Topics in algebra Chapter IV: Commutative rings and modules I - 1

M ′ /Imβ g′ 1 ′ ′ −→ N /Imγ −→ 0 is exact as g is onto.
L ′ /Imα f ′ 1 ′
g
−→ M /Imβ ′ 1 ′ −→ N /Imγ is exact:
Let x + Imβ ∈ kerg ′ 1; then g ′ (x) ∈ Imγ, so that there is an element
y ∈ N such that g ′ (x) = γ(y). Since g is onto, there is an element x ′ ∈ M
such that g(x ′ ) = y. It follows that g ′ (x − β(x ′ )) = g ′ (x) − (γ ◦ g)(x ′ ) = 0.
Therefore there is an element z ∈ L ′ such that f ′ (z) = x − β(x ′ ). Hence
f ′ 1(z + Imα) = x + Imβ.
To finish the proof, we need to know the definition of d. The map d :
kerγ −→ L ′ /Imα is given as follows: Let z ∈ kerγ. Since g is onto and
z ∈ N, there is an element y ∈ M such that g(y) = z. Furthermore,
(g ′ ◦ β)(y) = (γ ◦ g)(y) = 0, therefore β(y) ∈ kerg ′ = Imf ′ . Hence, there
is an element x ∈ L ′ such that f ′ (x) = β(y). Now, d(z) = x + Imα.
From the above, we need to show that d is well-defined, i.e., the definition is
independent of the choice of y. Suppose there is another element y ′ ∈ M such
that g(y ′ ) = z. Then β(y ′ ) = f ′ (x ′ ) for some x ′ ∈ L ′ . On the other hand,
since y−y ′ ∈ kerg = Imf, there is an element x ′′ ∈ L such that f(x ′′ ) = y−y ′ .
Since f ′ is 1-1, α(x ′′ ) = x − x ′ . This shows that x + Imα = x ′ + Imα.
kerβ g1
−→ kerγ d
−→ L ′ /Imα is exact:
Let z ∈ kerγ and z ∈ kerd. Let x ∈ L ′ and y ∈ M be such that g(y) = z
and f ′ (x) = β(y). Since d(z) = 0, x ∈ Imα, so there is an element x ′ ∈ L
such that x = α(x ′ ). It follows that β(y) = (f ′ ◦ α)(x ′ ) = (β ◦ f)(x ′ ).
Therefore y − f(x ′ ) ∈ kerβ. Since g(y − f(x ′ )) = g(y) = z, z ∈ Img1.
−→ L ′ /Imα f ′ 1
Let x ∈ L ′ such that f ′ (x) ∈ Imβ; then there is an element y ∈ M such
that β(y) = f ′ (x). Since (g ′ ◦ f ′ ) = 0, (γ ◦ g)(y) = (g ′ ◦ β)(y) = 0. Let
z = g(y); then z ∈ kerγ, so that by the definition of d, d(z) = x + Imα and
kerf ′ 1 ⊆ Im(d).
kerγ d
−→ M ′ /Imβ is exact:
Corollary 4.32 (The short five lemma) Let
0 −→ L
f
−→ M g
−→ N −→ 0
↓ α ↓ β ↓ γ
0 −→ L ′ f ′
−→ M ′ g ′
−→ N ′ −→ 0
be a commutative diagram of R-modules and R-homomorphisms such that
each row is a s.e.s.. Then the following hold:
(a) If α and γ are monomorphism then so is β.
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