Commutative algebra - Department of Mathematical Sciences - old ...

3.5. PROJECTIVE MODULES 51
projections, then g ◦ pN = h ◦ pF . Now pF is surjective since g is. Let v : F → M
be a section of pF , then h ′ = pN ◦ v satisfies h = g ◦ h ′ .
pN
N g
L
h
M
F
′
h
v
3.5.3. Corollary. A short exact sequence
0
f
M
pF
g
N
0
F
where F is a projective module is a split exact sequence.
3.5.4. Corollary. A free module is projective. Over a field every module is projective.
3.5.5. Example. Let I ⊂ R be an ideal. If R/I is projective, then there is a ring
decomposition R/I × R ′ R.
3.5.6. Proposition. A direct summand in a projective module is projective.
Proof. Let F ⊕ F ′ be projective and g : M → F surjective. By 3.5.2 there is a
section v ′ : F ⊕ F ′ → M ⊕ F ′ to (g, 1F ′). Then v(y) = pM ◦ v ′ (y, 0) is a section
to g and F is projective.
3.5.7. Proposition. A module is projective if and only if it is a direct summand in
a free module.
Proof. By 2.4.12 any module is a factor module of a free module. By 3.5.2 a
projective factor module has a section, and is therefore by 3.1.10 a direct summand.
3.5.8. Proposition. Let Fα be a family of projective modules, then the direct sum
α Fα is a projective module.
Proof. Let N → L be surjective. Then by 2.5.8
is the product
0
HomR( Fα, N) → HomR( Fα, L)
HomR(Fα, N) → HomR(Fα, L)
which is surjective by 3.1.6. So Fα is projective.
3.5.9. Proposition. Let F, F ′ be projective modules. Then F ⊗R F ′ is projective.
Proof. F ⊗R F ′ is clearly a direct summand in a free module.
3.5.10. Proposition. Let R → S be a ring homomorphism and F a projective
module. The change of ring module F ⊗R S is a projective S-module.
Proof. A direct summand of a free R-module is changed to a direct summand of a
free S-module.