代数数论教程-习题解答(Solutions)
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6. The sequence 0 → N → M → M/N → 0 is exact, so by Problem 5, the sequence
0 → NS → MS → (M/N )S → 0 is exact. (If f is one of the maps of the first sequence, the corresponding map in the second sequence is S−1f .) It follows from the definition of
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R is a local ring with maximal ideal M. 4. S−1(g ◦ f ) takes m/s to g(f (m))/s, as does (S−1g) ◦ (S−1f ). If f is the identity on M , then S−1f is the identity on S−1M . 5. By hypothesis, g ◦f = 0, so (S−1g)◦(S−1f ) = S−1(g ◦f ) = S−10 = 0. Thus im S−1f ⊆ ker S−1g. Conversely, let y ∈ N, s ∈ S, with y/s ∈ ker S−1g. Then g(y)/s = 0/1, so for
xn + an−1xn−1 + · · · + a1x + a0 = 0, ai ∈ A.
By Problem 4, b + Q satisfies the same equation with ai replaced by ai + P for all i. Thus B/Q is integral over A/P . 6. By Problems 1-3, A/P is a field if and only if B/Q is a field, and the result follows. (Note that B/Q is an integral domain (because Q is a prime ideal), as required in the hypothesis of the result just quoted.)
Solutions to Problems
Chapter 1
Section 1.1
1. Multiply the equation by an−1 to get
a−1 = −(cn−1 + · · · + c1an−2 + c0an−1) ∈ A.
2. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0, then z = 0. 3. Any linear transformation on a finite-dimensional vector space is injective iff it is surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1. Therefore B is a field. 4. Since P is the preimage of Q under the inclusion map of A into B, P is a prime ideal. The map a + P → a + Q is a well-defined injection of A/P into B/Q, since P = Q ∩ A. Thus A/P can be viewed as a subring of B/Q. 5. If b + Q ∈ B/Q, then b satisfies an equation of the form
localization of a module that NS ≤ MS, and by exactness of the second sequence we have (M/N )S ∼= MS/NS.
some t ∈ S we have tg(y) = g(ty) = 0. Therefore ty ∈ ker g = im f , so ty = f (x) for some x ∈ M . We now have y/s = ty/st = f (x)/st = (S−1f )(x/st) ∈ im S−1f .
Section 1.2
1. If x ∈/ M, then by maximality of M, the ideal generated by M and x is R. Thus there exists y ∈ M and z ∈ R such that y + zx = 1. By hypothesis, zx, hence x, is a unit. Take the contrapositive to conclude that M contains all nonunits, so R is a local ring by (1.2.8). 2. Any additive subgroup of the cyclic additive group of Z/pnZ must consist of multiples of some power of p, and it follows that every proper ideal is contained in (p), which must therefore be the unique maximal ideal. 3. The set of nonunits is M = {f /g : g(a) = 0, f (a) = 0}, which is an ideal. By (1.2.8),
0 → NS → MS → (M/N )S → 0 is exact. (If f is one of the maps of the first sequence, the corresponding map in the second sequence is S−1f .) It follows from the definition of
1源自文库
2
R is a local ring with maximal ideal M. 4. S−1(g ◦ f ) takes m/s to g(f (m))/s, as does (S−1g) ◦ (S−1f ). If f is the identity on M , then S−1f is the identity on S−1M . 5. By hypothesis, g ◦f = 0, so (S−1g)◦(S−1f ) = S−1(g ◦f ) = S−10 = 0. Thus im S−1f ⊆ ker S−1g. Conversely, let y ∈ N, s ∈ S, with y/s ∈ ker S−1g. Then g(y)/s = 0/1, so for
xn + an−1xn−1 + · · · + a1x + a0 = 0, ai ∈ A.
By Problem 4, b + Q satisfies the same equation with ai replaced by ai + P for all i. Thus B/Q is integral over A/P . 6. By Problems 1-3, A/P is a field if and only if B/Q is a field, and the result follows. (Note that B/Q is an integral domain (because Q is a prime ideal), as required in the hypothesis of the result just quoted.)
Solutions to Problems
Chapter 1
Section 1.1
1. Multiply the equation by an−1 to get
a−1 = −(cn−1 + · · · + c1an−2 + c0an−1) ∈ A.
2. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0, then z = 0. 3. Any linear transformation on a finite-dimensional vector space is injective iff it is surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1. Therefore B is a field. 4. Since P is the preimage of Q under the inclusion map of A into B, P is a prime ideal. The map a + P → a + Q is a well-defined injection of A/P into B/Q, since P = Q ∩ A. Thus A/P can be viewed as a subring of B/Q. 5. If b + Q ∈ B/Q, then b satisfies an equation of the form
localization of a module that NS ≤ MS, and by exactness of the second sequence we have (M/N )S ∼= MS/NS.
some t ∈ S we have tg(y) = g(ty) = 0. Therefore ty ∈ ker g = im f , so ty = f (x) for some x ∈ M . We now have y/s = ty/st = f (x)/st = (S−1f )(x/st) ∈ im S−1f .
Section 1.2
1. If x ∈/ M, then by maximality of M, the ideal generated by M and x is R. Thus there exists y ∈ M and z ∈ R such that y + zx = 1. By hypothesis, zx, hence x, is a unit. Take the contrapositive to conclude that M contains all nonunits, so R is a local ring by (1.2.8). 2. Any additive subgroup of the cyclic additive group of Z/pnZ must consist of multiples of some power of p, and it follows that every proper ideal is contained in (p), which must therefore be the unique maximal ideal. 3. The set of nonunits is M = {f /g : g(a) = 0, f (a) = 0}, which is an ideal. By (1.2.8),