<<

. 14
( 14)



of V (f ). [See the problems in Section 8.1, along with the Nullstellensatz and the
fact that every prime ideal is a radical ideal (Section 8.3, Problem 2).] No other
decomposition is possible, for if V (fi ) ⊆ V (fj ), then (fi ) ⊇ (fj ). This is impossible
if fi and fj are distinct irreducible factors of f .
19

2. By the Nullstellensatz, IV (f ) = (f ), and we claim that (f ) = (f1 · · · fr ). For if
g ∈ (f1 · · · fr ), then a su¬ciently high power of g will belong to (f). Conversely, if
g m = hf , then each fi divides g m , and since the fi are irreducible, each fi divides g,
so (f1 · · · fr ) divides g.
3. By Problem 1, f is irreducible if and only if V (f ) is an irreducible hypersurface. If f
and g are irreducible and V (f ) = V (g), then as in Problem 1, (f ) = (g), so f = cg
for some nonzero constant c (Section 2.1, Problem 2). Thus f ’ V (f ) is a bijection
between irreducible polynomials and irreducible hypersurfaces, if the polynomials f
and cf, c = 0, are identi¬ed.
4. This follows from the de¬nition of I(X) in (8.1.3), and the observation that a function
vanishes on a union of sets i¬ it vanishes on each of the sets.
5. By Section 8.1, Problem 5, every variety V is the union of ¬nitely many irreducible
subvarieties V1 , . . . , Vr . By Problem 4, I(V ) = ©r I(Vi ). By the Problems in Sec-
i=1
tion 8.1, each I(Vi ) is a prime ideal. By (8.4.3), every radical ideal is I(V ) for some
variety V , and the result follows.
6. By Section 8.1, Problem 6, and the inclusion-reversing property of I (part (4) of
(8.1.3)), the decomposition is unique if we discard any prime ideal that properly
contains another one. In other words, we retain only the minimal prime ideals.
7. If f is any irreducible factor of any of the fi , then f does not divide g. Thus for some
j = i, f does not divide fj . By Problem 7 of Section 8.3, the simultaneous equations
f = fj = 0 have only ¬nitely many solutions, and consequently X is a ¬nite set.
8. With notation as in Problem 7, fi = ghi , where the gcd of the hi is constant. Thus
X is the union of the algebraic curve de¬ned by g = 0 and the ¬nite set de¬ned by
h1 = · · · = hm = 0. (This analysis does not apply when X is de¬ned by the zero
polynomial, in which case X = A2 .)
9. If k = R, the zero-set of x2 + y 2n is {(0, 0)} for all n = 1, 2, . . . . If k is algebraically
closed, then as a consequence of the Nullstellensatz, V (f ) = V (g) with f and g
irreducible implies that f = cg for some constant c. (See Problem 3).
10. Let k = F2 , and let I be the ideal of k[X] generated by f (X) = X 2 + X + 1. Since
f is irreducible, I is a maximal ideal (Section 3.1, Problem 8), in particular, I is
proper. But f (0) and f (1) are nonzero, so V (I) is empty, contradicting the weak
Nullstellensatz.


Section 8.5
1. If x ∈ M , then the ideal generated by M and x is R, by maximality of M . Thus there
/
exists y ∈ M and z ∈ R such that y + zx = 1. By hypothesis, zx, hence x, is a unit.
Take the contrapositive to conclude that every nonunit belongs to M .
2. Any additive subgroup of the cyclic additive group of Zpn must consist of multiples
of some power of p, and it follows that every ideal is contained in (p), which must
therefore be the unique maximal ideal.
3. No. A can be nilpotent, that is, some power of A can be 0. The set will be multiplicative
if A is invertible.
20

4. S ’1 (gf ) takes m/s to g(f (m))/s, as does S ’1 gS ’1 f . If f is the identity on M , then
S ’1 f is the identity on S ’1 M .
5. By hypothesis, gf = 0, so S ’1 gS ’1 f = S ’1 gf = S ’1 0 = 0. Thus im S ’1 f ⊆ ker S ’1 g.
Conversely, let x ∈ N, s ∈ S, with x/s ∈ ker S ’1 g. Then g(x)/s = 0/1, so for some
t ∈ S we have tg(x) = g(tx) = 0. Therefore tx ∈ ker g = im f , so tx = f (y) for some
y ∈ M . We now have x/s = f (y)/st = (S ’1 f )(y/st) ∈ im S ’1 f .
6. The set of nonuits is M = {f /g : g(a) = 0, f (a) = 0}, which is an ideal. By (8.5.9), R
is a local ring with maximal ideal M .
7. The sequence 0 ’ N ’ M ’ M/N ’ 0 is exact, so by Problem 5, 0 ’ NS ’
MS ’ (M/N )S ’ 0 is exact. (If f is one of the maps in the ¬rst sequence, the
corresponding map in the second sequence is S ’1 f .) It follows from the de¬nition of
localization of a module that NS ¤ MS , and by exactness of the second sequence we
have (M/N )S ∼ MS /NS , as desired.
=


Section 8.6
√ √
1. If xm belongs to the intersection of the Ii , then x belongs to each I i , so x ∈ ©n
i=1 I i .

Conversely, if x ∈ ©i=1 I i , let x ∈ Ii . If m is the maximum of the mi , then
n mi

x ∈ ©i=1 Ii , so x ∈ ©i=1 i
m n n I.

2. We are essentially setting X = Z = 0 in R, and this collapses R down to k[Y ]. Formally,
map f + I to g + I, where g consists of those terms in f that do not involve X or Z.
Then R/P ∼ k[Y ], an integral domain. Therefore P is prime.
=

3. (X + I)(Y + I) = Z 2 + I ∈ P 2 , but X + I ∈ P 2 and Y + I ∈ P 2 = P .
/ /
4. P1 is prime because R/P1 ∼ k[Y ], an integral domain. P2 is maximal by (8.3.1), so P2 2
=
is P2 -primary by (8.6.6). The radical of Q is P2 , so by (8.6.5), Q is P2 -primary.
5. The ¬rst assertion is that
(X 2 , XY ) = (X) © (X, Y )2 = (X) © (X 2 , XY, Y 2 )
and the second is
(X 2 , XY ) = (X) © (X 2 , Y ).
In each case, the left side is contained in the right side by de¬nition of the ideals
involved. The inclusion from right to left follows because if f (X, Y )X = g(X, Y )Y 2
(or f (X, Y )X = g(X, Y )Y ), then g(X, Y ) must involve X and f (X, Y ) must involve Y .
Thus f (X, Y )X is a polynomial multiple of XY .
6. By (8.6.9), a proper ideal I can be expressed as the intersection of ¬nitely many primary
ideals Qi . If Qi is Pi -primary, then by Problem 1,

I = I = ©i Qi = ©i Pi .
√ √
have X, Y ∈ In , so (X, Y ) ⊆ In . By (8.3.1),
7. Since X 3 and Y n belong to In , we √
(X, Y ) is √ maximal ideal. Since In is proper (it does not contain 1), we have
a
(X, Y ) = In . By (8.6.5), In is primary.
21

Section 8.7
1. [(x, y + y ) + G] ’ [(x, y) + G] ’ [(x, y ) + G] = 0 since (x, y + y ) ’ (x, y) ’ (x, y ) ∈ G;
r[(x, y) + G] ’ [(rx, y) + G] = 0 since r(x, y) ’ (rx, y) ∈ G; the other cases are similar.
2. Let a and b be integers such that am + bn = 1. If x ∈ Zm and y ∈ Zn , then
x — y = 1(x — y) = a(mx — y) + b(x — ny) = 0 since z — 0 = 0 — z = 0.
3. Let A be a torsion abelian group, that is, each element of A has ¬nite order. If x ∈ A
and y ∈ Q, then nx = 0 for some positive integer n. Thus x — y = n(x — (y/n)) =
nx — (y/n) = 0 — (y/n) = 0.
4. We have h = g h = g gh and h = gh = gg h . But if P = T and f = h, then
g = 1T makes the diagram commute, as does g g. By the uniqueness requirement in
the universal mapping property, we must have g g = 1T , and similarly gg = 1T . Thus
T and T are isomorphic.
5. n — x = n(1 — x) = 1 — nx = 1 — 0 = 0.
6. nZ — Zn ∼ Z — Zn ∼ Zn by (8.7.6), with n — x ’ 1 — x ’ x, and since x = 0, n — x
= =
cannot be 0.
7. We have a bilinear map (f, g) ’ f — g from HomR (M, M ) — HomR (N, N ) to
HomR (M —R N, M —R N ), and the result follows from the universal mapping property
of tensor products.
8. In terms of matrices, we are to prove that Mm (R) — Mn (R) ∼ Mmn (R). This follows
=
2
because Mt (R) is a free R-module of rank t .


Section 8.8
1. y1 · · · (yi +yj ) · · · (yi +yj ) · · · yp = y1 · · · yi · · · yi · · · yp +y1 · · · yi · · · yj · · · yp +y1 · · · yj · · ·
yi · · · yp + y1 · · · yj · · · yj · · · yp . The left side, as well as the ¬rst and last terms on the
right, are zero by de¬nition of N . Thus y1 · · · yi · · · yj · · · yp = ’y1 · · · yj · · · yi · · · yp , as
asserted.
2. If π is any permutation of {a, . . . , b}, then by Problem 1,

xπ(a) · · · xπ(b) = (sgn π)xa · · · xb .

The left side will be ±xa · · · xb , regardless of the particular permutation π, and the
result follows.
3. The multilinear map f induces a unique h : M —p ’ Q such that h(y1 — · · · — yp ) =
f (y1 , . . . , yp ). Since f is alternating, the kernel of h contains N , so the existence and
uniqueness of the map g follows from the factor theorem.
4. By Problem 3, there is a unique R-homomorphism g : Λn M ’ R such that
g(y1 · · · yn ) = f (y1 , . . . , yn ). In particular, g(x1 · · · xn ) = f (x1 , . . . , xn ) = 1 = 0. Thus
x1 · · · xn = 0. If r is any nonzero element of R, then g(rx1 x2 · · · xn ) = f (rx1 , . . . , xn ) = r,
so rx1 · · · xn = 0. By Problem 2, {x1 · · · xn } is a basis for Λn M .
I aI xI = 0, xI ∈ R,
5. Fix the set of indices I0 and its complementary set J0 . If
multiply both sides on the right by xJ0 . If I = I0 , then xI xJ0 = 0 by de¬nition of N .
22

Thus aI0 xI0 xJ0 = ±aI0 x1 · · · xn = 0. By Problem 4, aI0 = 0. Since I0 is arbitrary, the
result follows.
6. We have R0 ⊆ S by de¬nition of S. Assume that Rm ⊆ S for m = 0, 1, . . . , n ’ 1, and
r
let a ∈ Rn (n > 0). Then a ∈ I, so a = i=1 ci xi where (since xi ∈ Rni and R is the
direct sum of the Rm ) ci ∈ Rn’ni . By induction hypothesis, ci ∈ S, and since xi ∈ S
by de¬nition of S, we have a ∈ S, completing the induction.
7. The “if” part follows from Section 8.2, Problem 5, so assume R Noetherian. Since
R0 ∼ R/I, it follows that R0 is Noetherian. Since R is Noetherian, I is ¬nitely
=
generated, so by Problem 6, R = S, a ¬nitely generated R0 -algebra.


Section 9.1
1. Assume R is simple, and let x ∈ R, x = 0. Then Rx coincides with R, so 1 ∈ Rx. Thus
there is an element y ∈ R such that yx = 1. Similarly, there is an element z ∈ R such
that zy = 1. Therefore
z = z1 = zyx = 1x = x, so xy = zy = 1
and y is a two-sided inverse of x. Conversely, assume that R is a division ring, and
let x be a nonzero element of the left ideal I. If y is the inverse of x, then 1 = yx ∈ I,
so I = R and R is simple.
I is proper because f (1) = x = 0, and R/I ∼ Rx by the ¬rst isomorphism theorem.
2. =
3. The “if” part follows from the correspondence theorem, so assume that M is simple.
If x is a nonzero element of M , then M = Rx by simplicity. If I = ker f as in
Problem 2, then M ∼ R/I, and I is maximal by the correspondence theorem.
=
4. The “only if” part was done in Problem 3, so assume that M is not simple. Let N be a
submodule of M with 0 < N < M . If x is a nonzero element of N , then Rx ¤ N < M ,
so x cannot generate M .
By Problem 3, a simple Z-module is isomorphic to Z/I, where I is a maximal ideal
5.
of Z. By Section 2.4, Problems 1 and 2, I = (p) where p is prime.
6. As in Problem 5, a simple F [X]-module is isomorphic to F [X]/(f ), where f is an
irreducible polynomial in F [X]. (See Section 3.1, Problem 8.)
7. If x is a nonzero element of V and y an arbitrary element of V , there is an endomor-
phism f such that f (x) = y. Therefore V = (Endk V )x. By Problem 4, V is a simple
Endk (V )-module.
By (4.7.4), every such short exact sequence splits i¬ for any submodule N ¤ M ,
8.
M ∼ N • P , where the map N ’ M can be identi¬ed with inclusion and the map
=
M ’ P can be identi¬ed with projection. In other words, every submodule of M is a
direct summand. Equivalently, by (9.1.2), M is semisimple.


Section 9.2
1. Unfortunately, multiplication by r is not necessarily an R-endomorphism of M , since
r(sx) = (rs)x, which need not equal s(rx) = (sr)x.
23

2. Let x be a generator of M , and de¬ne f : R ’ M by f (r) = rx. By the ¬rst isomor-
phism theorem, M ∼ R/ ann M . The result follows from the correspondence theorem.
=
3. Let M = Zp •Zp where p is prime. Then M is not a simple Z-module, but ann M = pZ
is a maximal ideal of Z.
4. The computation given in the statement of the problem shows that (1,0) is a generator
of V , hence V is cyclic. But N = {(0, b) : b ∈ F } is a nontrivial proper submodule of
V . (Note that T (0, b) = (0, 0) ∈ N .) Therefore V is not simple.
5. Since F is algebraically closed, f has an eigenvalue » ∈ F . Thus the kernel of f ’ »I
is not zero, so it must be all of M . Therefore f (m) = »m for all m ∈ M .
6. If r ∈ I and s + I ∈ R/I, then r(s + I) = rs + I. But if I is not a right ideal, we cannot
guarantee that rs belongs to I.


Section 9.3
1. For each j = 1, . . . , n, there is a ¬nite subset I(j) of I such that xj belongs to the
direct sum of the Mi , i ∈ I(j). If J is the union of the I(j), then M ⊆ i∈J Mi ⊆ M ,
so M is the direct sum of the Mi , i ∈ J.
2. Each simple module Mi , i = 1, . . . , n, is cyclic (Section 9.1, Problem 4), and therefore
can be generated by a single element xi . Thus M is generated by x1 , . . . , xn .
3. A left ideal is simple i¬ it is minimal, so the result follows from (9.1.2).
4. No. If it were, then by Section 9.1, Problem 5, Z would be a direct sum of cyclic groups
of prime order. Thus each element of Z would have ¬nite order, a contradiction.
5. By (4.6.4), every ¬nite abelian group is the direct sum of various Zp , p prime. If p and
q are distinct primes, then Zp • Zq ∼ Zpq by the Chinese remainder theorem. Thus Zn
=
can be assembled from cyclic groups of prime order as long as no prime appears more
than once in the factorization of n. (If Zp • Zp is part of the decomposition, the group
cannot be cyclic.) Consequently, Zn is semisimple if and only if n is square-free.
6. This follows from Section 9.1, Problem 8. (In the ¬rst case, B is semisimple by hy-
pothesis, and in the second case A is semisimple. The degenerate case M = 0 can be
handled directly.)
7. Conditions (a) and (b) are equivalent by (9.3.2) and the de¬nition of semisimple ring.
By Problem 6, (b) implies both (c) and (d). To show that (c) implies (b) and (d)
implies (b), let M be a nonzero R-module, with N a submodule of M . By hypothesis,
the sequence 0 ’ N ’ M ’ M/N ’ 0 splits. (By hypothesis, M/N is projective in
the ¬rst case and N is injective in the second case.) By Section 9.1, Problem 8, M is
semisimple.


Section 9.4
1. If there is an in¬nite descending sequence I1 ⊃ I2 ⊃ · · · of left ideals, we can proceed
exactly as in (9.4.7) to reach a contradiction.
24

2. Let I1 be any nonzero left ideal. If I1 is simple, we are ¬nished. If not, there is
a nonzero left ideal I2 such that I1 ⊃ I2 . If we continue inductively, the Artinian
hypothesis implies that the process must terminate in a simple left ideal It .
3. By Problem 2, the ring R has a simple R-module M . The hypothesis that R has no
nontrivial two-sided ideals implies that we can proceed exactly as in (9.4.6) to show
that M is faithful.
4. If V is in¬nite-dimensional over D, then exactly as in (9.4.7), we ¬nd an in¬nite
descending chain of left ideals, contradicting the assumption that R is Artinian.
5. By Problem 4, V is a ¬nite-dimensional vector space over D, so we can reproduce the
discussion preceding (9.4.7) to show that R ∼ EndD (V ) ∼ Mn (Do ).
= =
6. The following is a composition series:
0 < M1 < M1 • M2 < · · · < M1 • M2 • · · · • Mn = M.

7. By (9.1.2), M is a direct sum of simple modules. If the direct sum is in¬nite, then we
can proceed as in Problem 6 to construct an in¬nite ascending (or descending) chain
of submodules of M , contradicting the hypothesis that M is Artinian and Noetherian.


Section 9.5
1. If g ∈ ker ρ, then gv = v for every v ∈ V . Take v = 1G to get g1G = 1G , so g = 1G
and ρ is injective.
2. (gh)(v(i)) = v(g(h(i))) and g(h(v(i))) = g(v(h(i))) = v(g(h(i))). Also, 1G (v(i)) =
v(1G (i)) = v(i).
3. We have g(v(1)) = v(4), g(v(2)) = v(2), g(v(3)) = v(1), g(v(4)) = v(3), so
® 
0010
0 1 0 0
[g] =  
°0 0 0 1» .
1000

4. We have gv1 = v2 , gv2 = v3 = ’v1 ’ v2 ; hv1 = v1 , hv2 = v3 = ’v1 ’ v2 . Thus
’1 1 ’1
0
[g] = , h= .
’1 0 ’1
1

5. We have v1 = e1 and v2 = ’ 1 e1 + 1 3e2 . Thus
2 2
√
®  ®
1 ’1 113
2 3
P ’1 = ° », P = ° »
√ √
013 023
2 3

and
√
®
’1 ’1 3
2 2
[g] = P ’1 [g]P = ° √ »,
’1
1
3
2 2
25

1 0
[h] = P ’1 [h]P = .
’1
0

6. Check by direct computation that the matrices A and B satisfy the de¬ning relations
of D8 : A4 = I, B 2 = I, AB = BA’1 . (See Section 5.8.)
7. Yes. Again, check by direct computation that the matrices Ai B j , i = 0, 1, 2, 3, j = 0, 1,
are distinct. Thus if g ∈ D8 and ρ(g) = I, then g is the identity element of D8 .


Section 9.6
1. Let W be the one-dimensional subspace spanned by v1 +v2 +v3 . Since any permutation
in S3 permutes the vi , v ∈ W implies gv ∈ W .
1 0
2. Multiplying [ar ] by , we have ar v1 = v1 and ar v2 = rv1 + v2 . Since W is
and
0 1
spanned by v1 , it is closed under the action of G and is therefore a kG-submodule.
3. If
x x
[ar ] =c
y y

then x + ry = cx and y = cy. If y = 0 then c = 1, so ry, hence y, must be 0. Thus
c = 1 and x is arbitrary, so that any one-dimensional kG-submodule must coincide
with W .
4. If V = W • U , where U is a kG-submodule, then U must be one-dimensional. By
Problem 3, W = U , and since W = 0, this is impossible.
5. If M is semisimple and either Noetherian or Artinian, then M is the direct sum of
¬nitely many simple modules. These simple modules are the factors of a composition
series, and the result follows from (7.5.12).
6. Let e be the natural projection on M1 . Then e is an idempotent, e = 0 since M1 = 0,
and e = 1 since e = 0 on M2 and M2 = 0.
7. Let e be a nontrivial idempotent, and de¬ne e1 = e, e2 = 1 ’ e. By direct computation,
e1 and e2 are nontrivial idempotents that are orthogonal. Take M1 = e1 (M ), M2 =
e2 (M ). Then M1 and M2 are nonzero submodules with M = M1 + M2 . To show that
the sum is direct, let z = e1 x = e2 y ∈ M1 © M2 , with x, y ∈ M . Then e1 z = e1 e2 y = 0,
and similarly e2 z = 0. Thus z = 1z = e1 z + e2 z = 0.


Section 9.7
1. If M = Rx, de¬ne f : R ’ M by f (r) = rx. By the ¬rst isomorphism theorem,
M ∼ R/ ker f . Moreover, ker f = ann(M ). Conversely, R/I is cyclic since it is
=
generated by 1 + I.
2. If N is a maximal submodule of M , then N is the kernel of the canonical map of M
onto the simple module M/N . Conversely, if f : M ’ S, S simple, then f (M ) is 0
26

or S, so f = 0 or S ∼ M/ ker f . Thus ker f is either M or a maximal submodule of M .
=
The intersection of all kernels therefore coincides with the intersection of all maximal
submodules. [If there are no maximal submodules, then the intersection of all kernels
is M .]
3. By the correspondence theorem, the intersection of all maximal left ideals of R/I is 0.
This follows because the intersection of all maximal left ideals of R containing I is
the intersection of all maximal left ideals of R. [Note that J(R) is contained in every
maximal left ideal, and I = J(R).]
4. Let g be an R-module homomorphism from N to the simple R-module S. Then
gf : M ’ S, so by Problem 2, J(M ) ⊆ ker(gf ). But then f (J(M )) ⊆ ker g. Take the
intersection over all g to get f (J(M )) ⊆ J(N ).
Suppose a ∈ J(R). If 1+ab is a nonunit, then it belongs to some maximal ideal M . But
5.
then a belongs to M as well, and therefore so does ab. Thus 1 ∈ M , a contradiction.
Now assume a ∈ J(R), so that for some maximal ideal M , a ∈ M . By maximality,
/ /
M + Ra = R, so 1 = x + ra for some x ∈ M and r ∈ R. Since x belongs to M , it
cannot be a unit, so if we set b = ’r, it follows that 1 + ab is a nonunit.
6. By the correspondence theorem, there is a bijection, given by ψ(A) = A/N , between
maximal submodules of M containing N and maximal submodules of M/N . Since
N ¤ J(M ) by hypothesis, a maximal submodule of M containing N is the same
thing as a maximal submodule of M . Thus J(M ) corresponds to J(M/N ), that is,
ψ(J(M )) = J(M/N ). Since ψ(J(M )) = J(M )/N , the result follows.


Section 9.8
1. at ∈ (at+1 ), so there exists b ∈ R such that at = bat+1 . Since R is an integral domain
we have 1 = ba.
2. Let a be a nonzero element of the Artinian integral domain R. The sequence (a) ⊇
(a2 ) ⊇ . . . stabilizes, so for some t we have (at ) = (at+1 ). By Problem 1, a has an
inverse, proving that R is a ¬eld.
3. If P is a prime ideal of R, then R/P is an Artinian integral domain by (7.5.7) and
(2.4.5). By Problem 2, R/P is a ¬eld, so by (2.4.3), P is maximal.
4. We have P © I ∈ S and P © I ⊆ I, so by minimality of I, P © I = I. Thus
P ⊇ I = ©n Ij , so by (7.6.2), P ⊇ Ij for some j. But P and Ij are maximal ideals,
j=1
hence P = Ij .
5. If z ∈ F , with z = x + y, x ∈ M , y ∈ N , de¬ne h : F ’ M/JM • N/JN by
h(z) = (x + JM ) + (y + JN ). Then h is an epimorphism with kernel JM + JN = JF ,
and the result follows from the ¬rst isomorphism theorem.
6. By (9.8.5), M/JM is an n-dimensional vector space over the residue ¬eld k. Since F ,
hence F/JF , is generated by n elements, F/JF has dimension at most n over k.
Thus N/JN must have dimension zero, and the result follows.
7. Multiplication of an element of I on the right by a polynomial f (X, Y ) amounts to
multiplication by the constant term of f . Thus I is ¬nitely generated as a right R-
module i¬ it is ¬nitely generated as an abelian group. This is a contradiction, because
27

as an abelian group,
I = •∞ ZX n Y.
n=0

8. By the Hilbert basis theorem, Z[X] is a Noetherian ring and therefore a Noetherian
left Z[X]-module. The isomorphic copy Z[X]Y is also a Noetherian left Z[X]-module,
hence so is R, by (7.5.8). A left ideal of R that is ¬nitely generated as a Z[X]-module
is ¬nitely generated as an R-module, so R is left-Noetherian.
9. The set {x1 , . . . , xn } spans V , and therefore contains a basis. If the containment is
proper, then by (9.8.5) part (ii), {x1 , . . . , xn } cannot be a minimal generating set, a
contradiction.
10. In vector-matrix notation we have y = Ax and therefore y = Ax, where aij = aij + J.
By Problem 9, x and y are bases, so that det A = 0. But under the canonical map of
R onto k, det A maps to det A, and therefore det A cannot belong to the kernel of
the map, namely J. But by (8.5.9), J is the ideal of nonunits of R, so det A is a unit.


Section 10.1
1. This is exactly the same as the proof for groups in (1.1.1).
2. Any ring homomorphism on Q is determined by its values on Z (write m = (m/n)n
and apply the homomorphism g to get g(m/n) = g(m)/g(n). Thus if gi = hi, then g
coincides with h on Z, so g = h, proving that i is epic.
3. Let AZB be a shorthand notation for the composition of the unique morphism from
A to the zero object Z, followed by the unique morphism from Z to B. If Z is another
zero object, then AZB = (AZ Z)B = A(Z ZB) = A(Z B) = AZ B, as claimed.
4. If is = it, then f is = f it = 0, so is(= it) = ih where h is unique. Thus s and t must
coincide with h, hence i is monic.
5. A kernel of a monic f : A ’ B is 0, realized as the zero map from a zero object Z
to A. For f 0 = 0, and if f g = 0, then f g = f 0; since f is monic, g = 0. But then g
can be factored through 0ZA . Similarly, a cokernel of the epic f : A ’ B is the zero
map from B to a zero object.
6. If j = ih and i = jh , then i = ihh , so by uniqueness in part (2) of the de¬nition of
kernel (applied when g = i), hh , and similarly h h, must be the identity.
7. De¬ne h : K ’ A by h(x) = 1 for all x. Since K is nontrivial, h cannot be injective,
so that g = h. But f g = f h, since both maps take everything in K to the identity
of B.
8. The kernel of a ring homomorphism is an ideal, but not a subring (since it does not
contains the multiplicative identity).
9. Let f : A ’ B be a noninjective ring homomorphism. Let C be the set of pairs (x, y)
in the direct product A — A such that f (x) = f (y). Since f is not injective, there is
an element (x, y) of C with x = y. Thus if D = {(x, x) : x ∈ A}, then D ‚ C. If g is
the projection of A — A on the ¬rst coordinate, and h is the projection on the second
coordinate, then f (g(x, y)) = f (x) and f (h(x, y)) = f (y), so f g = f h on the ring C.
But g and h disagree on the nonempty set C \ D, so f is not monic.
28

10. Let g be the canonical map of N onto N/f (M ), and let h : N ’ N/f (M ) be identically
zero. Since gf sends everything to 0, we have gf = hf with g = h. Thus f is not epic.


Section 10.2
1. Suppose that y is the product of the xi . By de¬nition of product, if fi : x ’ xi for
all i, there is a unique f : x ’ y such that pi f = fi . Since pi : y ’ xi , we have y ¤ xi .
Moreover, if x ¤ xi for all i, then x ¤ y. Therefore y is a greatest lower bound of
the xi .
2. No. For example, consider the usual ordering on the integers.
3. By duality, a coproduct of the xi , if it exists, is a least upper bound.
4. If x has order r and y has order s, then rs(x + y) = s(rx) + r(sy) = 0. Thus the sum
of two elements of ¬nite order also has ¬nite order, and the result follows.
5. The key point is that if f is a homomorphism of a torsion abelian group S, then f (S)
is a also torsion [since nf (x) = f (nx)]. Thus in diagram (1) with A = Ai , we have
f (S) ⊆ T (A). Since Ai is the product in the category of abelian groups, it follows
that T (A) satis¬es the universal mapping property and is therefore the product in the
category of torsion abelian groups.
6. Given homomorphisms fj : Gj ’ H, we must lift the fj to a homomorphism from
the free product to H. This is done via f (a1 · · · an ) = f1 (a1 ) · · · fn (an ). If ij is the
inclusion map from Gj to —i Gi , then f (ij (aj )) = f (aj ) = fj (aj ), as required.
7. We have pi f = fi , where fi : G ’ Ci . The fi can be chosen to be surjective (e.g., take
G to be the direct product of the Ci ), and it follows that the pi are surjective.
8. Since f : C1 ’ C, we have f (a1 ) = na for some positive integer n. Thus

a1 = f1 (a1 ) = p1 f (a1 ) = p1 (na) = np1 (a) = na1 ;
0 = f2 (a1 ) = p2 f (a1 ) = p2 (na) = np2 (a) = na2 .

9. By Problem 8, the order of C1 divides n ’ 1, and the order of C2 divides n. There are
many choices of C1 and C2 for which this is impossible. For example, let C1 and C2
be nontrivial p-groups for a ¬xed prime p.


Section 10.3
1. If f : x ’ y, then F f : F x ’ F y. By de¬nition of the category of preordered sets, this
statement is equivalent to x ¤ y =’ F x ¤ F y. Thus functors are order-preserving
maps.
2. F must take the morphism associated with xy to the composition of the morphism
associated with F x and the morphism associated with F y. In other words, F (xy) =
F (x)F (y), that is, F is a homomorphism.
3. If β ∈ X — , then (gf )— (β) = βgf and f — g — (β) = f — (βg) = βgf .
29

4. To verify the functorial property, note that

(gf )—— (v —— ) = v —— (gf )— = v —— f — g — (by Problem 3)

and

g —— f —— v —— = g —— (v —— f — ) = v —— f — g — .

Thus (gf )—— = g —— f —— . If f is the identity, then so is f — , and consequently so is f —— .
5. f —— tV (v) = f —— (v) = vf — , and if β ∈ W — , then (vf — )(β) = v(f — β) = (f — β)(v) = βf (v).
But tW f (v) = f (v) where f (v)(β) = βf (v).
6. Groups form a subcategory because every group is a monoid and every group homo-
morphism is, in particular, a monoid homomorphism. The subcategory is full because
every monoid homomorphism from one group to another is also a group homomor-
phism.
7. (a) If two group homomorphisms are the same as set mappings, they are identical as
homomorphisms as well. Thus the forgetful functor is faithful. But not every map
of sets is a homomorphism, so the forgetful functor is not full.
(b) Since (f, g) is mapped to f for arbitrary g, the projection functor is full but not
faithful (except in some degenerate cases).


Section 10.4
1. If a homomorphism from Z2 to Q takes 1 to x, then 0 = 1 + 1 ’ x + x = 2x. But 0
must be mapped to 0, so x = 0.
2. A nonzero homomorphism can be constructed with 0 ’ 0, 1 ’ Then 1 + 1 ’
1
2.
2 + 2 = 1 = 0 in Q/Z.
1 1

3. Since a trivial group cannot be mapped onto a nontrivial group, there is no way that
F g can be surjective.
4. Let f be a homomorphism from Q to Z. If r is any rational number and m is a positive
integer, then
r r r
+ ··· +
f (r) = f = mf
m m m
so
r f (r)
f = .
m m
But if f (r) = 0, we can choose m such that f (r)/m is not an integer, a contradiction.
Therefore f = 0.
5. By Problem 4, Hom(Q, Z) = 0. But Hom(Z, Z) = 0, so as in Problem 3, Gf cannot be
surjective.
30

6. We have Z2 — Z ∼ Z2 and Z2 — Q = 0
=

m 2m m m
1— =1— =2— =0— = 0.
n 2n 2n 2n

Thus the map Hf cannot be injective.
7. Since f— = F f is injective, f ± = 0 implies ± = 0, so f is monic and hence injective.
Since g— f— = 0, we have gf ± = 0 for all ± ∈ Hom(M, A). Take M = A and ± = 1A to
conclude that gf = 0, so that im f ⊆ ker g. Finally, take M = ker g and ± : M ’ B
the inclusion map. Then g— ± = g± = 0, so ± ∈ ker g— = im f— . Thus ± = f β for some
β ∈ Hom(M, A). Thus ker g = M = im ± ⊆ im f .
8. If (3) is exact for all possible R-modules N , then (1) is exact. This is dual to the result
of Problem 7, and the proof amounts to interchanging injective and surjective, monic
and epic, inclusion map and canonical map, kernel and cokernel.


Section 10.5
1. If x ∈ P , then f (x) can be expressed as i ti ei (a ¬nite sum), and we de¬ne fi (x) = ti
and xi = π(ei ). Then

x = π(f (x)) = π( ti ei ) = ti π(ei ) = fi (x)xi .
i i i


2. π(f (x)) = fi (x)π(ei ) = fi (x)xi = x.
i i
3. By Problem 2, the exact sequence 0 ’ ker π ’ F ’ P ’ 0 (with π : F ’ P ) splits,
and therefore P is a direct summand of the free module F and hence projective.
4. Since Rn is free, the “if” part follows from (10.5.3), part (4). If P is projective, then
by the proof of (3) implies (4) in (10.5.3), P is a direct summand of a free module of
rank n. [The free module can be taken to have a basis whose size is the same as that
of a set of generators for P .]
5. This follows from Problem 1 with F = Rn .
6. If P is projective and isomorphic to M/N , we have an exact sequence 0 ’ N ’
M ’ P ’ 0. Since P is projective, the sequence splits by (10.5.3), part (3), so P is
isomorphic to a direct summand of M . Conversely, assume that P is a direct summand
of every module of which it is a quotient. Since P is a quotient of a free module F , it
follows that P is a direct summand of F . By (10.5.3) part (4), P is projective.
7. In the diagram above (10.5.1), take M = R1 • R2 , with R1 = R2 = R. Take P =
N = R1 , f = 1R1 , and let g be the natural projection of M on N . Then we can take
h(r) = r + s, r ∈ R1 , s ∈ R2 , where s is either 0 or r. By replacing M by an arbitrary
direct sum of copies of R, we can produce two choices for the component of h(r) in each
Ri , i = 2, 3, . . . (with the restriction that only ¬nitely many components are nonzero).
Thus there will be in¬nitely many possible choices for h altogether.
31

Section 10.6
1. f g(a) = (g(a), 0) + W and g f (a) = (0, f (a)) + W , with (g(a), 0) ’ (0, f (a)) =
(g(a), ’f (a)) ∈ W . Thus f g = g f .
2. If (b, c) ∈ W , then b = g(a), c = ’f (a) for some a ∈ A. Therefore

g (c) + f (b) = ’g f (a) + f g(a) = 0

and h is well-de¬ned.
3. hg (c) = h((0, c) + W ) = g (c) + 0 = g (c) and hf (b) = h((b, 0) + W ) = 0 + f (b) =
f (b).
4. h ((b, c) + W ) = h ((0, c) + W + (b, 0) + W ) = h (g (c) + f (b)) = h g (c) + h f (b) =
g (c) + f (b) = h((b, c) + W ).
5. If f (b) = 0, then by de¬nition of f , (b, 0) ∈ W , so for some a ∈ A we have b = g(a)
and f (a) = 0. Since f is injective, a = 0, hence b = 0 and f is injective.
6. If b ∈ B, c ∈ C, then surjectivity of f gives c = f (a) for some a ∈ A. Thus
f (b + g(a)) = (b + g(a), 0) + W = (b + g(a), 0) + W + (’g(a), f (a)) + W [note
that (’g(a), f (a)) ∈ W ] = (b, f (a)) + W = (b, c) + W , proving that f is surjective.
7. If (a, c) ∈ D, then f (a) = g(c) and f g (a, c) = f (a), gf (a, c) = g(c). Thus f g = gf .
8. If x ∈ E, then f g (x) = gf (x), so (g (x), f (x)) ∈ D. Take h(x) = (g (x), f (x)),
which is the only possible choice that satis¬es g h = g and f h = f .
9. If (a, c) ∈ D and f (a, c) = 0, then c = 0, so f (a) = g(c) = 0. Since f is injective,
a = 0. Consequently, (a, c) = 0 and f is injective.
10. If c ∈ C, then there exists a ∈ A such that f (a) = g(c). Thus (a, c) ∈ D and
f (a, c) = c, proving that f is surjective.
11. If x, y ∈ I, then f (xy) = xf (y) and f (xy) = f (yx) = yf (x). Thus xf (y) = yf (x),
and if x and y are nonzero, the result follows upon division by xy.
12. We must extend f : I ’ Q to h : R ’ Q. Let z be the common value f (x)/x, x ∈ I,
x = 0. De¬ne h(r) = rz, r ∈ R. Then h is an R-homomorphism, and if x ∈ I, x = 0,
then h(x) = xz = xf (x)/x = f (x). Since h(0) = f (0) = 0, h is an extension of f and
the result follows from (10.6.4).


Section 10.7
1. The only nonroutine veri¬cation is the check that sf ∈ HomR (M, N ):

(sf )(rm) = f (rms) = rf (ms) = r[(sf )(m)].

2. (f r)(ms) = f (r(ms)) = f ((rm)s) = f (rm)s = [(f r)(m)]s.
3. (sf )(mr) = s(f (mr)) = s(f (m)r) = (sf (m)r = [(sf )(m)]r.
4. (f r)(sm) = (f (sm))r = (sf (m))r = s(f (m)r) = s[(f r)(m)].
32

5. In Problem 2, M and N are right S-modules, so we write f on the left: (f r)m =
f (rm). In Problem 3, M and N are right R-modules, and we write f on the left:
(sf )m = s(f m). In Problem 4, M and N are left S-modules, and we write f on the
right: m(f r) = (mf )r.
6. Let y ∈ M , r ∈ R with r = 0. By hypothesis, x = 1 y ∈ M , so we have x ∈ M such
r
that y = rx, proving M divisible.
7. If y ∈ M , r ∈ R with r = 0, we must de¬ne 1 y. Since M is divisible, there exists x ∈ M
r
such that y = rx, and we take 1 y = x. If x ∈ M and y = rx , then r(x ’ x ) = 0,
r
and since M is torsion-free, x = x . Thus x is unique and scalar multiplication is
well-de¬ned.
8. Let f be a nonzero R-homomorphism from Q to R. Then f (u) = 1 for some u ∈ Q. [If
f (x) = r = 0, then rf (x/r) = f (rx/r) = f (x) = r, so we can take u = x/r.] Now if s
is a nonzero element of R, then sf (u/s) = f (su/s) = f (u) = 1, so f (u/s) is an inverse
of s. Consequently, R is a ¬eld, contradicting the hypothesis.


Section 10.8
1. Let A = R[X] where R is any commutative ring. As an R-algebra, A is generated
by X, but A is not ¬nitely generated as an R-module since it contains polynomials of
arbitrarily high degree.
2. The bilinear map determined by (X i , Y j ) ’ X i Y j induces an R-homomorphism of
R[X] —R R[Y ] onto R[X, Y ], with inverse determined by X i Y j ’ X i — Y j .
3. Abbreviate X1 , . . . , Xn by X and Y1 , . . . , Ym by Y . Let A be a homomorphic image
of R[X] under f , and B a homomorphic image of R[Y ] under g. Then A —R B is a
homomorphic image of R[X] — R[Y ](∼ R[X, Y ] by Problem 2) under f — g.
=
4. If f : A ’ B is an injective R-module homomorphism, then by hypothesis, (1—f ) : S—R
A ’ S —R B is injective. Also by hypothesis,
(1 — (1 — f )) : M —S S —R A ’ M —S S —R B
is injective. Since M —S S ∼ M , the result follows.
=
5. Let f : A ’ B be injective. Since A —S S ∼ A and B —S S ∼ B, it follows from the
= =
hypothesis that (f — 1) : A —S (S —R M ) ’ B —S (S —R M ) is injective. Thus S —R M
is a ¬‚at S-module.
6. ± is derived from the bilinear map S ’1 R — M ’ S ’1 M given by (r/s, x) ’ rx/s. We
must also show that β is well-de¬ned. If x/s = y/t, then there exists u ∈ S such that
utx = usy. Thus
1 ut 1 1 1
—x= —x= — utx = — usy = — y
s sut sut sut t
as required. By construction, ± and β are inverses of each other and yield the desired
isomorphism.
7. We must show that S ’1 R —R is an exact functor. But in view of Problem 6, an
equivalent statement is the localization functor S ’1 is exact, and this has already been
proved in Section 8.5, Problem 5.
33

Section 10.9
1. The proof of (10.9.4) uses the fact that we are working in the category of modules. To
simply say “duality” and reverse all the arrows, we would need an argument that did
not depend on the particular category.
2. Let N be the direct limit of the Ni . The direct system {Ni , h(i, j)} induces a direct
system {M — Ni , 1 — h(i, j)}. Compatibility in the new system reduces to compatibility
in the old system; tensoring with 1 is harmless. Since compatible maps fi : Ni ’ B
can be lifted to f : N ’ B, it follows that compatible maps gi : M — Ni ’ B can be
lifted to g : M — N ’ B. Thus M — N satis¬es the universal mapping property for
{M — Ni }.
3. The direct limit is A = ∪∞ An , with ±n : An ’ A the inclusion map
n=1
5. Each R-homomorphism f from the direct sum of the Ai to B induces an
R-homomorphism fi : Ai ’ B. [fi is the injection of Ai into the direct sum, fol-
lowed by f ]. Take ±(f ) = (fi , i ∈ I) ∈ i HomR (Ai , B). Conversely, given such a
family (fi , i ∈ I), the fi can be lifted uniquely to β(fi , i ∈ I) = f . Since ± and β are
inverse R-homomorphisms, the result follows.
6. If f : A ’ i Bi , de¬ne ±(f ) = (pi f, i ∈ I) ∈ i HomR (A, Bi ), where pi is the
projection of the direct product onto the ith factor. Conversely, given (gi , i ∈ I), where
gi : A ’ Bi , the gi can be lifted to a unique g : A ’ i Bi such that pi g = gi for all i.
If we take β(gi , i ∈ I) = g, then ± and β are inverse R-homomorphisms, and the result
follows.
7. There is a free module F such that F = M • M , and since F is torsion-free, so is M .
Since M is injective, it is divisible, so by Problem 7 of Section 10.7, M is a vector
space over the quotient ¬eld Q, hence a direct sum of copies of Q. Therefore, using
Problem 5 above and Problem 8 of Section 10.7,

HomR (M, R) = HomR (•Q, R) ∼ Hom(Q, R) = 0.
=

8. By Problem 7, HomR (M, R) = 0. Let M be a direct summand of the free module F
with basis {ri , i ∈ I}. If x is a nonzero element of M , then x has some nonzero
coordinate with respect to the basis, say coordinate j. If pj is the projection of F on
coordinate j (the j th copy of R), then pj restricted to M is a nonzero R-homomorphism
from M to R. (Note that x does not belong to the kernel of pj .) Thus the assumption
that M = 0 leads to a contradiction.
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Cambridge, 1994
List of Symbols

Throughout the text, ⊆ means subset, ‚ means proper subset
Zn integers modulo n 1.1
Z integers 1.1
A subgroup generated by A 1.1
Sn symmetric group 1.2
An alternating group 1.2
D2n dihedral group 1.2
• Euler phi function 1.1, 1.3
normal subgroup 1.3
proper normal subgrouad 1.3
ker kernel 1.3, 2.2
∼ isomorphism 1.4
=
Z(G) center of a group 1.4
H —K direct product 1.5
Q rationals 2.1
Mn (R) matrix ring 2.1
R[X] polynomial ring 2.1
R[[X]] formal power series ring 2.
End endomorphism ring 2.1
X ideal generated by X 2.2
UFD unique factorization domain 2.6
PID principal ideal domain 2.6
ED Euclidean domain 2.7
min(±, F ) minimal polynomial 3.1
i Ki composite of ¬elds 3.1
Gal(E/F ) Galois group 3.5
the module {0} and the ideal {0}
0 4.1
•i Mi direct sum of modules 4.3
i Mi sum of modules 4.3
HomR (M, N ) set of R-module homomorphisms from M to N 4.4
EndR (M ) endomorphism ring 4.4
g•x group action 5.1
G commutator subgroup 5.7
G(i) derived subgroups 5.7
S|K presentation of a group 5.8
F(H) ¬xed ¬eld 6.1
G(K) ¬xing group 6.1
GF (pn ) ¬nite ¬eld with pn elements 6.4
nth cyclotomic polynomial
Ψn (X) 6.5
∆ product of di¬erences of roots 6.6
D discriminant 6.6, 7.4
N [E/F ] norm 7.3
T [E/F ] trace 7.3
char characteristic polynomial 7.3
nP (I) exponent of P in the factorization of I 7.7
vp p-adic valuation 7.9
| |p p-adic absolute value 7.9
V (S) variety in a¬ne space 8.1
I(X) ideal of a set of points 8.1
√ 1 , . . . , Xn ]
k[X polynomial ring in n variables over the ¬eld k 8.1
I radical of an ideal 8.3
k(X1 , . . . , Xn ) rational function ¬eld over k 8.4
S ’1 R localization of the ring R by S 8.5
S ’1 M localization of the module M by S 8.5
N (R) nilradical of the ring R 8.6
M —R N tensor product of modules 8.7
UMP universal mapping property 8.7
A—B tensor (Kronecker) product of matrices 8.7
kG group algebra 9.5
RG group ring 9.5
J(M ), J(R) Jacobson radical 9.7
HomR (M, ), HomR ( , N ) hom functors 10.3
M —R , —R N tensor fuctors 10.3
Q/Z additive group of rationals mod 1 10.6
(also 1.1, Problem 7)
Z(p∞ ) quasicyclic group A10
G[n] elements of G annihilated by n A10
Hn homology functor S1
fg chain homotopy S1
‚ connecting homomorphism S2, S3
P— ’ M projective resolution S4
M ’ E— injective resolution S4
Ln F left derived functor S5
Rn F right derived functor S5
derived functor of —
Tor S5
Ext derived functor of Hom S5
INDEX
abelian category 10.4 character 6.1
characteristic of a ring or ¬eld 2.1
abelian group 1.1
characteristic polynomial 7.3
absolute value 7.9
characteristic subgroup 5.7
action of a group on a set 5.1
chief series 5.6
adjoint associativity 10.7
Chinese remainder theorem 2.3
adjoint functors 10.7
class equation 5.2
a¬ne n-space 8.1
cokernel 10.1
a¬ne variety 8.1
colorings 5.3
AKLB setup 7.3
commutative diagram 1.4
algebra 4.1
commutative ring 2.1
algebraic closure 3.3, 10.9
commutator 5.7
algebraic curve 8.3
compatible morphisms 10.9
algebraic element 3.1
complete ring of fractions 2.8
algebraic extension 3.1
composite of ¬elds 3.1, 6.2
algebraic function ¬eld 6.9
composition factors 5.6, 7.5
algebraic geometry 8.1¬
composition length 5.6, 7.5
algebraic integers 7.1
composition of morphisms 10.1
algebraic number 3.3, 7.1
composition series 5.6, 7.5
algebraic number theory 7.1¬, 7.3
conjugate elements 5.1, 5.2
algebraically closed ¬eld 3.3
conjugate sub¬elds 6.2
algebraically independent set 6.9
conjugate subgroups 5.1, 6.2
algebraically spanning set 6.9
conjugates of a ¬eld ement 3.5
alternating group 1.2
conjugation 5.1, 5.2-1
annihilator 4.2, 9.2, 9.7
connecting homomorphism S2, S3
archimedian absolute value 7.9
constructible numbers and points 6.8
Artin-Schreier theorem 6.7
content 2.9
Artinian modules 7.5
contravariant functor 10.3
Artinian rings 7.5
coproduct 10.2
ascending chain condition (acc) 2.6, 7.5
core 5.1
associates 2.6
correspondence theorem for
associative law 1.1, 2.1
groups 1.4
automorphism 1.3
correspondence theorem for
Baer™s criterion 10.6 modules 4.2
base change 10.8 correspondence theorem for
basis 4.3 rings 2.3
bilinear mapping 8.7 coset 1.3
binomial expansion modulo p 3.4 counting two ways 5.3
binomial theorem 2.1 covariant functor 10.3
boundary S1 cycle 1.2, S1
cyclic extension 6.7
canonical map 1.3
cyclic group 1.1
category 10.1 cyclic module 4.2, 9.1, 9.2, 9.7
Cauchy™s theorem 5.4 cyclotomic extension.5
Cayley™s theorem 5.1 cyclotomic ¬eld 6.5, 7.2
center of a group 1.4 cyclotomic polynomial 6.5
center of a ring 4.1
central series 5.7 decomposable module 9.6
centralizer 5.2 Dedekind domain 7.6, 7.7
chain complex S1 Dedekind™s lemma 6.1, 6.7, 7.3, 7.4
chain homotopy S1 degree 2.5
chain map S1 deleted projective (or injective)
chain rule 1.3, 3.1 resolution S4
derivative of a polynomial 3.4 faithful action 5.1
derived functors S5 faithful module 7.1, 9.2, 9.4
derived length 5.7 faithful representation 9.5
derived series 5.7 Fermat primes 6.8
descending chain condition 7.5 Fermat™s little theorem 1.3
diagram chasing 4.7 ¬eld 2.1
di¬erential S1 ¬eld discriminant 7.4
dihedral group 1.2, 5.8, 5.8 ¬nite abelian groups 4.6
(in¬nite dilhedral group), 9.5 ¬nite extension 3.1
direct limit 10.9 ¬nite ¬elds 6.4
direct product of groups 1.5 ¬nitely cogenerated module 7.5
direct product of modules 4.3 ¬nitely generated algebra 10.8
direct product of rings 2.3 ¬nitely generated module 4.4
direct sum of modules 4.3 ¬nitely generated submodule 7.5
direct system 10.9 ¬ve lemma 4.7
directed set 10.9 ¬xed ¬eld 6.1
discriminant 6.6, A6, 7.4 ¬xing group 6.1
divides means contains 2.6, 7.7 ¬‚at modules 10.8
divisible abelian group A10 forgetful functor 10.3
divisible module 10.6 formal power series 2.1, 8.2
division ring 2.1, 9.1 four group 1.2, 1.5, A6
double centralizer 9.2 four lemma 4.7
double dual functor 10.3 fractional ideal 7.6
dual basis 7.4 Frattini argument 5.8-2
duality 10.1 free abelian gup functor 10.3
duplicating the cube 6.8 free group 5.8-1
free module 4.3,15
Eisenstein™s irreducibility criterion 2.9
free product 10.2-2
elementary divisors 4.6
Frobenius automorphism 3.4, 6.4
elementary symmetric functions 6.1
full functor 10.3
embedding 3.3, 3.5
full ring of fractions 2.8
embedding in an injective module 10.7
full subcategory 10.3
endomorphism 1.3, 4.4
functor 10.3
epic 10.1
fundamental decomposition theorem
epimorphism 1.3
(for ¬nitely generated modules
equivalent absolute values 7.9
over a PID) 4.6
equivalent matrices 4.4
fundamental theorem of Galois theory
equivalent matrix
6.2-1
representations 9.5
Euclidean domain 2.7
Galois extension 3.5, 6.1¬.
Euler™s identity 2.1
Galois group 3.5, 6.1¬
Euler™s theorem 1.3
Galois group of a cubic, 6.6
evaluation map 2.1
Galois group of a polynomial 6.3
exact functor 8.5, 10.4
Galois group of a quadratic 6.3
exact sequence 4.7
Galois group of a quartic A6
exponent of a group 1.1, 6.4
Gauss™ lemma 2.9
Ext S5
Gaussian integers 2.1, 2.7
extension of a ¬eld 3.1
general equation of degree n 6.8
extension of scalars 8.7, 10.8
general linear group 1.3
exterior algebra 8.8
generating set 4.3
generators and relations 1.2, 4.6, 5.8
F-isomorphism, etc. 3.2
greatest common divisor 2.6, 7.7
factor theorem for groups 1.4
group 1.1
factor theorem for modules 4.2
factor theorem for rings 2.3 group algebra 9.5
group representations 9.5 Jacobson radical 9.7
group ring 9.5 Jacobson™s theorem 9.2
Jordan-Holder theorem 5.6, 7.5
Hermite normal form 4.5
Hilbert basis theorem 8.2 kernel 1.3, 2.2, 10.1
Hilbert™s Nullstellensatz 8.3, 8.4 kernel of an action 5.1
Hilbert™s Theorem 90 7.3 Kronecker product of matrices 8.7
hom functors 10.3-1 Krull-Schmidt theorem 9.6
homology functors S1 Kummer extension 6.7
homology group S1
Lagrange interpolation formula 2.5
homology module S1
Lagrange™s theorem 1.3
homomorphism from R to M deter-
Laurent series 7.9
mined by what it does to the
leading coe¬cient 2.5
identity, 9.4, S6
least common multiple 2.6, 7.7
homomorphism of algebras 4.1
left adjoint 10.7
homomorphism of groups 1.3
left cancellable 10.1
homomorphism of modules 4.1
left derived functors S5
homomorphism of rings 2.2
left exact functor 10.4
Hopkins-Levitzki theorem 9.8
left ideal 2.2
hypersurface 8.2
left resolution S4
ideal 2.2, 8.1 left-Noetherian ring 9.8
ideal class group 7.8 left-quasiregular element 9.7
idempotent linear transformation 9.5 left-semisimple ring 9.6
image 2.3, 4.1 length of a module 7.5
indecomposable module 9.6 lifting of a map 4.3, 10.2
index 1.3 linearly indepdent set 4.3
inductive limit 10.9 local ring 2.4, 7.9, 8.5
initial object 10.1 localization 2.8, 8.5
injection (inclusion) 4.7
long division 6.4
injective hull 10.7
long exact homology sequence S3
injective modules 10.6
injective resolution S4 Maschke™s theorem 9.6
inner automorphism 1.4, 5.7 matrices 2.1, 4.4
integral basis 7.2, 7.4 maximal ideal 2.4, 8.3
integral closure 7.1 maximal submodule 9.7
integral domain 2.1 metric 7.9
integral extensions 7.1 minimal generating set 9.8
integral ideal 7.6 minimal left ideal 9.3
integrally closed 7.1 minimal polynomial 3.1
invariant factors 4.5 minimal prime ideal 8.4
inverse limit 10.9 modding out 5.7
inverse system 10.9 modular law 4.1
inversions 1.2 module 4.1
irreducible element 2.6 modules over a principal ideal domain
irreducible ideal 8.6 4.6, 10.5
irreducible polynomial 2.9 monic 10.1
irreducible variety 8.1 monoid 1.1
isomorphic groups 1.1 monomorphism 1.3
isomorphism 1.3 morphism 10.1
isomorphism extension theorem 3.2
Nakayama™s lemma 9.8
isomorphism theorems for groups 1.4
natural action 5.3, 6.3
isomorphism theorems for modules 4.2
isomorphism theorems for rings 2.3 natural map 1.3
natural projection 4.7, 7.5, 9.2, 9.4, 9.5, principal ideal domain 2.6
10.5 product 10.2
natural transformation 10.3 product of an ideal and a module 9.3
naturality S3 product of ideals 2.3, 8.5, 7.6
Newton™s identities A6 projection 4.7, 9.2, 9.4, 9.5, 10.5
nil ideal 9.7 projection functor 10.3
nilpotent element 8.6, 9.7 projective basis lemma 10.5
nilpotent group 5.7 projective limit 10.9
nilpotent ideal 9.7 projective modules 9.8, 10.5
nilradical 8.6 projective resolution S4
Noetherian modules 4.6, 7.5 proper ideal 2.2
Noetherian rings 4.6, 7.5 Prufer group A10
nonarchimedian absolute value 7.9 pullback 10.6
noncommuting indeterminates 9.8 purely inseparable 3.4
nontrivial ideal 2.2 pushout 10.6,3
norm, 7.1, 7.3
quadratic extensions 6.3, 7.2
normal closure, 3.5
quasicyclic group A10
normal extension 3.5
quasi-regular element 9.7
normal series 5.6
quaternion group 2.1
normal Sylow p-subgroup 5.5
quaternions 2.1
normalizer 5.2
quotient ¬eld 2.8
Nullstellensatz 8.3, 8.4
quotient group 1.3
number ¬eld 7.1
quotient ring 2.2
objects 10.1
R-homomorphism on R 9.4
opposite category 10.1
Rabinowitsch trick 8.4
opposite ring 4.4
radical extension 6.8
orbit 5.2
radical of an ideal 8.3
orbit-counting theorem 5.3
rank of a free module 4.4
orbit-stabilizer theorem 5.2
rational integer 7.2
order 1.1
rational root test 2.9
order ideal 4.2
rationals mod 1 1.1 (Problem 7), 10.4,
orthogonal idempotents 9.6
10.6, A10
Ostrowski™s theorem 7.9
re¬nement 5.6
p-adic absolute value 7.9 regular action 5.1, 5.2
p-adic integers 7.9 regular n-gon 6.8
p-adic numbers 7.9 regular representation 9.5
p-adic valuation 7.9 relatively prime ideals 2.3
p-group 5.4 remainder theorem 2.5
perfect ¬eld 3.4 representation 9.5
permutation 1.2 residue ¬eld 9.8
permutation group 1.2 resolvent cubic A6
permutation module 9.5 restriction of scalars 8.7, 10.8
polynomial rings 2.1, 2.5 right adjoint 10.7
polynomials over a ¬eld 3.1 right cancellable 10.1
power sums, A6 right derived functors S5
preordered set 10.2 right exact functor 10.4
primary component A10 right ideal 2.2
primary decomposition 8.6 right resolution S4
primary ideal 8.6 right-Noetherian ring 9.8
prime element 2.6hbn right-quasiregular element 9.7
prime ideal 2.4 right-semisimple ring 9.6
primitive element, theorem of 3.5, 6.6 ring 2.1
primitive polynomial 2.9 ring of fractions 2.8, 8.5
torsion subgroup A10
Schreier re¬nement theorem 5.6
torsion submodule 4.6
Schur™s lemma 9.2
torsion-free module 4.6
semidirect product 5.8
trace 7.3
semigroup 1.1
transcendence basis 6.9
semisimple module 9.1
transcendence degree 6.9
semisimple ring 9.3
transcendental element 3.1
separable element 3.4
transcendental extension 6.9
separable extension 3.4
transcendental number 3.3
separable polynomial 3.4
transitive action 5.2
series for a module 7.5
transitive subgroup of Sn 6.3
simple group 5.1, 5.5
transitivity of algebraic extensions 3.3
simple left ideal 9.3
transitivity of separable extensions 3.4
simple module 7.5, 9.1, 9.2
transitivity of trace and norm 7.3
simple ring 9.3, 9.5
transposition 1.2
simplicity of the alternating group 5.6
trisecting the angle 6.8
simultaneous basis theorem 4.6
trivial absolute value 7.9
skew ¬eld 2.1
trivial action 5.1, 5.2
Smith normal form 4.5
twisted cubic 8.3
snake diagram S2
two-sided ideal 2.2
snake lemma S2
solvability by radicals 6.8 ultrametric inequality 7.9
solvable group 5.7 underlying functor 10.3
spanning set 4.3, 6.9 unimodular matrix 7.4
special linear group 1.3 unique factorization domain 2.6
split exact sequence 4.7, 5.8 unique factorization of ideals 7.7
splitting ¬eld 3.2 unit 2.1
squaring the circle 6.8 universal mapping property (UMP)
standard representation of a p-adic 8.7, 10.2
integer 7.9 upper central series 5.7
Steinitz exchange 6.9
valuation 7.9
Stickelberger™s theorem 7.4
valuation ideal 7.9
subcategory 10.3
valuation ring 7.9
subgroup 1.1
Vandermonde determinant A6, 7.4
submodule 4.1
vector space as an F [X]-module 4.1
subnormal series 5.6
Von Dyck™s theorem 5.8
subring 2.1
sum of ideals 2.2
weak Nullstellensatz 8.3
sum of modules 4.3
Wedderburn structure theorem 9.5
Sylow p-subgroup 5.4
Wedderburn-Artin theorem 9.4
Sylow theorems 5.4
symmetric group 1.2, 6.6 Zariski topology 8.1
symmetric polynomial 6.1 Zassenhaus lemma 5.6
zero object 10.1
tensor functors 10.3
tensor product of matrices 8.7
tensor product of module homo-
morphisms 8.7
tensor product of modules 8.7
terminal object 10.1
Tor S5
torsion abelian group 8.7 (Problem 3),
10.2
torsion element 4.6
torsion module 4.6

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