cid:12)eld containing (cid:11)1) and that
(cid:11)2 2 Q((cid:11)1) (so that Q((cid:11)1) is a (cid:12)eld containing (cid:11)2). Explain why this is enough.
p p
(cid:15) When does Q( m) = Q( m0)? Given the above, equality holds if there are positive
2 2 0
integers k and ‘ such that k m = ‘ m . It follows that all quadratic extensions are of the
p
form Q( m) with m a squarefree integer and m 6= 1. Since m squarefree implies m is not
9
p
a spquare, these are all quadratic extensionps. Are thpese all di(cid:11)erent? Spuppose Q( m) =
Q( m0) with m and m0 squarefree. Then m 2 Q( m0) implies that m0m 2 Q (with a
0
tiny bit of work). It follows that m = m .
p
(cid:15) What are the algebraic integers in Q( m)? We suppose now that m is a squarefree
integer with m 6= 1. Note that m 6(cid:17) 0 (mod 4). The algebraic integers in Q((cid:11)) in general
form a ring. We show the following:
p
Theorem 10. The ring of algebraic integers in Q( m) (where m is a squarefree integer
with m 6= 1) is
p
R = Z[ m] if m (cid:17) 2 or 3 (mod 4)
and is
(cid:26) p (cid:27) (cid:20) p (cid:21)
a+b m 1+ m
R = : a 2 Z;b 2 Z;a (cid:17) b (mod 2) = Z if m (cid:17) 1 (mod 4):
2 2
p
Proof. Let R0 be the ring of algebraic integers in Q( m), and let R be de(cid:12)ned as in
p
the displayed equations above. Let (cid:12) 2 Q( m). The above implies we may write (cid:12) =
p
(a+b m)=dwherea, b, anddareintegerswithd > 0andgcd(a;b;d) = 1. From(d(cid:12)−a)2 =
b2m, we obtain that (cid:12) is a root of the quadratic f(x) = x2−(2a=d)x+(a2−b2m)=d2. We
easily deduce that R (cid:18) R0.
To show R0 (cid:18) R, we consider (cid:12) 2 R0 and show it must be in R. If (cid:12) is rational, then
Theorem 2 implies that (cid:12) 2 Z and, hence, (cid:12) 2 R. Suppose then that (cid:12) 62 Q. Since (cid:12) is
a root of the monic quadratic f(x), we deduce that f(x) is the minimal polynomial for
(cid:12). By Theorem 7, we obtain dj(2a) and d2j(a2 − b2m). The condition gcd(a;b;d) = 1
implies gcd(a;d) = 1 (otherwise, pjgcd(a;d) and d2j(a2 −b2m) implies pjb). Since d > 0
and dj(2a), we obtain that d is either 1 or 2.
If d = 1, then (cid:12) 2 R as desired. So suppose d = 2. Since gcd(a;b;d) = 1, at least one of
a and b is odd. Since d2j(a2 −b2m), we obtain a2 (cid:17) b2m (mod 4). Now, m 6(cid:17) 0 (mod 4)
implies that a and b are both odd. Therefore, a2 (cid:17) b2 (cid:17) 1 (mod 4). The congruence
a2 (cid:17) b2m (mod 4) gives that m (cid:17) 1 (mod 4). The theorem follows. (cid:4)
Good Rational Approximations and Units:
(cid:15) Given a real number (cid:11), what does it mean to have a good rational approximation
to it? As we know, it is possible to obtain an arbitrary good approximation to (cid:11) by using
rational numbers. In other words, given an " > 0, we can (cid:12)nd a rational number a=b
(here, a and b denote integers with b > 0) such that j(cid:11)−(a=b)j < ". So how much better
can \good" be? A proof of this " result is helpful. Of course, one can appeal to the fact
that the rationals are dense on the real line, but we consider a di(cid:11)erent approach. Let
b > 1=(2") and divide the number line into disjoint intervals I = (k=b;(k+1)=b] of length
1=b. The number (cid:11) will lie in one of them. Consider the endpoints of this interval, and let
a=b be the endpoint which is nearest to (cid:11) (either endpoint will do if (cid:11) is the midpoint).
Then j(cid:11)−(a=b)j (cid:20) 1=(2b) < ".
10
(cid:15) Answering the question. We can view the a=b we constructed as a rational approxi-
mation of (cid:11), but it is possible to have better approximations in the following sense. The
above choice for a=b satis(cid:12)es j(cid:11)−(a=b)j (cid:20) 1=(2b). Let’s prove now that there are a=b satis-
fying j(cid:11)−(a=b)j < 1=b2 (but we note here a di(cid:11)erence: for in(cid:12)nitely many positive integers
2
b, there is an a for which a=b is within 1=b of being (cid:11); for every positive integer b, there
is an a for which a=b is within 1=(2b) of being (cid:11)). We use the Dirichlet drawer principle
as Dirichlet himself did. Fix a positive integer N. For each b 2 [0;N], consider a = [b(cid:11)] so
that b(cid:11)−a 2 [0;1). Two of these N +1 values must be within 1=N of each other. More
precisely, there are integers b1 and b2 in [0;N] with (b2(cid:11)−a2)−(b1(cid:11)−a1) 2 [0;1=N) for
some integers a1 and a2. By taking b = jb2 −b1j 2 [1;N], and a = (cid:6)(a2 −a1), we deduce
(cid:12) (cid:12)
(cid:12) (cid:12)
jb(cid:11)−aj < 1 =) (cid:12)(cid:12)(cid:11)− a(cid:12)(cid:12) < 1 (cid:20) 1 :
N b bN b2
(cid:15) Avoiding the question further. What exactly \good" means is questionable. We
will see later that for everpy real number (cid:11), there are in(cid:12)nitely many positivepintegers b
such that j(cid:11)−(a=b)j < 1=( 5b2) for some integer a. Furthermore, the number 5 is best
possible. Perhaps then such a=b should be considered good rational approximations of (cid:11).
Or maybe those are great rational approximations and we should view any rational number
2
a=b within 1=b of (cid:11) or something close to that as being a good rational approximation.
I didn’t really intend to de(cid:12)ne good because what’s good in general tends to depend on
the individual asking the question, and whatever I tell you is good you might not believe
anyway.
(cid:15) Units and an example. A unit in a ring R is an element of R thpat has a multiplicative
inverse.pLet’s consider R to be the ring of algebraic integers in Q( 2). pBy Theorem 10,
R = Z[ 2]. Let (cid:12) 2 R, so there are integers a and b such that (cid:12) = a+b 2. We suppose
(cid:12) is a unit in R. Then p
1 1 a−b 2 p
= p = 2 Z[ 2]:
(cid:12) a+b 2 a2 −2b2
p
Thpus, (a2 −2b2)ja and (a2 −2b2)jb (use the uniqueness of the representation x+y 2 in
Q( 2) where x and y are rational). Let d = gcd(a;b). Then d2j(a2 − 2b2) implies that
d2ja and d2jb. But this means d2 (cid:20) d. We deduce that d = 1. On the other hand, a2−2b2
is a commonpdivisor of a and b. It follows that a2 −2b2 = (cid:6)1. Remember this for later.
If (cid:12) = a + b 2 is a unit in R, then a2 − 2b2 = (cid:6)1. The converse is easily seen to be
true as well. We consider now the case when a and b are positive (the other solutions of
a2 −2b2 = (cid:6)1 can be obtained from these).
We obtain that (cid:18) (cid:19)(cid:18) (cid:19)
p p
a a 1
− 2 + 2 = (cid:6)
b b b2
so that (cid:12) (cid:12)
(cid:12)p (cid:12)
(cid:12)(cid:12) 2− a(cid:12)(cid:12) = (cid:18) 1p (cid:19) < p1 :
b a 2b2
+ 2 b2
b