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Math 648 Final
Answer as many questions as you can! Make sure you state clearly any theorems from class that
you use.
Part I. Definitions.
1. For any (S,R)-bimodule P, (P⊗R?,HomS(P, ?)) is an adjoint pair of functors. Explain
carefully what this sentence means.
Solution. It means that the functors HomS(P⊗R?,?) and HomR(?,HomS(P, ?)) from R-modop ×
S-mod to Ab are isomorphic.
Part II. True or False. Justify your answers briefly.
1. Q⊗ZR∼
=Ras an abelian group.
Solution. TRUE. Define a map Q×R→R,(x, y)%→ xy. It is balanced, hence there is
induced a unique map Q⊗ZR→Runder which x⊗y%→ xy. Clearly this map is onto. Note
a
b⊗y=a
b⊗by
b=a
bb⊗y
b=a⊗y
b= 1 ⊗ay
b. Hence every element of Q⊗ZRis of the form
1⊗xfor x∈R. Now if 1 ⊗xis in the kernel, then x= 0 hence 1 ⊗x= 0. Hence the map is
1-1.
2. If Mis a free R-module and X⊆Mis a minimal spanning set, then Xis a basis for M.
Solution. FALSE. Take for example M=R=Zand X={2,3}. Since 1 = 3 −2 this is a
spanning set. It is minimal because 2 and 3 do not span by themselves. But Xis certainly
not a basis, because any basis would have just one element.
3. If Gis a finite abelian group and Z(p)is the localization of Zat the prime ideal (p), then
Z(p)⊗ZGis a p-group.
Solution. TRUE. By the primary decomposition theorem, Gis a direct sum of Zqr’s for
primes q. Since tensor commutes with direct sum, we just need to show that Z(p)⊗ZZqr= 0
if qis a prime different from p. But since qr/∈(p), qris a unit in Z(p), so there exists x∈Z(p)
such that qrx= 1. Then a⊗b=axqr⊗b=ax ⊗qrb=ax ⊗0 = 0 for any a∈Z(p), b ∈Zqr.
Hence it is zero.
4. Any R-module of finite length can be decomposed into a direct sum of cyclic submodules.
Solution. FALSE. (But note it is true for finitely generated modules over a PID...) To find
a counterexample, try R=F[x, y], Fa field. Look at the R-module F[x, y ]/(x2, y2). That
is four dimensional, with basis (the images of) 1, x, y, xy. Let Mbe the submodule of it
generated just by x, y. So Mhas basis x, y, xy . I claim that Mis indecomposable but not
cyclic, so gives a counterexample...
To prove that it is indecompsable, just note that any non-zero submodule of Mnecessarily
contains the vector xy.
To prove that it is not cyclic, suppose that it is cyclic. Say ax +by +cxy generates the
whole module. Then we can write x= (p+qx +ry)(ax +by +cxy) for some p, q, r. Hence
ap = 1, bp = 0, so b= 0 and a)= 0. Similarly you can write ylike this and get b)= 0 and
a= 0. Contradiction.
5. For any ring R, the map f%→ f(1R) is a ring isomorphism EndR(RR)∼
→R.
Solution. FALSE. Just take Rto be non-commutative. Pick x, y ∈Rsuch that xy )=
yx. Let f:R→Rbe the ring homomorphism mapping 1 to x, let g:R→Rbe the
ring homomorphism mapping 1 to y. If it was a ring isomorphism, we would have that
(f◦g)(1) = f(1)g(1), i.e. f(g(1)) = xy. But f(g(1)) = f(y) = f(y1) = yf (1) = yx. So its
not the case as xy )=yx.
