Commutative algebra - Department of Mathematical Sciences - old ...
Commutative algebra - Department of Mathematical Sciences - old ...
Commutative algebra - Department of Mathematical Sciences - old ...
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32 2. MODULES<br />
Pro<strong>of</strong>. By 2.5.3 the construction on homomorphisms are homomorphisms. Given<br />
also homomorphisms f ′ : M ′ → M ′′ and g ′ : N ′ → N ′′ . Then<br />
and<br />
Hom(1M, g ′ ◦ g) = Hom(1M, g ′ ) ◦ Hom(1M, g)<br />
Hom(f ′ ◦ f, 1N) = Hom(f, 1N) ◦ Hom(f ′ , 1N)<br />
showing the conditions on compositions.<br />
2.5.6. Corollary. Let a ∈ R give scalar multiplications aM, aN.<br />
(1)<br />
Hom(aM, 1N) = Hom(1M, aN) : HomR(M, N) → HomR(M, N) f ↦→ af<br />
is scalar multiplication a HomR(M,N).<br />
(2) The map R → HomR(M, M), a ↦→ aM is a homomorphism.<br />
2.5.7. Example. Let R = R1 × R2 be the product ring. The constructions in 2.2.9<br />
is: (1) A functor which to a pair <strong>of</strong> an R1-module and an R2-module associates an<br />
R-module.(2) A functor which to an R-module associates a pair <strong>of</strong> an R1-module<br />
and an R2-module.<br />
2.5.8. Proposition. Let R be a ring and Mα a family <strong>of</strong> modules. For any module<br />
N there are natural isomorphisms<br />
(1) HomR( <br />
α Mα, N) <br />
α HomR(Mα, N)<br />
(2) HomR(N, <br />
α Mα) <br />
α HomR(N, Mα)<br />
Pro<strong>of</strong>. This is 2.4.3 reformulated. (1) A homomorphism g : <br />
α Mα → N is<br />
uniquely determined by the family gα = g ◦ iα : Mα → N. (2) A homomorphism<br />
f : N → <br />
α Mα is uniquely determined by the family fα = pα ◦f : N → Mα.<br />
2.5.9. Lemma. Let R be a ring and M, N modules. For x ∈ M there is a homomorphism<br />
HomR(M, N) → N, f ↦→ f(x).<br />
Pro<strong>of</strong>. Calculate according to 2.1.1 (f + g) ↦→ (f + g)(x) = f(x) + g(x) and<br />
(af) ↦→ (af)(x) = af(x).<br />
2.5.10. Definition. The natural homomorphism 2.5.9<br />
is the evaluation at x.<br />
evx : HomR(M, N) → N, f ↦→ f(x)<br />
2.5.11. Lemma. There is a natural homomorphism<br />
M → HomR(HomR(M, N), N), x ↦→ evx<br />
Pro<strong>of</strong>. Calculate according to 2.1.1 evx+y(f) = f(x + y) = f(x) + f(y) =<br />
evx(f) + evy(f). and evax(f) = f(ax) = af(x) = aevx(f) to see that the map<br />
is a homomorphism.<br />
2.5.12. Proposition. Let R be a ring and M a module. The evaluation<br />
ev1 : HomR(R, M) M, f ↦→ f(1)<br />
is a natural isomorphism. x ↦→ 1x 2.3.11 is the inverse.<br />
Pro<strong>of</strong>. Calculate the composite ev1(x ↦→ 1x) = 1x(1) = x and 1 f(1)(a) =<br />
af(1) = f(a) proving the claims.