Techniques of abstract algebra have been applied to problems in number theory for a long

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 = Î±Î² where neither Î± nor Î² is a unit.

Define the norm of Î³ = a + bi as N(Î³) = a2 + b2. Then N(Î³) = 1 iff Î³ is 1,-1,i or −i,

equivalently, iff Î³ is a unit. Thus

p2 = N(p) = N(Î±)N(Î²) with N(Î±) > 1 and N(Î²) > 1,

so N(Î±) = N(Î²) = p. If Î± = 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 Î± ∈ E, then Î± is algebraic over F iff Î± 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 Î²1, . . . , Î²n generate B over A, then xÎ²i is a linear combination of the

Î²j, say xÎ²i = n

j=1 cijÎ²j. Thus if Î² is a column vector whose components are the Î²i, I

is an n by n identity matrix, and C = [cij ], then

(xI − C)Î² = 0,and if we premultiply by the adjoint matrix of xI − C (as in Cramer’s rule), we get

[det(xI − C)]IÎ² = 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 xÎ²i ∈ 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.

Next

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 = Î±Î² where neither Î± nor Î² is a unit.

Define the norm of Î³ = a + bi as N(Î³) = a2 + b2. Then N(Î³) = 1 iff Î³ is 1,-1,i or −i,

equivalently, iff Î³ is a unit. Thus

p2 = N(p) = N(Î±)N(Î²) with N(Î±) > 1 and N(Î²) > 1,

so N(Î±) = N(Î²) = p. If Î± = 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 Î± ∈ E, then Î± is algebraic over F iff Î± 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 Î²1, . . . , Î²n generate B over A, then xÎ²i is a linear combination of the

Î²j, say xÎ²i = n

j=1 cijÎ²j. Thus if Î² is a column vector whose components are the Î²i, I

is an n by n identity matrix, and C = [cij ], then

(xI − C)Î² = 0,and if we premultiply by the adjoint matrix of xI − C (as in Cramer’s rule), we get

[det(xI − C)]IÎ² = 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 xÎ²i ∈ 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.

Next