MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #3

JOHN VOIGHT

Some solutions are omitted or sketched.

Problem 6. Let M, N be flat. Show that M ⊗ N is flat.

Solution. If E′ ↪ E is injective, then since M is flat, so is the map E′⊗M ↪ E⊗M; since N is also flat, so too is the map (E′⊗M)⊗N ↪ (E⊗M)⊗N. By associativity of the tensor product, we obtain the injection

E′ ⊗ (M ⊗ N) ↪ E ⊗ (M ⊗ N),

which is to say, M ⊗ N is flat.

Problem 7. Let F be a flat R-module, and let a ∈ R be an element which is not a zerodivisor. Show that if ax = 0 for some x ∈ F then x = 0.

Solution. The map R ⟶^{a} R which is multiplication by a is injective since a is not a zerodivisor, by definition. Since M is flat, the map R ⊗ F ⟶^{a⊗1} R ⊗ F is also injective. Since R ⊗ F ≅ F, we see that the map F ⟶^{a} F is injective, which is what we were to show.

Problem 8. Prove the following:

(i) Let S be a multiplicative subset of R. Then S^{−1}R is flat over R.

(ii) A module M is flat over R if and only if the localization M_{𝔭} is flat over R_{𝔭}

for each prime ideal 𝔭 of R. (iii) Let R be a principal (entire) ring. A module F is flat if and only if F is

torsion free.

Solution. Statement (i) follows from the fact that S^{−1}E ≅ S^{−1}R ⊗_{R} E, and if E′ → E is injective, then so is S^{−1}E′ → S^{−1}E.

For (ii), note that M_{𝔭} = R_{𝔭} ⊗_{R} M, so taking S = R \ 𝔭, we may apply (i) and Problem 6 to see that M_{𝔭} is flat for every prime ideal 𝔭. For the converse, if M is not flat, then there is an injection E′ → E of R-modules such that E′⊗M → E⊗M is no longer an injection; let N be the kernel of this map. There exists a prime 𝔭 such that N_{𝔭} ≠ 0 (for example, choose a maximal ideal containing all elements which annihilate N). Then M_{𝔭} is clearly not flat.

For (iii), one direction follows from Problem 7, and for the converse, if F is torsion free then since R is a principal entire ring, F is the direct limit of its finitely generated submodules which are free by III.7.3, hence flat (using Exercise 12 below).

Date: February 13, 2003.

XVI: 6–9, 12, *.

Problem 9. Let M be an A-module. We say that M is faithfully flat if M is flat, and if the functor

T_{M} : E ↦ M ⊗_{A} E

is faithful, that is E ≠ 0 implies M ⊗_{A} E ≠ 0. Prove that the following conditions are equivalent:

(i) M is faithfully flat; (ii) M is flat, and if u : F → E is a homomorphism of A-modules, u ≠ 0, then

T_{M}(u) : M ⊗_{A} F → M ⊗_{A} E is also nonzero; (iii) M is flat, and for all maximal ideals 𝔪 of A, we have 𝔪M ≠ M; and (iv) A sequence of A-modules N′ → N → N′′ is exact if and only if the sequence

tensored with M is exact.

Solution. First, we show (i) ⇒ (ii). Let E′ be the image of u. Since u ≠ 0 we know E′ ≠ 0. Then the image of T_{M}(u) is M ⊗_{A} E′ which is nonzero, since M is faithful.

For (ii) ⇒ (iii), take the nonzero homomorphism u : A → A/𝔪A; we have T_{M}(u) : M → M/𝔪M, so M/𝔪M ≠ 0, i.e. 𝔪M ≠ M.

Now we show (iii) ⇒ (i). Suppose that E ≠ 0; let x ∈ E be a nonzero element, and let 𝔪 be a maximal ideal containing the annihilator of x; then we have an injection of R-modules

x(R/𝔪) ↪ E/𝔪E.

(If ax = 0 then a is in the annihilator of x.) Tensoring with M, we obtain

x(R/𝔪) ⊗ M ↪ (E/𝔪E) ⊗ M

which becomes

x(M/𝔪M) ↪ (E ⊗ M) ⊗ R/𝔪.

By (iii), M/𝔪M ≠ 0, so E ⊗ M ≠ 0.

To show (iv) ⇒ (i), taking N′ = 0 we see that M is flat; if E ⊗ M = 0 then 0 ⊗ M → E ⊗ M → 0 ⊗ M is exact, so 0 → E → 0 is exact, so E = 0.

To conclude, we show that (i) ⇒ (iv). If M is flat, then N′ → N → N′′ exact implies N′ ⊗ M → N ⊗ M → N′′ ⊗ M exact. Conversely, if

N′ ⊗ M ⟶^{f⊗1} N ⊗ M ⟶^{g⊗1} N′′ ⊗ M

is exact, then (img g/ ker f) ⊗ M = 0, so by (i) we have img g/ ker f = 0, i.e.

N′ ⟶^{f} N ⟶^{g} N′′

is exact.

Problem 12. Show that the tensor product commutes with direct limits. In other words, if {E_{i}} is a direct family of modules, and M is any module, then there is a natural isomorphism

\underrightarrow{lim}(E_{i} ⊗_{A} M) ≅ (\underrightarrow{lim} E_{i}) ⊗_{A} M.

Solution (sketch). Show that (\underrightarrow{lim} E_{i}) ⊗_{A} M satisfies the universal property of the direct limit; this is clear on any finite level.

Problem *. Let R = k[x, y] be the indicated polynomial ring in two variables over a field k. Show that the maximal ideal (x, y) of R is not flat over R.

Solution. Using Proposition 3.7, we note that it suffices to show that the multiplication map 𝔪 ⊗ 𝔪 → 𝔪^{2} is not an isomorphism. This follows because

x ⊗ y − y ⊗ x = xy − yx = 0,

so the map even fails to be injective.

Note that x ⊗ y − y ⊗ x ≠ 0, because there is a R-linear map

f : 𝔪 × 𝔪 → k[x, y]/(x, y)
(a, b) ↦ (∂a/∂x)(∂b/∂y)

with f(x, y) = 1 ≠ 0 = f(y, x).