Unknown
time, notably in the effort to prove Fermat’s last theorem. As an introductory example,
we will sketch a problem for which an algebraic approach works very well. If p is an odd
prime and p ≡ 1 mod 4, we will prove that p is the sum of two squares, that is, p can
expressed as x2 + y2 where x and y are integers. Since p−1
2 is even, it follows that −1
is a quadratic residue (that is, a square) mod p. To see this, pair each of the numbers
2, 3, . . . , p−2 with its inverse mod p, and pair 1 with p−1 ≡ −1 mod p. The product of
the numbers 1 through p − 1 is, mod p,
1 × 2 × ·· ·× p − 1
2
×−1×−2 × ··· × −p − 1
2
and therefore
[(p − 1
2
)!]2 ≡ −1 modp.
If −1 ≡ x2 mod p, then p divides x2 +1. Now we enter the ring Z[i] of Gaussian integers
and factor x2 + 1 as (x + i)(x − i). Since p can divide neither factor, it follows that p is
not prime in Z[i]. Since the Gaussian integers form a unique factorization domain, p is
not irreducible, and we can write p = αβ where neither α nor β is a unit.
Define the norm of γ = a + bi as N(γ) = a2 + b2. Then N(γ) = 1 iff γ is 1,-1,i or −i,
equivalently, iff γ is a unit. Thus
p2 = N(p) = N(α)N(β) with N(α) > 1 and N(β) > 1,
so N(α) = N(β) = p. If α = x + iy, then p = x2 + y2.
Conversely, if p is an odd prime and p = x2 + y2, then p is congruent to 1 mod 4. [If
x is even, then x2 ≡ 0 mod 4, and if x is odd, then x2 ≡ 1 mod 4. We cannot have x and
y both even or both odd, since p is odd.]
It is natural to conjecture that we can identify those primes that can be represented as
x2 + |m|y2, where m is a negative integer, by working in the ring Z[
√
m]. But the above
argument depends critically on unique factorization, which does not hold in general. Astandard example is 2 × 3 = (1 +
√
−5)(1 −
√
−5) in Z[
√
−5]. Difficulties of this sort led
Kummer to invent “ideal numbers”, which became ideals at the hands of Dedekind. We
will see that although a ring of algebraic integers need not be a UFD, unique factorization
of ideals will always hold.
1.1 Integral Extensions
If E/F is a field extension and α ∈ E, then α is algebraic over F iff α is a root of
a nonconstant polynomial with coefficients in F. We can assume if we like that the
polynomial is monic, and this turns out to be crucial in generalizing the idea to ring
extensions.
1.1.1 Definitions and Comments
All rings are assumed commutative. Let A be a subring of the ring R, and let x ∈ R. We
say that x is integral over A if x is a root of a monic polynomial f with coefficients in
A. The equation f(X) = 0 is called an equation of integral dependence for x over A. If x
is a real or complex number that is integral over Z, then x is called an algebraic integer.
Thus for every integer d,
√
d is an algebraic integer, as is any nth root of unity. (The
monic polynomials are, respectively, X2 − d and Xn − 1.) The next results gives several
conditions equivalent to integrality.
1.1.2 Theorem
Let A be a subring of R, and let x ∈ R. The following conditions are equivalent:
(i) The element x is integral over A;
(ii) The A-module A[x] is finitely generated;
(iii) The element x belongs to a subring B of R such that A ⊆ B and B is a finitely
generated A-module;
(iv) There is a subring B of R such that B is a finitely generated A-module and x stabilizes
B, that is, xB ⊆ B. (If R is a field, the assumption that B is a subring can be dropped,
as long as B = 0);
(v) There is a faithful A[x]-module B that is finitely generated as an A-module. (Recall
that a faithful module is one whose annihilator is 0.)
Proof.
(i)implies (ii): If x is a root of a monic polynomial of degree n over A, then xn and all
higher powers of x can be expressed as linear combinations of lower powers of x. Thus
1, x, x2, . . . , xn−1 generate A[x] over A.
(ii) implies (iii): Take B = A[x].
(iii) implies (i): If β1, . . . , βn generate B over A, then xβi is a linear combination of the
βj, say xβi = n
j=1 cijβj. Thus if β is a column vector whose components are the βi, I
is an n by n identity matrix, and C = [cij ], then
(xI − C)β = 0,and if we premultiply by the adjoint matrix of xI − C (as in Cramer’s rule), we get
[det(xI − C)]Iβ = 0
hence det(xI −C)b = 0 for every b ∈ B. Since B is a ring, we may set b = 1 and conclude
that x is a root of the monic polynomial det(XI − C) in A[X].
If we replace (iii) by (iv), the same proofs work. If R is a field, then in (iv)⇒(i), x is
an eigenvalue of C, so det(xI − C) = 0.
If we replace (iii) by (v), the proofs go through as before. [Since B is an A[x]-module,
in (v)⇒(i) we have xβi ∈ B. When we obtain [det(xI − C)]b = 0 for every b ∈ B, the
hypothesis that B is faithful yields det(xI − C) = 0.] ♣
We are going to prove a transitivity property for integral extensions, and the following
result will be helpful.
1.1.3 Lemma
Let A be a subring of R, with x1, . . . , xn ∈ R. If x1 is integral over A, x2 is integral
over A[x1], . . . , and xn is integral over A[x1, . . . , xn−1], then A[x1, . . . , xn] is a finitely
generated A-module.
Proof. The n = 1 case follows from (1.1.2), condition (ii). Going from n−1 to n amounts
to proving that if A,B and C are rings, with C a finitely generated B-module and B a
finitely generated A-module, then C is a finitely generated A-module. This follows by a
brief computation:
C =
r
j=1
Byj, B =
s
k=1
Axk, so C =
r
j=1
s
k=1
Ayjxk. ♣
1.1.4 Transitivityof Integral Extensions
Let A, B and C be subrings of R. If C is integral over B, that is, every element of C is
integral over B, and B is integral over A, then C is integral over A.
Proof. Let x ∈ C, with xn + bn−1xn−1 + · · · + b1x + b0 = 0, bi ∈ B. Then x is integral
over A[b0, . . . , bn−1]. Each bi is integral over A, hence over A[b0, . . . , bi−1]. By (1.1.3),
A[b0, . . . , bn−1, x] is a finitely generated A-module. It follows from condition (iii) of (1.1.2)
that x is integral over A. ♣
1.1.5 Definitions and Comments
If A is a subring of R, the integral closure of A in R is the set Ac of elements of R that
are integral over A. Note that A ⊆ Ac because each a ∈ A is a root of X −a. We say that
A is integrally closed in R if Ac = A. If we simply say that A is integrally closed without
reference to R, we assume that A is an integral domain with fraction field K, and A is
integrally closed in K.
If x and y are integral over A, then just as in the proof of (1.1.4), it follows from
(1.1.3) that A[x, y] is a finitely generated A-module. Since x + y, x − y and xy belong tothis module, they are integral over A by (1.1.2), condition (iii). The important conclusion
is that
Ac is a subring of R containing A.
If we take the integral closure of the integral closure, we get nothing new.
1.1.6 Proposition
The integral closure Ac of A in R is integrally closed in R.
Proof. By definition, Ac is integral over A. If x is integral over Ac, then as in the proof
of (1.1.4), x is integral over A, and therefore x ∈ Ac. ♣
We can identify a large class of integrally closed rings.
1.1.7 Proposition
If A is a UFD, then A is integrally closed.
Proof. If x belongs to the fraction field K, then we can write x = a/b where a, b ∈ A,
with a and b relatively prime. If x is integral over A, then there is an equation of the form
(a/b)n + an−1(a/b)n−1 + · · · + a1(a/b) + a0 = 0
with all ai belonging to A. Multiplying by bn, we have an + bc = 0, with c ∈ A. Thus b
divides an, which cannot happen for relatively prime a and b unless b has no prime factors
at all, in other words, b is a unit. But then x = ab−1 ∈ A. ♣
Problems For Section 1.1
Let A be a subring of the integral domain B, with B integral over A. In Problems 1-3,
we are going to show that A is a field if and only if B is a field.
1. Assume that B is a field, and let a be a nonzero element of A. Then since a−1 ∈ B,
there is an equation of the form
(a
−1)n + cn−1(a
−1)n−1 + · · · + c1a
−1 + c0 = 0
with all ci belonging to A. Show that a−1 ∈ A, proving that A is a field.
2. Now assume that A is a field, and let b be a nonzero element of B. By condition
(ii) of (1.1.2), A[b] is a finite-dimensional vector space over A. Let f be the A-linear
transformation on this vector space given by multiplication by b, in other words, f(z) =
bz, z ∈ A[b]. Show that f is injective.
3. Show that f is surjective as well, and conclude that B is a field.
In Problems 4-6, let A be a subring of B, with B integral over A. Let Q be a prime
ideal of B and let P = Q ∩ A.
4. Show that P is a prime ideal of A, and that A/P can be regarded as a subring of B/Q.
5. Show that B/Q is integral over A/P.
6. Show that P is a maximal ideal of A if and only if Q is a maximal ideal of B.Let S be a subset of the ring R, and assume that S is multiplicative, in other words,
0 /∈
S, 1 ∈ S, and if a and b belong to S, so does ab. In the case of interest to us, S will
be the complement of a prime ideal. We would like to divide elements of R by elements
of S to form the localized ring S−1R, also called the ring of fractions of R by S. There is
no difficulty when R is an integral domain, because in this case all division takes place in
the fraction field of R. Although we will not need the general construction for arbitrary
rings R, we will give a sketch. For full details, see TBGY, Section 2.8.
1.2.1 Construction of the Localized Ring
If S is a multiplicative subset of the ring R, we define an equivalence relation on R × S
by (a, b) ∼ (c, d) iff for some s ∈ S we have s(ad − bc) = 0. If a ∈ R and b ∈ S, we define
the fraction a/b as the equivalence class of (a, b). We make the set of fractions into a ring
in a natural way. The sum of a/b and c/d is defined as (ad + bc)/bd, and the product of
a/b and c/d is defined as ac/bd. The additive identity is 0/1, which coincides with 0/s for
every s ∈ S. The additive inverse of a/b is −(a/b) = (−a)/b. The multiplicative identity
is 1/1, which coincides with s/s for every s ∈ S. To summarize:
S−1R is a ring. If R is an integral domain, so is S−1R. If R is an integral domain and
S = R \ {0}, then S−1R is a field, the fraction field of R.
There is a natural ring homomorphism h : R → S−1R given by h(a) = a/1. If S
has no zero-divisors, then h is a monomorphism, so R can be embedded in S−1R. In
particular, a ring R can be embedded in its full ring of fractions S−1R, where S consists
of all non-divisors of 0 in R. An integral domain can be embedded in its fraction field.
Our goal is to study the relation between prime ideals of R and prime ideals of S−1R.
1.2.2 Lemma
If X is any subset of R, define S−1X = {x/s : x ∈ X, s ∈ S}. If I is an ideal of R, then
S−1I is an ideal of S−1R. If J is another ideal of R, then
(i) S−1(I + J) = S−1I + S−1J;
(ii) S−1(IJ) = (S−1I)(S−1J);
(iii) S−1(I ∩ J) = (S−1I) ∩ (S−1J);
(iv) S−1I is a proper ideal iff S ∩ I = ∅.
Proof. The definitions of addition and multiplication in S−1R imply that S−1R is an
ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse
inclusions in (i) and (ii) follow from
a
s
+ b
t
= at + bs
st
,
a
s
b
t
= ab
st
.
To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s, t ∈ S. There exists u ∈ S such that
u(at − bs) = 0. Then a/s = uat/ust = ubs/ust ∈ S−1(I ∩ J).
Finally, if s ∈ S ∩ I, then 1/1 = s/s ∈ S−1I, so S−1I = S−1R. Conversely, if
S−1I = S−1R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that
t(s − a) = 0, so at = st ∈ S ∩ I. ♣
1.2.3 Lemma
Let h be the natural homomorphism from R to S−1R [see (1.2.1)]. If J is an ideal of
S−1R and I = h−1(J), then I is an ideal of R and S−1I = J.
Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S−1I, with
a ∈ I and s ∈ S. Then a/1 = h(a) ∈ J, so a/s = (a/1)(1/s) ∈ J. Conversely, let a/s ∈ J,
with a ∈ R, s ∈ S. Then h(a) = a/1 = (a/s)(s/1) ∈ J, so a ∈ I and a/s ∈ S−1I. ♣
Prime ideals yield sharper results.
1.2.4 Lemma
If I is any ideal of R, then I ⊆ h−1(S−1I). There will be equality if I is prime and disjoint
from S.
Proof. If a ∈ I, then h(a) = a/1 ∈ S−1I. Thus assume that I is prime and disjoint from
S, and let a ∈ h−1(S−1I). Then h(a) = a/1 ∈ S−1I, so a/1 = b/s for some b ∈ I, s ∈ S.
There exists t ∈ S such that t(as − b) = 0. Thus ast = bt ∈ I, with st /∈
I because
S ∩ I = ∅. Since I is prime, we have a ∈ I. ♣
1.2.5 Lemma
If I is a prime ideal of R disjoint from S, then S−1I is a prime ideal of S−1R.
Proof. By part (iv) of (1.2.2), S−1I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S−1I,
with a, b ∈ R, s, t ∈ S. Then ab/st = c/u for some c ∈ I, u ∈ S. There exists v ∈ S such
that v(abu − cst) = 0. Thus abuv = cstv ∈ I, and uv /∈
I because S ∩ I = ∅. Since I is
prime, ab ∈ I, hence a ∈ I or b ∈ I. Therefore either a/s or b/t belongs to S−1I. ♣
The sequence of lemmas can be assembled to give a precise conclusion.
1.2.6 Theorem
There is a one-to-one correspondence between prime ideals P of R that are disjoint from
S and prime ideals Q of S−1R, given by
P → S
−1P and Q → h
−1(Q).
Proof. By (1.2.3), S−1(h−1(Q)) = Q, and by (1.2.4), h−1(S−1P) = P. By (1.2.5), S−1P
is a prime ideal, and h−1(Q) is a prime ideal by the basic properties of preimages of sets.
If h−1(Q) meets S, then by (1.2.2) part (iv), Q = S−1(h−1(Q)) = S−1R, a contradiction.
Thus the maps P → S−1P and Q → h−1(Q) are inverses of each other, and the result
follows. ♣
S−1R, and call it the localization of R at P. We are going to show that RP is
a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions
equivalent to the definition of a local ring.
1.2.8 Proposition
For a ring R, the following conditions are equivalent.
(i) R is a local ring;
(ii) There is a proper ideal I of R that contains all nonunits of R;
(iii) The set of nonunits of R is an ideal.
Proof.
(i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the
unique maximal ideal I.
(ii) implies (iii): If a and b are nonunits, so are a + b and ra. If not, then I contains a
unit, so I = R, contradicting the hypothesis.
(iii) implies (i): If I is the ideal of nonunits, then I is maximal, because any larger ideal J
would have to contain a unit, so J = R. If H is any proper ideal, then H cannot contain
a unit, so H ⊆ I. Therefore I is the unique maximal ideal. ♣
1.2.9 Theorem
RP is a local ring.
Proof. Let Q be a maximal ideal of RP . Then Q is prime, so by (1.2.6), Q = S−1I
for some prime ideal I of R that is disjoint from S = R \ P. In other words, I ⊆ P.
Consequently, Q = S−1I ⊆ S−1P. If S−1P = RP = S−1R, then by (1.2.2) part (iv), P
is not disjoint from S = R \ P, which is impossible. Therefore S−1P is a proper ideal
containing every maximal ideal, so it must be the unique maximal ideal. ♣
1.2.10 Remark
It is convenient to write the ideal S−1I as IRP . There is no ambiguity, because the
product of an element of I and an arbitrary element of R belongs to I.
1.2.11 Localization of Modules
If M is an R-module and S a multiplicative subset of R, we can essentially repeat the
construction of (1.2.1) to form the localization of M by S, and thereby divide elements
of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t) equivalent if for
some u ∈ S, we have u(tx − sy) = 0. The equivalence class of (x, s) is denoted by x/s,
and addition is defined by
x
s
+ y
t
= tx + sy
st
.
R, denote S−1M by MS. If
N is a submodule of M, show that (M/N)S
∼=
MS/NS.
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