MATH 612 Due Wednesday February 10, 1993

1. Two more universal (and couniversal) constructions.

Consider a square \begin{matrix}L & ⟶^{φ′} & N \\ ↓^{ψ′} & ^{ψ}↓ \\ M & ⟶^{φ} & P \end{matrix}.

Define σ: L ⟶ M ⊕ N and τ: M ⊕ N ⟶ P by

σ(ℓ) = (ψ′(ℓ), −φ′(ℓ)) and τ(m, n) = φ(m) + ψ(n).

a) Prove that the square commutes if and only if τσ = 0.

b) Prove that the following conditions are equivalent:

(1) 0 → L ⟶^{σ} M ⊕ N ⟶^{τ} P is exact.

(2) The square commutates and (∀m ∈ M, n ∈ N) [ φ(m) = ψ(n) ⟺ (∃! ℓ ∈ L) n = φ′(ℓ), m = ψ′(ℓ) ].

(3) The square commutes and whenever α: X → M and β: X → N are maps (for any R-module X) such that φα = ψβ, then there exists a unique map θ: X → L such that α = ψ′θ and β = φ′θ.

\begin{aligned}X \\ & L & ⟶^{φ′} & N \\ & ^{ψ′}↓ & & ^{ψ}↓ \\ & M & ⟶^{φ} & P \end{aligned}  \begin{aligned}L & ⟶^{φ′} & N \\ ^{ψ′}↓ & & ^{ψ}↓ \\ M & ⟶^{φ} & P \\ & & & Y \end{aligned}

c) Prove that the following conditions are equivalent:

(1) L ⟶^{σ} M ⊕ N ⟶^{τ} P → 0 is exact.

(2) The square commutes, P = φ(M) + ψ(N) and (∀m ∈ M, n ∈ N) [ φ(m) = ψ(n) ⟺ (∃ℓ ∈ L) n = φ′(ℓ), m = ψ′(ℓ) ]. (Note that in this case ℓ need not be unique.)

(3) The square commutes and whenever γ: M → Y and δ: N → Y are maps (for any R-module Y ) such that γψ′ = δφ′, then there exists a unique map ζ: P → Y such that γ = ζφ and δ = ζψ.

Definition. If the equivalent conditions in b) are satisfied, we say that the square above is a pull-back (Hungerford, p. 484). If the conditions in c) are satisfied, we say that it is a push-out.

2. a) Prove that if the square in problem 1 is a pull-back, then Ker φ′ ≈ Ker φ.

b) Prove that if the square in problem 1 is a push-out, then Coker φ′ ≈ Coker φ. (note: Coker φ = P/φ(M). )

c) Show that Noether’s Second Isomorphism Theorem (Hungerford, Theorem 1.9 (i), p. 173) is a special case of part b).

3. Consider the following commutative diagram with exact rows.

0 ⟶ K ⟶ L ⟶^{φ′} N ⟶ 0
∥  ^{ψ′}↓  ^{ψ}↓
0 ⟶ K ⟶ M ⟶^{φ} P ⟶ 0

Prove that the right hand square is both a pull-back and a push-out.

4. Let S be a multiplicative set in a commutative noetherian ring R and let M be an R-module. Prove that

Supp_{S^{−1}R} S^{−1}M = { 𝔭S^{−1}R | 𝔭 ∈ Supp M & 𝔭 ∩ S = ∅ }
Ass_{S^{−1}R} S^{−1}M = { 𝔭S^{−1}R | 𝔭 ∈ Ass M & 𝔭 ∩ S = ∅ }
Ass_{R} S^{−1}M = { 𝔭 | 𝔭 ∈ Ass M & 𝔭 ∩ S = ∅ }.

5. Let M be a module over a commutative noetherian ring R such that Ass M consists of maximal ideals.

a) Prove that Ass M = Supp M.

b) Prove for every 𝔭 ∈ Ass M, the canonical map M → M_{𝔭} is a surjection and

M_{𝔭} ≈ {m ∈ M | (∃k) 𝔭^{k}m = 0}.

c) Prove that the family of maps M → M_{𝔭} for 𝔭 ∈ Ass M induces an isomorphism

M ⟶^{≈} ⨁_{Ass M} M_{𝔭}.

February 10 Answers

1.

Consider a square \begin{matrix}L & ⟶^{φ′} & N \\ ↓^{ψ′} & ^{ψ}↓ \\ M & ⟶^{φ} & P \end{matrix}.

Define σ: L ⟶ M ⊕ N and τ: M ⊕ N ⟶ P by

σ(ℓ) = (ψ′(ℓ), −φ′(ℓ)) and τ(m, n) = φ(m) + ψ(n).

c) (1) L ⟶^{σ} M ⊕ N ⟶^{τ} P → 0 is exact.

(3) The square commutes and whenever γ: M → Y and δ: N → Y are maps (for any R-module Y ) such that γψ′ = δφ′, then there exists a unique map ζ: P → Y such that γ = ζφ and δ = ζψ.

proof: (3) ⇒ (1): Proof that Ker τ ⊆ σ(L):

Note that σ(L) = {(ψ′(ℓ), −φ′(ℓ)) | ℓ ∈ L}.

Consider the following square:

L ⟶^{φ′} N
ψ′↓  ^{ψ}↓
M ⟶^{φ} P

\frac{M ⊕ N}{σ(L)} ,

where γ(m) = (m, 0) + σ(L) and δ(n) = (0, n) + σ(L). Note that

γψ′(ℓ) − δφ′(ℓ) = (ψ′(ℓ), −φ′(ℓ)) + σ(L) = 0 ∈ (M ⊕ N)/σ(L) ,

so by hypothesis there exists ζ making the diagram commute. Now suppose that τ(m, n) = φ(m) + ψ(n) = 0. Then

(m, n) + σ(L) = γ(m) + δ(n) = ζφ(m) + ζψ(n) = ζ(φ(m) + ψ(n)) = 0

so (m, n) ∈ σ(L). ✓

2. b)

L ⟶^{φ′} N ⟶^{γ′} C ⟶ 0
ψ′↓  ^{ψ}↓  ^{μ}↓
M ⟶^{φ} P ⟶^{ξ} D ⟶ 0

Let C = Coker φ′ and D = Coker φ. Now (ξψ)φ′ = ξφψ′ = 0 so by the Induced Homomorphism Theorem there exists a unique map μ: C → D making the above diagram commute. On the other hand, since γ′φ′ = 0 = 0ψ′, by the categorical definition of a push-out there exists a unique map ζ: P → C with ζψ = γ′ and ζφ = 0. Again by the Induced Homomorphism Theorem ζ induces a map η: D → C such that ηξψ = ζψ = γ′. Then (μηξ)ψ = μζψ = μγ′ = ξψ and (μηξ)φ = 0 = ξφ, so by the definition of a push-out it follows that μηξ = ξ, and thus μη = 1_{D} because ξ is an epimorphism. Also (ημ)γ′ = ηξψ = ζψ = γ′ so ημ = 1_{C} because γ′ is an epimorphism. Thus C ≈ D. ✓

3.

0 ⟶ K ⟶^{η′} L ⟶^{φ′} N ⟶ 0
∥  ^{ψ′}↓  ^{ψ}↓
0 ⟶ K ⟶^{η} M ⟶^{φ} P ⟶ 0

Proof that the square is a push-out: Let γ: M → Y and δ: N → Y be such that γψ′ = δφ′. Then γη = γψ′η′ = δφ′η′ = 0, so by the Induced Homomorphism Theorem there exists a unique ζ: P → Y with ζφ = γ. Furthermore (ζψ)φ′ = ζφψ′ = γψ′ = δφ′. Since φ′ is an epimorphism, we conclude that ζψ = δ. Thus the square in question satisfies the categorical definition of a push-out. Proof that the square is a pull-back: (Actually, knowing that the square is a push-out, we are already half-way to proving it is a pull-back. But we will start from scratch.) Suppose m ∈ M and n ∈ N with φ(m) = ψ(n). Since φ′ is epic,

(∃ℓ ∈ L) n = φ′(ℓ).

Then φ(m − ψ′(ℓ)) = φ(m) − φψ′(ℓ) = ψ(n) − ψφ′(ℓ) = 0. Therefore m − ψ′(ℓ) ∈ Ker φ so by exactness there exists a unique k ∈ K with m − ψ′(ℓ) = η(k) = ψ′η′(k). Thus

m = ψ′(η′(k) + ℓ) and n = φ′(η′(k) + ℓ).

Furthermore, η′(k) + ℓ is the unique element in L that works. In fact if m = ψ′(ℓ′) and n = φ′(ℓ′), then φ′(η′(k) + ℓ − ℓ′) = 0 so by exactness

(∃k′ ∈ K) η′(k) + ℓ − ℓ′ = η′(k′)

and so 0 = ψ′(η′(k) + ℓ − ℓ′) = ψ′η′(k′) = η(k′) and so k′ = 0 because η is monic, so ℓ′ = η′(k) + ℓ.

4. There are three relevant observations:

(1) The primes of S^{−1}R are precisely the ideals of the form S^{−1}𝔭, where 𝔭 is a prime ideal of R such that 𝔭 ∩ S = ∅. (If 𝔭 ∩ S ≠ ∅ then S^{−1}𝔭 = R.)

(2) If 𝔭 is a prime in R then S^{−1}𝔭 ⊆ S^{−1}R and in fact S^{−1}𝔭 = 𝔭S^{−1}R.

(3) If S ∩ 𝔭 = ∅ then S ⊆ R ∖ 𝔭 and so S^{−1}M_{𝔭} ≈ M_{𝔭} ≈ S^{−1}M_{S^{−1}𝔭} (where the LHS can be interpreted in two ways and these two interpretations agree). We now see immediately that if 𝔭 ∩ S = ∅ then S^{−1}M_{S^{−1}𝔭} ≠ 0 and if only if M_{𝔭} ≠ 0.

Another observation: If m ∈ M, then the annihilator in S^{−1}R of m/1 is S^{−1}(ann m) ⊆ S^{−1}R. In fact, if r ∈ ann m then rm = 0 and for any s ∈ S, (r/s)(m/1) = rm/s = 0, so r/s ∈ ann_{S^{−1}R}(m/1). Conversely, if (r/s)(m/1) = 0 then s′rm = 0 for some s′ ∈ S. Then s′r ∈ ann m and r/s = s′r/s′s ∈ S^{−1}(ann m). Now let 𝔭 ∈ Ass M and 𝔭 ∩ S = ∅. Then 𝔭 = ann m for some m ∈ M and so S^{−1}𝔭 = ann(m/1) ∈ Ass_{S^{−1}R} S^{−1}M. Conversely, a prime 𝔓 = ann_{S^{−1}R}(m/s) in Ass_{S^{−1}R} S^{−1}M must have the form S^{−1}𝔭, where 𝔭 = θ_{R}^{−1}(𝔓). Thus for each p ∈ 𝔭 there exists s′ ∈ S with s′pm = 0. Since 𝔭 is finitely generated, there exists s′′ ∈ S with s′′𝔭 m = 𝔭s′′m = 0. Thus 𝔭 ⊆ ann s′′m. On

the other hand, if r ∈ ann s′′m then rs′′m = 0 and so \frac{rs′′}{s′′} \frac{m}{s} = 0 so r/1 ∈ S^{−1}𝔭 and

thus r ∈ θ_{R}^{−1}(S^{−1}𝔭) = 𝔭. Therefore 𝔭 = ann s′′m.

Now suppose 𝔭 ∩ S = ∅ and 𝔭 ∈ Ass R and 𝔭 = ann m. We claim that 𝔭 = ann(m/1). In fact, for r ∈ R, r ∈ ann(m/1) if and only if srm = 0 for some s ∈ S, and this holds if and only if sr ∈ ann m = 𝔭 for some s ∈ S, but this is true if and only if r ∈ 𝔭 since 𝔭 is prime and s ∉ 𝔭. Thus 𝔭 ∈ Ass_{R} S^{−1}M. Conversely, suppose 𝔭 is a prime in R and 𝔭 ∈ Ass S^{−1}M. Then 𝔭 = ann(m/1) for m/1 ∈ S^{−1}M. Since m/1 ≠ 0 (otherwise ann(m/1) = R), it follows that sm ≠ 0 for all s ∈ S and thus 𝔭 ∩ S = ∅. Furthermore for each p ∈ 𝔭, pm/1 = 0 so spm = 0 for some s ∈ S. Since 𝔭 is finitely generated it follows that there exists s ∈ S such that s𝔭m = 𝔭sm = 0. Therefore 𝔭 ⊆ ann(sm). On the other hand, clearly

ann(sm) ⊆ ann(\frac{sm}{s}) = 𝔭. So 𝔭 = ann sm ∈ Ass M. ✓

5. b) To see that M → M_{𝔭} is a surjection it suffices to see that for all prime ideals 𝔮, the induced map M_{𝔮} → (M_{𝔭})_{𝔮} is a surjection. This is clear if 𝔭 ⊈ 𝔮 since in that case by problem 4, 𝔭 ∉ Ass M_{𝔮} = Supp M_{𝔮} and so (M_{𝔭})_{𝔮} = (M_{𝔮})_{𝔭} = 0. Since 𝔭 is maximal, this leaves only the case 𝔮 = 𝔭, which is trivial. A More Straightforward Proof: Let m/s ∈ M_{𝔭}, where s ∉ 𝔭. By Problem 3, Ass M_{𝔭} = {𝔭}, i. e. M_{𝔭} is 𝔭-primary, so by a previous homework problem 𝔭^{k}(m/1) = 0 for some k ≥ 1. Now since 𝔭 is maximal, the only prime containing 𝔭^{k} is 𝔭. It follows that R/𝔭^{k} is a local ring with maximal ideal 𝔭/𝔭^{k}. Since s ∉ 𝔭, thus the image of s in R/𝔭^{k} is invertible. Hence there exists s′ ∉ 𝔭 such that ss′ ≡ 1 (mod 𝔭^{k}). Then (ss′ − 1) m = 0 so that in M_{𝔭},

\frac{m}{s} = \frac{s′m}{1} = θ(s′m),

showing that θ: M → M_{𝔭} is surjective.

Now let M(𝔭) = {m ∈ M | (∃k) 𝔭^{k}m = 0}. As seen in the previous paragraph, if s ∉ 𝔭 then s is invertible modulo 𝔭^{k} for all k, so if m ∈ M(𝔭) then sm ≠ 0 for all s ∉ 𝔭 and therefore m/1 ≠ 0 ∈ M_{𝔭}. This shows that θ: M → M_{𝔭} restricts to a monomorphism from M(𝔭) into M_{𝔭}. Now let m/1 ∈ M_{𝔭}. As previously noted, there exists k ≥ 1 such that 𝔭^{k}(m/1) = 0. Thus for each r ∈ 𝔭^{k}, rm/1 = 0 so there exists s ∉ S such that srm = 0. Since 𝔭 is finitely generated, it follows that there exists s ∉ 𝔭 with s𝔭^{k}m = 𝔭^{k}sm = 0, showing that sm ∈ M(𝔭). . Furthermore, as previously seen, there exists s′ ∉ 𝔭 such that ss′ ≡ 1 (mod 𝔭^{k}). Then ss′m ∈ M(𝔭) and

\frac{m}{1} = \frac{ss′m}{1}.

Therefore θ(M(𝔭)) = M_{𝔭} and so M_{𝔭} ≈ M(𝔭).

c) The family of maps M → M_{𝔭} induces

ζ: M → ∏_{Ass M} M_{𝔭}

(where each coordinate of ζ(m) is given by m/1 ∈ M_{𝔭}). Now note that for any m ∈ M, Supp Rm = Ass Rm is finite, i. e. there are only finitely many prime ideals 𝔭 such that m/1 ≠ 0 ∈ M_{𝔭}. This shows that the image of ζ is in fact contained in ⨁_{Ass M} M_{𝔭}. It now suffices to see that for every prime ideal 𝔮, the induced map

M_{𝔮} → ⨁_{Ass M}(M_{𝔭})_{𝔮}

is an isomorphism. But as seen in part b), this reduces to the identity map M_{𝔮} → M_{𝔮}.

A more conventional proof: It is easy to see that the family of submodules M(𝔭) of M is independent, so ⨁_{Ass M} M(𝔭) ⊆ M. Now let m ∈ M. For each of the finitely many primes 𝔭_{i} such that m/1 ≠ 0 ∈ M_{𝔭_{i}} then exists k_{i} such that 𝔭_{i}^{k_{i}}m/1 = 0 ∈ M_{𝔭_{i}}. Then 𝔭_{1}^{k_{1}} ⋯ 𝔭_{n}^{k_{n}}m = 0. Now since 𝔭_{i} + 𝔭_{j} = R for i ≠ j, no maximal ideal can contain all the ideals 𝔞_{1}, …, 𝔞_{n}, where

𝔞_{i} = 𝔭_{1}^{k_{1}} ⋯ 𝔭̂_{i}^{k_{i}} ⋯ 𝔭_{n}^{k_{n}},

so 𝔞_{1} + ⋯ + 𝔞_{n} = R and there exist elements a_{i} ∈ 𝔞_{i} with a_{1} + ⋯ + a_{n} = 1. Furthermore, 𝔭^{k_{i}}𝔞_{i}m = 0 so a_{i}m ∈ 𝔞_{i}m ⊆ M(𝔭_{i}). Thus m = a_{1}m + ⋯ + a_{n}M ∈ ⨁_{Ass M} M(𝔭), showing that M = ⨁ M(𝔭). ✓