Download Solutions to Homework 9 46. (Dummit

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Solutions to Homework 9
46. (Dummit-Foote 10.3 #2) Suppose Rn ' Rm . Let I be a maximal ideal of R, then Rn /IRn '
Rm /IRm . By the exercise 12 of section 2, this implies that (R/I)n ' (R/I)m , and as these are
vector spaces over the field R/I, we have n = m.
47. (a) Suppose that Q were a free Z-module. Let a/b, c/d ∈ Q. Then bc(a/b) + (−ad)(c/d) = 0,
which says that there is a non-trivial linear relation between any two basis elements of Q. Hence
Q is a free Z-module of rank at most 1. But a/2b is not in the Z-span of a/b for any a/b ∈ Q,
which says that Q cannot have rank 1 either. Thus Q is not a free Z-module.
(b) Suppose that I were a free R-module. If f and g are basis elements, then the non-trivial
relation −gf + f g = 0 forces I to have rank at most 1. Thus I = Rf for some f , but this implies
that I is a principal ideal.
To see that I cannot be principal, suppose we had (X, Y ) = I = (f ). Then there exist g, h ∈ R
such that gf = X, hf = Y . The first equation says that the Y -degree of f (the degree of f when
viewed as a polynomial in Y with coefficients in F [X]) must be zero. Similarly, the X-degree of
f must be zero by the second equation. But this implies that f is a constant and thus I = R, a
48. (a) (Dummit-Foote 10.3 #11) Let φ : M1 → M2 be a non-zero R-module homomorphism. Since
M1 is irreducible, we have ker(φ) = 0. Since M2 is irreducible, we have im(φ) = M2 . Therefore
φ is injective and surjective, and hence an isomorphism. If M1 = M = M2 , then we have shown
that every element of EndR (M ) is an isomorphism. In particular, every element of EndR (M ) has
an inverse with respect to composition, which proves that EndR (M ) is a division ring.
(b) Since M is a simple R-module, we have M = Rm for some m ∈ M (if m 6= 0 then Rm is a
submodule of M and thus must be all of M ). Then for any R-module homomorphism φ : M → M ,
we have φ(m) = rm for some r ∈ R, and this formula completely determines φ. Let ψ ∈ EndR (M )
be another endomorphism of M , and suppose that ψ(m) = sm. Then
φ(ψ(m)) = φ(sm) = sφ(m) = srm = rsm = rψ(m) = ψ(rm) = ψ(φ(m))
which shows that EndR (M ) is commutative. By part (a), it is a division ring, and thus a field.
49. Let N be a submodule of M , and let S = {Ni } be the set of all finitely generated submodules
of N . By assumption, S has a maximal element, call it N 0 . Then for any Ni ∈ S, we have Ni ⊂ N 0 .
If this were not the case, then we could pick x ∈ Ni , x ∈
/ N 0 . The submodule N 0 + Rx would
be a finitely generated submodule properly containing N 0 , contradicting the maximality of N 0 .
Ni ⊂ N 0 ⊂ N
N 0.
which says that N =
that M is Noetherian.
We conclude that every submodule of M is finitely generated, which says
50. (a) The chain
· · · ker(un ) ⊆ ker(un+1 ) ⊆ ker(un+2 ) · · ·
of submodules of M must terminate by the Noetherian assumption. So pick n such that ker(un ) =
ker(un+1 ). Let m ∈ ker(u). The maps uk are all surjective by the surjectivity of u. Pick x ∈ M
such that un (x) = m. Applying u, we get the equation
un+1 (x) = u(m) = 0
which says that x ∈ ker(un+1 ) = ker(un ). This implies that m = 0, and hence u is injective.
(b) The chain
· · · im(un ) ⊇ im(un+1 ) ⊇ im(un+2 ) ⊇ · · ·
of submodules must terminate by the Artinian condition. So pick n such that im(un ) = im(un+1 ).
Let m ∈ M . We can find an x ∈ M such that un (m) = un+1 (x). Then un (m − u(x)) = 0, but un
is injective because u is injective. Hence m = u(x), and it follows that u is surjective and thus an
51. Let {ei } be a basis of V , indexed by Z. We define a map R ⊕ R → R, by sending (1, 0) 7→ f ,
and (0, 1) 7→ g, where f (e2i ) = ei and f (e2i−1 ) = 0, and g(e2i ) = 0 and g(e2i−1 ) = ei . We extend
the map by R-linearity to get an R-module map. Note that 1 here really means idV . Checking
surjectivity and injectivity is straightforward, and thus our map is an isomorphism.