MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #9

JOHN VOIGHT

Problem R1. Show that −1 is a square in ℤ_{5}.

Solution. Although one can prove this directly, it is worth noting the following general result:

Let f(x) ∈ ℤ[x], let p be a prime, and suppose that there exists α_{1} ∈ ℤ/pℤ such that f(α_{1}) ≡ 0 (mod p) and f′(α_{1}) ≢ 0 (mod p). Then there exists a unique root α ∈ ℤ_{p} of f satisfying α ≡ α_{1} (mod p).

Since ℤ_{p} = \underleftarrow{lim}_{n} ℤ/p^{n}ℤ, giving a root α of f in ℤ_{p} is equivalent to giving for each n ≥ 1 a root α_{n} ∈ ℤ/p^{n}ℤ of f such that α_{n+1} ≡ α_{n} (mod p^{n}): then lim_{n→∞} α_{n} = α is a root of f in ℤ_{p}.

We prove the existence of α_{n} by induction. We are given α_{1} ∈ ℤ/pℤ. Given α_{n}, we lift it to α_{n+1} = α_{n} + tp^{n} ∈ ℤ/p^{n+1}ℤ as follows: since

f(α_{n+1}) = f(α_{n} + tp^{n}) ≡ f(α_{n}) + tp^{n}f′(α_{n}) (mod p^{n+1})

by Taylor expansion, and f(α_{n}) ≡ 0 (mod p^{n}), it is enough to solve

f′(α_{n})t ≡ −f(α_{n})/p^{n} (mod p)

for t which is possible (uniquely) because f′(α_{n}) ≡ f′(α_{1}) ≢ 0 (mod p). In our case, we may take α_{1} = 2 since f′(2) = 4 (mod 5).

Problem R2. Let E be the set consisting of all nonnegative integers, together with an extra element ∞. A supernatural number is a formal product ∏_{p} p^{e_{p}} where the product runs over all primes p and where the exponents e_{p} are elements of E. If m is a supernatural number, let mℤ̂ be the intersection of the groups nℤ̂, taken over positive integers n that divide m.

Show that the set of closed subgroups of ℤ̂ corresponds bijectively with the set of supernatural numbers under the map m ↦ mℤ̂.

Solution. Let m be a supernatural number. We note that mℤ̂ is indeed a closed subgroup of ℤ̂ since it is the intersection of the open (by definition) hence closed subgroups nℤ̂ for n ∈ ℤ.

For each prime p and n ∈ ℤ_{>0}, we have the projection map ℤ̂ → ℤ/p^{n}ℤ. These maps are continuous (by definition) and are compatible in the sense that if n′ ≥ n, the diagram

Date: May 16, 2003.

R1–R4, VIII: 2–7.

\xymatrix{ℤ̂ \ar[r] \ar[dr] & ℤ/p^{n′}ℤ \ar[d] \\  & ℤ/p^{n}ℤ}

commutes. By the property of projective limits, we obtain a continuous map ℤ̂ → \underleftarrow{lim}_{n} ℤ/p^{n}ℤ = ℤ_{p}. (In fact, the map ℤ̂ → ∏_{p} ℤ_{p} is an isomorphism of topological rings if the product is given the product topology.)

From this we see that the map m ↦ mℤ̂ is injective: if m ≠ m′, then there exists a prime p such that e_{p} ≠ e′_{p} (where m′ = ∏_{p} p^{e′_{p}}), hence the images of mℤ̂ and m′ℤ̂ in ℤ_{p} are different.

Let H be a closed subgroup of ℤ̂. Since ℤ̂ is compact (it is a closed subset of ∏_{m} ℤ/mℤ in the product topology and each factor is discrete), the image of H in ℤ_{p} is compact; since ℤ_{p} is Hausdorff (it is a subset of ∏_{n} ℤ/p^{n}ℤ, each factor again discrete), this image is closed. Let e_{p} be the largest element of E such that H ⊂ p^{e_{p}}ℤ_{p} (we take e_{p} = ∞ e.g. if H = {0}). This gives a map from the set of closed subgroups of ℤ̂ to the set of supernatural numbers by H ↦ ∏_{p} p^{e_{p}(H)}. It is now easy to see that

mℤ̂ ↦ _{∏p}p^{e_{p}(mℤ̂)} = m

under this map; therefore the map m ↦ mℤ̂ has a right inverse, so it is surjective as well.

Problem R3. Let K and L be extensions of a field k inside a large field Ω (as in Chapter VIII, §3). Is it true that K and L are linearly disjoint over k if and only if the natural map K ⊗_{k} L → Ω is injective?

Solution. Yes, this statement is true. First suppose K and L are linearly disjoint. Let

α = ∑_{i}x_{i} ⊗ y_{i} ∈ K ⊗_{k} L

be any element. By k-bilinearity of the tensor product, we may assume that the x_{i} are linearly independent over k, for e.g. if x_{j} = ∑_{i} α_{i}x_{i}, then

x_{j} ⊗ y_{j} = ∑_{i}x_{i} ⊗ (α_{i}y_{j}).

Then since K and L are linearly disjoint, the x_{i} are linearly independent over L, so if α ↦ 0 = ∑_{i} x_{i}y_{i}, we have y_{i} = 0 for all i, hence α = 0.

Conversely, let x_{1}, …, x_{n} ∈ K be linearly independent over k. Suppose that there exist y_{i} ∈ L such that ∑_{i} x_{i}y_{i} = 0 ∈ Ω; then _{∑i}x_{i} ⊗ y_{i} ↦ 0, so by injectivity, we have ∑_{i} x_{i} ⊗ y_{i} = 0. Since x_{i} ∈ K are linearly independent over k, the elements x_{i} ⊗ 1 ∈ K ⊗ L are linearly independent over L, a contradiction.

Problem R4. At the beginning of the proof of Theorem VIII.4.13, Lang says, “From the hypotheses, we deduce that K is free from the algebraic closure L^{a} of L over k.” How do we deduce this?

Solution. Let x_{1}, …, x_{n} ∈ K be algebraically independent over k, in other words, trdeg(k(x)/k) = n. We know that K is free from L over k, so trdeg(L(x)/L) = n. We have the following diagram of fields:

[figure: \overline{L}(x); L(x); \overline{L}; L]

By problem 3 (proven below, without using this result),

trdeg(\overline{L}(x)/L) = trdeg(\overline{L}(x)/L(x)) + trdeg(L(x)/L) = 0 + n = n

since trdeg(\overline{L}(x)/L(x)) ≤ trdeg(\overline{L}/L) = 0. So x_{1}, …, x_{n} are also algebraically independent over \overline{L}.

Problem 2. A subfield k of a field K is said to be algebraically closed in K if every element of K which is algebraic over k is contained in k. Prove: If k is algebraically closed in K, and K, L are free over k, and L is separable over k or K is separable over k, then L is algebraically closed in KL.

Solution. If K is separable over k, then K/k is a regular extension, so by Theorem 4.13, KL/L is regular and in particular, L is algebraically closed in L.

So we suppose that L is separable over k. We may assume that L is finitely generated over k. Let x_{1}, …, x_{n} be a separating transcendence base for the extension L over k, so that L is a finite separable extension of k(x) = k(x_{1}, …, x_{n}). Since the x_{i} are algebraically independent over k and L is free from K, we know that k is algebraically closed in K(x), so k(x) is algebraically closed in K(x). We therefore reduce to the case where L is a finite separable extension of k.

[figure: KL; K(x); L; K; k(x); k]

Suppose that α ∈ KL is algebraic over L. By the proof of Lemma 4.10, the minimal polynomial of α over k remains irreducible over K, and hence is minimal. But since L is separable over k, we have by Corollary 4.5 that KL is separable over K, therefore the minimal polynomial of α must be separable over K, so α is in fact separable over k. Therefore L(α) is finite and separable, so it is primitively generated; by Lemma 4.10 we have

[L(α) : k] = [KL(α) : K] = [KL : K] = [L : k]

so L = L(α).

Note that we do in fact need the assumption that L is separable over k. For example, take k = 𝔽_{p}(x, y), K = k(u, v) where u and v are independent transcendentals related by the equation xu^{p} − yv^{p} = 1, and L = k(x^{1/p}). One can check that k is algebraically closed in K. Since L is algebraic over k, L is free from K.

Then in KL we have the equation

y = \frac{xu^{p} − 1}{v^{p}} = (\frac{x^{1/p}u − 1}{v})^{p}

and KL ⊃ L(y^{1/p}) ⊃ L.

Problem 3. Let k ⊂ E ⊂ K be extension fields. Show that

trdeg(K/k) = trdeg(K/E) + trdeg(E/k).

Show if {x_{i}} is a transcendence base of E/k, and {y_{j}} is a transcendence base of K/E, then {x_{i}, y_{j}} is a transcendence base of K/k.

Solution. We prove the second statement; the first statement follows. Since {x_{i}} is a transcendence basis of E/k, by definition E is algebraic over k(x); the class of algebraic extensions is distinguished (§V.1), so E(y) is algebraic over k(x, y) and K is algebraic over E(y) so K is algebraic over k(x, y). Thus

trdeg(K/k) ≤ #{x_{i}, y_{j}},

and it suffices to show that the set {x_{i}, y_{j}} ⊂ K is algebraically independent, for then

trdeg(K/k) ≥ #{x_{i}, y_{j}} = #{x_{i}} + #{y_{j}} = trdeg(E/k) + trdeg(K/E).

Suppose that

f(x_{i}, y_{j}) = ∑_{I,J}a_{IJ}x^{I}y^{J} = 0

is an algebraic dependence relation with a_{IJ} ∈ k and x^{I}, y^{J} monomials in the x_{i}, y_{j}, respectively. Since y_{j} are algebraically independent over E, viewing f as a polynomial in the y_{j} we see that f must be of the form

f(x_{i}) = ∑_{I}a_{I}x^{I} = 0;

but then again the x_{i} are algebraically independent over k, so this polynomial is identically zero.

Problem 4. Let K/k be a finitely generated extension, and let K ⊃ E ⊃ k be a subextension. Show that E/k is finitely generated.

Solution. Let {x_{i}} be a transcendence base for E/k and {y_{j}} for K/E. By the previous exercise, we see that each of these sets is finite and that K is algebraic over k(x, y). Since K is finitely generated, K is finitely generated algebraic over k(x, y), hence [K : k(x, y)] < ∞.

[figure: K; k(x, y); E; k(x); k]

It suffices to show that E is finitely generated over k(x); we will show it is finite. From Proposition 3.3, the field k(x, y) is linearly disjoint from E (since y_{j} are algebraically independent). Thus if {u_{m}} ⊂ E is linearly independent over k(x) it remains so over k(x, y), hence #{u_{m}} ≤ [K : k(x, y)] < ∞ and the claim follows.

Problem 5. Let k be a field and k(x_{1}, …, x_{n}) = k(x) be a finite separable extension. Let u_{1}, …, u_{n} be algebraically independent over k. Let

w = u_{1}x_{1} + ⋯ + u_{n}x_{n}.

Let k(u) = k(u_{1}, …, u_{n}). Show that k(u)(w) = k(u)(x).

Solution. The inclusion k(u)(w) ⊂ k(u)(x) is clear.

Let K be the normal closure of k(x) in a fixed algebraic closure; by §V.4, the extension K/k is also finite, separable. Let d = [k(x) : k], and let σ_{i} : k ↪ K be the d distinct embeddings of k into K. By Proposition 3.3 and Lemma 4.10, the extensions k(u) (pure transcendental) and K (finite, separable, hence singly generated) are linearly disjoint over k hence free, and [k(u)(x) : k(u)] = d. But we have σ_{i}(w) ≠ σ_{j}(w) for all i ≠ j, since otherwise

∑_{m} (σ_{i}(x_{m}) − σ_{j}(x_{m})) u_{m} = 0 ∈ K(u),

so by freeness σ_{i}(x_{m}) = σ_{j}(x_{m}) for all m, a contradiction. Therefore the minimal polynomial of w is degree ≥ d [k(x)(u) : k(u)] ≥ d, which completes the proof.

Problem 6. Let k(x) = k(x_{1}, …, x_{n}) be a separable extension of transcendence degree r ≥ 1. Let u_{ij} (with i = 1, …, r, j = 1, …, n) be algebraically independent over k(x). Let

y_{i} = ∑_{j=1}^{n}u_{ij}x_{j}.

Let k(u) = k(u_{ij})_{i,j}.

(a) Show that k(u)(x) is separable algebraic over k(u)(y_{1}, …, y_{r}) = k(u)(y).

(b) Show that there exists a polynomial P (u) ∈ k[u] having the following property: Let (c) = (c_{ij}) be elements of k such that P (c) ≠ 0. Let

y′_{i} = ∑_{j=1}^{n}c_{ij}x_{j}.

Then k(x) is separable algebraic over k(y′).

Solution. We have the following diagram of fields:

[figure: k(u)(x); k(x); k(u)(y); k]

The extension k(u)(x) of k(x) is separable since the u_{ij} are algebraically independent over k(x). By assumption, k(x) is separable over k, so by Corollary 4.3, k(u)(x) is separable over k. Therefore by Corollary 4.2, k(u)(x) is separable over k(y).

By Problem 3, we have

trdeg(k(u)(x)/k) = trdeg(k(u)(x)/k(x)) + trdeg(k(x)/k) = rn + r

since the u_{ij} are algebraically independent. Therefore

rn + r = trdeg(k(u)(x)/k(u)(y)) + trdeg(k(u)(y)/k(u)) + trdeg(k(u)/k);

since u_{ij} are algebraically independent over k(x) they are so over k, and we conclude that trdeg(k(u)/k) = rn, and it suffices to prove that trdeg(k(u)(y)/k(u)) = r. If not, there exists an algebraic dependence ∑_{I} a_{I}y^{I} = 0 with the a_{I} ∈ k(u) and y^{I} a monomial in the y_{j}. By clearing denominators, we can write this as ∑_{I} b_{I}u^{I} = 0 with b_{I} ∈ k(y) and u^{I} monomials in u_{j}. Expanding this relation in the x_{i} gives a relation ∑_{I} b′_{I}u^{I} = 0 with b_{I} ∈ k(x), a contradiction as the x_{i} are algebraically independent over k(u).

Part (b) follows from Corollary 2.3 and part (a).

Problem 7. Let k be a field and k[x_{1}, …, x_{n}] = R a finitely generated entire ring over k with quotient field k(x). Let L be a finite extension of k(x). Let I be the integral closure of R in L. Show that I is a finite R-module. [Hint: Use Noether Normalization, and deal with the inseparability problem and the separable case in two steps.]

Solution. By Proposition V.6.6, we have L ⊃ L_{0} ⊃ k(x) where L is purely inseparable over L_{0} and L_{0} is separable over k(x). By Noether normalization (Theorem VIII.2.1), there exist y_{1}, …, y_{r} ∈ R such that R is integral over k[y_{1}, …, y_{r}]. Let k(y) = k(y_{1}, …, y_{r}) and k[y] = k[y_{1}, …, y_{r}]. We have the following diagram:

I  L I_{0} L_{0} R k(x) k[y] k(y)

First, assume that R is integrally closed in k(x). Then the fact that I_{0} is a finite R-module follows from Exercise VII.3. For the inseparable extension, it suffices to treat the case where L = L_{0}(t^{1/p}) where t ∈ L_{0} \ L_{0}^{p}. Let α = a_{0} + ⋯ + a_{p−1}(t^{1/p})^{p−1} ∈ L be integral over L_{0}; the minimal polynomial of α is X^{p} = a_{0}^{p} + ⋯ + a_{p−1}t^{p−1} so a_{i}^{p} ∈ I_{0} for each i. Since I_{0} is integrally closed, a_{i} ∈ I_{0}, hence I = I_{0}[t^{1/p}].

The general case now follows by applying the statement with L = K (since k[y] is integrally closed in k(y)): if I is a finite k[y]-module and R is a finite k[y]-module, then I is a finite R-module.