MATH 831 HOMEWORK SOLUTIONS – ASSIGNMENT 8

Exercise 5.3. Let f : B → B′ be a homomorphism of A-algebras, and let C be an A-algebra. If f is integral, prove that f ⊗ 1 : B ⊗_{A} C → B′ ⊗_{A} C is integral.

Proof. Because of Corollary 5.3, it is enough to show that, for every x ∈ B′ and c ∈ C, the element x ⊗ c ∈ B′ ⊗_{A} C is integral over (f ⊗ 1)(B ⊗_{A} C) (since a typical element of B′ ⊗_{A} C is of the form ∑_{i=1}^{n} x_{i} ⊗ c_{i} with x_{i} ∈ B′, c_{i} ∈ C). Then x ∈ B′ satisfies a monic polynomial equation

x^{n} + b_{1}x^{n−1} + ⋯ + b_{n−1}x + b_{n} = 0, with b_{i} ∈ f(B)

for i = 1, 2, …, n. Now direct checking shows that

(x ⊗ c)^{n} + (b_{1} ⊗ c)(x ⊗ c)^{n−1} + ⋯ + (b_{n−1} ⊗ c^{n−1})(x ⊗ c) + (b_{n} ⊗ c^{n}) = 0,

in which b_{i} ⊗ c^{n−i} ∈ (f ⊗ 1)(B ⊗_{A} C) for i = 1, 2, …, n. □

Exercise 5.5. Let A ⊆ B be rings, B integral over A.

i) If x ∈ A is a unit in B, then it is a unit in A.

ii) The Jacobson radical in A is the contraction of the Jacobson radical of B.

Proof. i): Since B is integral over A, the element x^{−1} ∈ B satisfies

(x^{−1})^{n} + a_{1}(x^{−1})^{n−1} + ⋯ + a_{n−1}(x^{−1}) + a_{n} = 0, with a_{i} ∈ A

for i = 1, 2, …, n. Multiply the above equation by x^{n−1} to get an equation

x^{−1} + a_{1} + ⋯ + a_{n−1}x^{n−2} + a_{n}x^{n−1} = 0,

which implies that

x^{−1} = −(a_{1} + ⋯ + a_{n−1}x^{n−2} + a_{n}x^{n−1}) ∈ A,

which means x is a unit in A.

Alternatively, we can use lying-over to prove part i): If x ∈ A is not a unit in A, then x in contained a some maximal ideal 𝔪_{A} of A. By lying-over (Theorem 5.10), there exists a prime ideal (which has to be maximal by Corollary 5.8) 𝔪_{B} of B such that 𝔪_{B} ∩ A = 𝔪_{A}. Hence x ∈ 𝔪_{B}, which contradicts the assumption that x is a unit in B.

ii): Denote the Jacobson radicals of A and B by ℜ_{A} and ℜ_{B} respectively. We need to show that ℜ_{A} = ℜ_{B} ∩ A. Choose x ∈ ℜ_{B} ∩ A and an arbitrary a ∈ A. Then 1 + ax ∈ A is a unit in B by Proposition 1.9 and hence is a unit in A by part i). Therefore x ∈ ℜ_{A} by Proposition 1.9 again. Hence ℜ_{B} ∩ A ⊆ ℜ_{A}. (You may also use Corollary 5.8 to prove ℜ_{B} ∩ A ⊆ ℜ_{A}, which is very similar to the next portion of the proof). On the other hand, choose x ∈ ℜ_{A} and an arbitrary maximal ideal 𝔪_{B} of B. Then 𝔪_{B} ∩A is a maximal ideal of A (by Corollary 5.8 since B is integral over A). Therefore x ∈ 𝔪_{B} ∩ A ⊆ 𝔪_{B}. Hence x ∈ ℜ_{B}, which implies that x ∈ ℜ_{B} ∩ A. Therefore ℜ_{A} ⊆ ℜ_{B} ∩ A. Thus ℜ_{A} = ℜ_{B} ∩ A. □

Exercise 5.7. Let A be a subring of B, such that the set B − A = B \ A is closed under multiplication. Show that A is integrally closed in B.

Proof. Suppose, on the contrary, that A is not integrally closed in B. Then let C be the integral closure of A in B and let n be the minimal degree of all monic polynomial equations satisfied by some elements of C − A = C \ A. Say x ∈ C − A = C \ A satisfies

x^{n} + a_{1}x^{n−1} + ⋯ + a_{n−1}x + a_{n} = 0, with a_{i} ∈ A

for i = 1, 2, …, n. Then clearly n ≥ 2 since x ∉ A. And also x^{n−1} + a_{1}x^{n−2} + ⋯ + a_{n−1} ∉ A by the minimality of n. But we have

x(x^{n−1} + a_{1}x^{n−2} + ⋯ + a_{n−1}) = −a_{n} ∈ A,

which contradicts the assumption that B − A = B \ A is closed under multiplication. □

Exercise 5.8. i) Let A be a subring of an integral domain B, and let C be the integral closure of A in B. Let f, g be monic polynomials in B[x] such that fg ∈ C[x]. Then f, g are in C[x].

ii) Prove the same result without assuming that B (or A) is an integral domain.

Proof. We are going to prove part ii) once and for all since it is more general than part i). We first claim that there is an extension ring D ⊇ B such that both f and g can be factorized as products of degree one monic polynomials. Assume the claim and say

f(x) = (x − d_{1})(x − d_{2}) ⋯ (x − d_{m})  and  g(x) = (x − d′_{1})(x − d′_{2}) ⋯ (x − d′_{n}),

in which d_{i}, d′_{j} ∈ D for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Then we have that the monic polynomial

f(x)g(x) = (x − d_{1})(x − d_{2}) ⋯ (x − d_{m})(x − d′_{1})(x − d′_{2}) ⋯ (x − d′_{n}) ∈ C[x],

which implies that d_{i}, d′_{j} are all integral over C and hence are integral over A by Corollary 5.4. Since the coefficients of f and g can be expressed as polynomials in d_{i} and d′_{j}, we conclude that the coefficients are all integral over A. Finally, since f, g ∈ B[x] and C is the integral closure of A in B, we conclude that the coefficients are all in C, i.e. f, g ∈ C[x].

To complete the proof, we need to prove the claim. It is enough to find an extension ring D ⊇ B such that f(x) ∈ B[x] can be factorized as a product of degree one monic polynomials. We prove it by induction on deg f(x). It is trivially true if deg f(x) = 1. If deg f(x) > 1, then let D_{1} := \frac{B[t]}{f(t)B[t]}, which is a ring. For any h(t) ∈ B[t], denote the coset h(t) + f(t)B[t] ∈ D_{1} :=

\frac{B[t]}{f(t)B[t]} by \overline{h(t)}. Then the ring B is naturally embedded into D_{1} by b ↦ \overline{b}. So we may as well consider B as a subring of D_{1} by identifying b ∈ B with \overline{b} ∈ D_{1}. Keeping this identification in mind, we can directly check that \overline{t} ∈ D_{1} is a root of f(x) = 0 since \overline{f(t)} = 0 ∈ D_{1}. Therefore f(x) = (x − \overline{t})f_{1}(x) ∈ D_{1}[x], in which deg f_{1}(x) < deg f(x). By induction hypothesis, there exists a extension ring D ⊇ D_{1} such that f_{1}(x) ∈ D_{1}[x] can be factorized as a product of degree one monic polynomials. Thus f(x) = (x − \overline{t})f_{1}(x) can be factorized as a product of degree one monic polynomials in D_{1}[x]. (Looks familiar? Find a algebra book containing field theory and look up the proof of existence of splitting fields.) □

Exercise 5.9. Let A be a subring of a ring B and let C be the integral closure of A in B. Prove that C[x] is the integral closure of A[x] in B[x].

Proof. By Exercise 5.3, C[x] = C ⊗_{A} A[x] is integral over A[x] = A ⊗_{A} A[x].

To show that C[x] is the the integral closure of A[x] in B[x], we follow the hint provided by the textbook. (Yon may look up Eisenbud’s book for another proof.) For any f ∈ B[x] that is integral over A[x], it satisfy

f^{m} + g_{1}f^{m−1} + ⋯ + g_{m−1}f + g_{m} = 0 with g_{i} ∈ A[x].

Let r be an integer larger than the degrees of f, g_{1}, g_{2}, …, g_{m}, and let f_{1} = f − x^{r}, so that

(f_{1} + x^{r})^{m} + g_{1}(f_{1} + x^{r})^{m−1} + ⋯ + g_{m−1}(f_{1} + x^{r}) + g_{m} = 0.

Expanding the powers (f_{1} + x^{r})^{i} out and then collecting the terms in terms of the powers of f_{1}, we get an equation of the following form

f_{1}^{m} + h_{1}f_{1}^{m−1} + ⋯ + h_{m−1}f_{1} + h_{m} = 0,

in which each h_{i} is a polynomial expression of x^{r} and the g_{j}. In particular,

h_{m} = (x^{r})^{m} + g_{1}(x^{r})^{m−1} + ⋯ + g_{m−1}x^{r} + g_{m} ∈ A[x],

which gives

−f_{1}(f_{1}^{m−1} + h_{1}f_{1}^{m−2} + ⋯ + h_{m−1}) = h_{m} ∈ A[x].

Notice that, by our choice of r, both −f_{1} = x^{r} − f and h_{m} are monic, which implies that f_{1}^{m−1} + h_{1}f_{1}^{m−2} + ⋯ + h_{m−1} is also monic. Now apply Exercise 5.8 to conclude that both −f_{1} = x^{r} − f and f_{1}^{m−1} + h_{1}f_{1}^{m−2} + ⋯ + h_{m−1} are in C[x]. But −f_{1} = x^{r} − f ∈ C[x] implies that f ∈ C[x]. □

Note: The exercises are from ‘Introduction to Commutative Algebra’ by M. F. Atiyah and I. G. Macdonald. All the quoted results are from the textbook unless different sources are quoted explicitly. For the convenience of the readers, the number of the chapter is included when a particular exercise is numbered. For example, Exercise m.n means the Exercise n from Chapter m.