Math Forum Discussions - Prerequisites for Hartshorne's Algebraic Geometry (ver. 1.0) [16]

Nobuo Saito

Posts: 608
Registered: 12/12/04

Prerequisites for Hartshorne's Algebraic Geometry (ver. 1.0) [16]
Posted: Jan 19, 2002 1:40 AM

Plain Text

Prerequisites for Hartshorne's Algebraic Geometry (version 1.0)
by Nobuo Saito
genkisaito@hotmail.com

[Installment 16]
Integral dependence (3)

We will show another proof of "Going-down Theorem"
(proposition 13, [15]) following Atiyah-MacDonald.
Moreover, we will introduce some important notions.

Definition 1
Let A be a subring of a ring B.
Let I be an ideal of A.
Let x be an element of B such that
x^n + a_1x^(n-1) + ..., a_(n_1)x + a_n = 0,
for some elements a_1,...,a_n of I.
Then x is said to be INTEGRAL over I.
The INTEGRAL CLOSURE of I in B is the set of all elements
of B which are integral over I.

Proposition 2
Let A be a subring of a ring B.
Let C be an integral closure of A in B.
Let I be an ideal of A.
Then the integral closure of I in B is equal
to the nilradical of IC in C.
Hence it is closed under addition and multiplication.

Proof
Let x be an element of B integral over I.
Then x^n + a_1x^(n-1) + ..., a_(n_1)x + a_n = 0,
for some elements a_1,...,a_n of I.
Since x is obviously integral over A,
x^n is contained in IC.
Hence x is contained in nil(IC).

Conversely, suppose x is contained in nil(IC).
Then x^m = a_1x_1 + ... + a_nx_n,
where a_i is contained in I and x_i is integral over A
for each i.
Let M = A[x_1,...,x_n] be the subring of C
generated by x_1,...,x_n over A.
By (proposition 10, [14]), M is a finite A-module.
Let y = x^m.
Since y is contained in IM,
yM is contained in IMM <= IM.
Let z_1,...,z_r be generators of M as an A-module.
Then we get

yz_1 = d_(1,1)z_1 + d_(1,2)z_2 + ... + d_(1,r)z_r
yz_2 = d_(2,1)z_1 + d_(2,2)z_2 + ... + d_(2,r)z_r
...
yz_r = d_(r,1)z_1 + d_(r,2)z_2 + ... + d_(r,r)z_r,

where d_(i,j) is an element of I for each pair (i,j).

By a similar argument as in the proof of (proposition 8, [14]),
det(yE - D) = 0, where E is the identity matrix,
and D is the matrix (d_(i,j)).
Hence y is integral over I.
Hence x is integral over I.
Q.E.D.

Proposition 3
Let B be an integral domain which contains
an integrally closed domain A.
Let K be the field of fractions of A.
Let x be an element of B integral over an ideal I of A.
Let x^n + a_1x^(n-1) + ..., a_(n_1)x + a_n
be the minimal polynomial of x over K.
Then a_i is contained in nil(I) for each i.

Proof
Let L be an algebraic extension field of K which contains
all the conjugates x_1,...,x_n of x.
Then x_i is integral over I for each i.
Since each a_i is expressed as a symmetric polynomial
of x_1,...,x_n, it is integral over I by proposition 2.
Hence, each a_i satisfies an equation
X^m + b_1X^(m-1) + ..., b_(m_1)X + b_m = 0,
where each b_j is contained in I.
Since A is integrally closed, a_i is contained in A.
Hence (a_i)^m is contained in I for each i.
Q.E.D.

Proposition 4
Let f:A -> B be a ring homomorphism.
Let P be a prime ideal of A.
Let S = f(A - P).
Let k(P) = A_P/PA_P.
Then (B(x)k(P))/A is isomorphic to B[1/S]/PB[1/S].

Proof
Since 0 -> PA_P -> A_P -> k(P) -> 0 is exact,
B(x)PA_P -> B(x)A_P ->B(x)k(P) -> 0 is exact
by (proposition 17, [4])
B(x)A_P can be identified with B[1/S]
by (proposition 14, [7]) and (proposition 18-2, [7]).
Im(B(x)PA_P -> B(x)A_P) can be identified with PB[1/S].
Hence the assertion follows.
Q.E.D.

Definition 5
Under the conditions and the notation of proposition,
Spec(B(x)k(P)) is called a fiber over P.
B(x)k(P) is called a fiber ring over P.

Proposition 6
Let f:A -> B be a ring homomorphism.
Let P be a prime ideal of A.
Let S = f(A - P).
Then there exists bijection between Spec(B[1/S]/PB[1/S]),
i.e. the fiber over P, and the set of prime ideals
of B lying over P.

Proof
The following diagram commutes.

A ---> B
| |
v v
A_P -> B[1/S]

Suppose P = Q|A for a prime ideal Q of B.
Since Q does not meet S, QB[1/S] is a prime ideal of B[1/S]
by (theorem 20, [7]).
Since PB <= Q, PB[1/S] <= QB[1/S].

Conversely, let Q' be a prime ideal of B[1/S]
such that PB[1/S] <= Q'.
Let Q = Q'|B.
Since Q|A does not meet A - P, P >= Q|A.
Since PB[1/S] <= Q', P <= Q|A.
Hence P = Q|A.
Q.E.D.

Corollary
Let f:A -> B be a ring homomorphism.
Let P be a prime ideal of A.
Then P = (PB)|A if and only if there exists a prime ideal Q of B
such that P = Q|A.

Proof
By proposition 6 and (theorem 27, [1]),
we get the following equivalences.

P = Q|A for some prime ideal of B <=>
Spec(B[1/S]/PB[1/S]) is not empty <=>
B[1/S]/PB[1/S] is not the zero ring <=>
PB[1/S] is a proper ideal of B[1/S] <=>
PB doesn't meet S = f(A - P) <=>
P = (PB)|A.
Q.E.D.

Proposition 7
Let A be an integrally closed domain.
Let B be an integral domain which contains A
and integral over A.
Let P be a prime ideal of A.
Then P = (PB)|A.

Proof
Let x be an element of PB.
By proposition 2, x is integral over P.
Let K be the field of fractions of A.
Let x^n + a_1x^(n-1) + ..., a_(n_1)x + a_n
be the minimal polynomial of x over K.
By proposition 3, each a_i is contained in P.
If x is contained in A, then x^n is contained in P.
Hence x is contained in P.
Q.E.D.

Proposition 8
Let A be an integrally closed domain.
Let B be an integral domain which contains A
and integral over A.
Let P > P' be prime ideals of A.
Let Q be a prime ideal of B lying over P.
Then there exist a prime ideal Q' of B
lying over P' and Q > Q'.

Proof
The following diagram commutes.

A --> B
| |
v v
A_P -> B_Q

By (proposition 15, [14]), A_P is integrally closed.
By (proposition 14, [14]), B_Q is integral over A_P.
Hence we can apply proposition 7 to A_P -> B_Q.
Hence P'A_P = (P'B_Q)|A_P.
Hence P' = (P'B_Q)|A.
By the corollary of proposition 6,
there exists a prime ideal Q''
of B_Q lying over P'.
Hence Q'= Q''|B lies over P'.
Since Q'' is a prime ideal of B_Q, Q' <= Q.
Since P' < P, Q' < Q.
Q.E.D.

Nobuo Saito