Download 201B Midterm 1

Math 201B Midterm 1
Edward Burkard
February 4, 2011
All modules are assumed to be left unless specified otherwise. Given a ring R, recall that an R-module M is said
to be simple if its only submodules are 0 and M .
Problem 1. Let C be a category with the zero object OC (that is, an object which is both universal and co-universal).
In particular, for any pair of objects M and N , we have a distinguished morphism 0M,N ∈ HomC (M, N ) , called
the zero morphism, which is the composition of the unique morphisms M → OC and OC → N . Let M and N be
objects in C and f ∈ HomC (M, N ) . Recall that a kernel of f is a pair (K, k) where K is an object of C and
k ∈ HomC (K, M ) such that (a) f ◦ k = 0M,N ; (b) if (K 0 , k 0 ) is another pair with the same property, then there
exists a unique h : K 0 → K such that k ◦ h = k 0 (in other words, the pair (K, k) is a universal object in a suitable
category). The dual concept is called a co-kernel of a morphism.
If (K, k) is a kernel of f : M → N , then k is a monomorphism.
If (C, c) is a cokernel of f : M → N then c is an epimorphism.
Kernel is unique up to a (necessarily unique) isomorphism.
Show that for any morphism of R-modules f : M → N , (ker f, i), where i : ker f → M is the canonical
inclusion, is a kernel of f in the above sense.
(v) Show that for any morphism of R-modules f : M → N , (N/ im f, π), where π is the canonical projection, is
a cokernel of f .
(vi) Suppose that any morphism in C has a kernel and a cokernel. What would be the appropriate definition of
im f for a morphism f in C ?
(i)
(ii)
(iii)
(iv)
Proof.
(i) Let M, N, K, L be objects of the category C . Let f ∈ HomC (M, N ) and let (K, k) be the kernel of f .
Suppose that g, j ∈ HomC (L, K) be morphisms such that k ◦ g = k ◦ j. Then since
f ◦ (k ◦ g) = (f ◦ k) ◦ g = 0K,N ◦ g = 0L,N
by the definition of the kernel of f , there is a unique morphism h : L → K such that k ◦ h = k ◦ g. By
assumption, both h = g and h = j satisfy this requirement. Thus by uniqueness we have that g = j.
Therefore k is a monomorphism.
(ii) Let M, N, C, D be objects of the category C . Let f ∈ HomC (M, N ) and let (C, c) be the cokernel of f .
Suppose that g, j ∈ HomC (C, D) be morphisms such that g ◦ c = j ◦ c. Then since
(g ◦ c) ◦ f = g ◦ (c ◦ f ) = g ◦ 0M,C = 0M,D
by the definition of the cokernel of f , there is a unique morphism d : C → D such that d ◦ c = g ◦ c.
By assumption, both d = g and d = j satisfy this requirement. Thus by uniqueness we have that g = j.
Therefore k is a epimorphism.
(iii) Let f ∈ HomC (M, N ) and let (K, k) and (L, l) be kernels of f . Then by the universality of the kernel, there
are uinique maps h1 : K → L and h2 : L → K such that k ◦ h2 = l and l ◦ h1 = k. Then
l ◦ (h1 ◦ h2 ) = (l ◦ h1 ) ◦ h2 = k ◦ h2 = l = l ◦ idL
and
k ◦ (h2 ◦ h1 ) = (k ◦ h2 ) ◦ h1 = l ◦ h1 = k = k ◦ idK .
Now since l and k are monomorphisms (by part (i)) we have that h1 ◦ h2 = idL and h2 ◦ h1 = idK . Thus h1
and h2 are unique isomorphisms, and hence the kernel is unique up to a unique isomorphism.
(iv) Let ι : ker f → M be the canonical inclusion (keep in mind that this is an injective map!). Then f ◦ ι = 0.
Let K be an R-module and k : K → M be an R-module homomorphism such that f ◦ k = 0. We are looking
1
2
for an h : K → ker f satisfying the commutative diagram:
kerO fE
EE
EEι
EE
E"
h
<M
yy
y
y
yy
yy k
K
f
/N
Since f ◦ k = 0 it must be that k(a) ∈ ker f for all a ∈ K, hence it makes sense to define h = ι−1 ◦ k. By
construction we have that ι ◦ h = k and it is clear that h is an R-module homomorphism (which is also true
by its construction). Now if η : K → ker f is another R-module homomorphism satisfying the commutative
diagram:
kerO OfE
EE
EEι
EE
E"
h η
<M
yy
y
y
yy
yy k
K
f
/N
Then ι ◦ η = ι ◦ h and since ι is injective we have for any a ∈ K, ι(η(a)) = ι(h(a)), thus η(a) = h(a) and
hence η and h agree on K, thus η = h.
∴ (ker f, ι) is the kernel of f in the categorical sense as well.
(v) Let π : N → N/ im f be the usual projection map. Then π ◦ f = 0 since the input to π will be an element
of im f . Let C be an R-module and c : N → C an R-module homomorphism such that c ◦ f = 0. Since
π is a surjective map, given x + im f ∈ N/ im f , there exists an n ∈ N such that π(n) = x + im f . Define
l : N/ im f → C by l(x + im f ) = c(n) where n is defined above. Suppose that n and n0 are two elements of
N such that π(n) = π(n0 ) = x + im f . Then n = n0 + f (m) for some m ∈ M and hence
c(n) = c(n0 + f (m)) = c(n0 ) + c(f (m)) = c(n0 ) + 0 = c(n0 ),
therefore l is independent of choice of n and hence is well defined.
By construction we have that c = l ◦ π and l is an R-module homomorphism since:
• if π(n) = x + im f , then π(rn) = rπ(x) = r(x + im f ) = rx + im f
• if π(n) = x + im f and π(m) = y + im f , then π(n + m) = π(n) + π(m) = (x + im f ) + (y + im f ) =
(x + y) + im f
• c is an R-module homomorphism
Let x + im f, y + im f ∈ N/ im f , r ∈ R, and m, n ∈ N such that π(m) = x + im f and π(n) = y + im f .
Then
l((x + im f ) + (y + im f )) = l((x + y) + im f ) = c(m + n) = c(m) + c(n) = l(x + im f ) + l(y + im f )
and
l(r(x + im f )) = l(rx + im f ) = c(rm) = rc(m) = rl(x + im f )
hence l is an R-module homomorphism.
Now suppose that ζ : N/ im f → C is another R-module homomorphism making the following diagram
commute:
M
f
N/ im f
w;
w
π ww
w
ww
ww
/N
ζ
l
HH
HH
HH
c HHH
H$ C
3
Then we have that l ◦ π = ζ ◦ π. But since π is surjective, the previous equality implies that l = ζ, hence l
is unique.
∴ (N/ im f, π) is a cokernel in the categorical sense as well.
(vi) Let f : M → N be a morphism in C . If f has both a kernel and cokernel, the appropriate definition of
im f would be the kernel of the cokernel morphism of f . For example, in the category of R-modules, the
cokernel of a morphism f : M → N is the object N/ im f and a morphism π : N → N/ im f which is the
usual projection. If we take the kernel of the map π we get the object ker π and the cannonical inclusion
ι : ker π → N . But the kernel of π is precisely those elements of im f .
Here is a more formal definition (probably not as formal as it could be made):
Definition 1 (Image of a Morphism). Suppose that C is a category in which every morphism has a kernel and
a cokernel. Let f ∈ HomC (M, N ) and let (C, c) be any cokernel of f (which is unique up to isomorphism).
The image of f is the kernel of the morphism c : N → C (which exists since every morphism in C has a
kernel), i.e. the image of f is a pair (I, i), I an object of C and i ∈ HomC (I, N ) , such that c ◦ i = 0I,C
and if (I 0 , i0 ) is another such pair there is a unique morphism h : I 0 → I such that i ◦ h = i0 . The image of a
morphism is unique up to isomorphism as both the kernel and cokernel of a map are.
Problem 2. Let R be a ring with unity, M a unitary R-module.
(i) Prove that EndR (M ) := HomR (M, M ) is a ring with respect to composition and point-wise defined addition.
(ii) Prove that EndR (M ) is isomorophic (as a ring) to the opposite ring, Rop , of R.
(iii) Suppose that M is simple. Prove that EndR (M ) is a division ring (this statement is a version of Schur’s
lemma).
(iv) Suppose that f ∈ EndR (M ) satisfies f 2 = f (that is, f is idempotent). Prove that M = ker f ⊕ im f .
Proof.
(i) Note that in this proof, I will freely use the R-module properties of M . Let 0 be the zero map from M to
itself. Let 0M be the additive identity of M . For f, g ∈ EndR (M ) define −f by (−f )(m) = −f (m) (the
additive inverse of f (m) in M ), (f + g)(m) = f (m) + g(m), and (f ◦ g)(m) = f (g(m)). Clearly EndR (M )
is not empty as 0 ∈ EndR (M ), and it is also clear that for any map f ∈ EndR (M ), (−f ) ∈ EndR (M ). Let
f, g, h ∈ EndR (M ), m, n ∈ M , and r ∈ R.
Since
(f +g)(m+n) = f (m+n)+g(m+n) = f (m)+f (n)+g(m)+g(n) = [f (m)+g(m)]+[f (n)+g(n)] = (f +g)(m)+(f +g)(n)
and
(f + g)(rm) = f (rm) + g(rm) = rf (m) + rg(m) = r(f (m) + g(m)) = r(f + g)(m),
we have that (f + g) ∈ EndR (M ), so + is a binary operation on EndR (M ). Now since
[f +(g+h)](m) = f (m)+[(g+h)(m)] = f (m)+[g(m)+h(m)] = [f (m)+g(m)]+h(m) = [(f +g)(m)]+h(m) = [(f +g)+h](m)
we also have that + is an associative binary operation on EndR (M ), so (EndR (M ), +) is a semigroup.
Now to show that (EndR (M ), +) is a group we only need to show left inverses and a left identity:
[(−f ) + f ](m) = (−f )(m) + f (m) = −f (m) + f (m) = 0M ,
and
(0 + f )(m) = 0(m) + f (m) = 0M + f (m) = f (m)
hence (EndR (M ), +) is a group, and moreover, is abelian since
(f + g)(m) = f (m) + g(m) = g(m) + f (m) = (g + f )(m).
Now to check the ring properties:
First:
(f ◦ g)(m + n) = f (g(m + n)) = f (g(m) + g(n)) = f (g(m)) + f (g(n)) = (f ◦ g)(m) + (f ◦ g)(n)
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and
(f ◦ g)(rm) = f (g(rm)) = f (rg(m)) = rf (g(m)) = r(f ◦ g)(m)
so (f ◦ g) ∈ EndR (M ). Next
[(f ◦ g) ◦ h](m) = (f ◦ g)(h(m)) = f (g[h(m)]) = f [g(h(m))] = f [(g ◦ h)(m)] = [f ◦ (g ◦ h)](m)
so ◦ is associative. Finally
[f ◦ (g + h)](m) = f [(g + h)(m)] = f (g(m) + h(m)) = f (g(m)) + f (h(m)) = (f ◦ g)(m) + (f ◦ h)(m)
and
[(f + g) ◦ h](m) = (f + g)(h(m)) = f (h(m)) + g(h(m)) = (f ◦ h)(m) + (g ◦ h)(m)
hence ◦ is left and right distributive.
∴ (EndR (M ), +, ◦) is a ring!
In fact, this ring has an identity, namely the identity map from M to itself, call it 1.
(1 ∗ f )(m) = 1(f (m)) = f (m)
and
(f ∗ 1)(m) = f (1(m)) = f (m).
(ii) Let 1 be the identity element of R. Define the map φ : EndR (R) → Rop by φ(f ) = f (1). I will use ∗ to denote
multiplication in Rop (i.e. a ∗ b = ba where the multiplication on the right is in R). Let f, g ∈ EndR (R), then
φ(f + g) = (f + g)(1) = f (1) + g(1) = φ(f ) + φ(g)
and
φ(f ◦ g) = (f ◦ g)(1) = f (g(1)) = f (g(1) · 1) = g(1)f (1) = f (1) ∗ g(1),
hence φ is a ring homomorphism.
Let f ∈ EndR (R) and suppose that φ(f ) = 0. Then φ(f ) = f (1) = 0 which implies that
f (r) = f (r · 1) = rf (1) = r · 0 = 0,
hence f ≡ 0 and φ is injective.
Now let r ∈ Rop and define fr : R → R by fr (1) = r and fr (a) = afr (1) for all a ∈ R. Let a, b ∈ R, then
since
fr (a + b) = fr [(a + b) · 1] = (a + b)fr (1) = (a + b)r = ar + br = fr (a) + fr (b)
and
fr (ab) = fr ((ab) · 1) = (ab)fr (1) = (ab)fr (1) = a(bfr (1)) = afr (b)
we have that fr is an R-module endomorphism of R. Thus φ(fr ) = fr (1) = r, hence φ is onto.
∴ φ is a ring isomorphism and EndR (R) ∼
= Rop .
(iii) Suppose that M is simple. Let f ∈ EndR (M ). Since ker f is a submodule of M it must be that ker f = M
or ker f = {0}. In the case that ker f = M , f must be the zero map. In the case that ker f = {0} we know
that f is injective. Since im f is also a submodule of M and we know that im f 6= {0}, it must be that
im f = M , hence f is an automorphism. Since the composition of two automorphisms is an automorphism
(and hence cannot be a zero map), if we have a composition f ◦ g = 0 for some f, g ∈ EndR (M ), it must be
that either f or g is the zero map. Thus EndR (M ) is a division ring!
(iv) Let f ∈ EndR (M ) be such that f 2 = f . Let x ∈ M . Then f (x) ∈ im f and f (x − f (x)) = f (x) − f (f (x)) =
f (x) − f (x) = 0. Thus x − f (x) ∈ ker f . Since x = x + (f (x) − f (x)) = (x − f (x)) + f (x), x ∈ ker f ⊕ im f .
Now let x ∈ ker f ∩ im f . Then f (x) = 0 (since x ∈ ker f ) and there is a y ∈ M such that f (y) = x (since
x ∈ im f ). So we know that f (f (y)) = f (x) = 0, but also that f (f (y)) = f (y) = x; hence x = 0. Thus
ker f ∩ im f = {0} so that by Theorem IV.1.15 of Hungerford:
M = ker f ⊕ im f.
Problem 3. Let R be a unital ring, I ⊂ R a left ideal. Given a left R-module M and m ∈ M , set
AnnR (m) = {x ∈ R | sm = 0} (that is clearly a left ideal), and let M [I] = {m ∈ M | AnnR (m) ⊃ I}.
(i) If M is a cyclic unitary module on m ∈ M , then M ∼
= R/ AnnR (m) .
(ii) HomR (R/I, M ) ∼
= M [I] as an abelian group.
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(iii) If I is a two-sided ideal, then R/I is an R − R − bimodule, M [I] is a left R-module (in fact, a submodule of
M ), and the isomorphism in part (ii) is a morphism of left R-modules.
(iv) If I is a two-sided ideal, then M [I] admits a natural R/I-module structure.
∗
(v) Assume that R has no zero divisors. Then (R/I) = 0. Is the converse true?
Proof.
(i) Recall that since M is cyclic and unitary, and R has an identity, M = {rm | r ∈ R}. Define the map
φ : R/ AnnR (m) → M by φ(r + AnnR (m) ) = rm. Let a + AnnR (m) , b + AnnR (m) ∈ R/ AnnR (m) and let
r ∈ R. Suppose that r + AnnR (m) = r0 + AnnR (m) . Then r = r0 + a for some a ∈ AnnR (m) . Hence
φ(r + AnnR (m) ) = φ((r0 + a) + AnnR (m) ) = (r0 + a)m = r0 m + am = r0 m + 0 = r0 m = φ(r0 + AnnR (m) )
so φ is well defined.
Claim. φ is an isomorphism.
φ([a+ AnnR (m) ]+[b+ AnnR (m) ]) = φ((a+b)+ AnnR (m) ) = (a+b)m = am+bm = φ(a+ AnnR (m) )+φ(b+ AnnR (m) )
φ(r(a + AnnR (m) )) = φ(ra + AnnR (m) ) = (ra)m = r(am) = rφ(a + AnnR (m) )
∴ φ is a homomorphism.
Suppose that φ(a + AnnR (m) ) = 0. Then φ(a + AnnR (m) ) = am = 0, which implies that a ∈ AnnR (m) ,
hence a + AnnR (m) = AnnR (m) . ∴ φ is injective.
Let rm ∈ M . Consider the element r + AnnR (m) ∈ R/ AnnR (m) . Since φ(r + AnnR (m) ) = rm we have
that φ is surjective (note that φ( AnnR (m) ) = 0m = 0).
∴ φ is an isomorphism.
(ii) First let’s show that M [I] is, in fact, an abelian group (in fact a subgroup of M ). It is obvious that M [I] ⊂ M .
Let m, n ∈ M [I], then i(m − n) = im − in = 0 − 0 = 0, thus (m − n) ∈ M [I]. Therefore M [I] / M and hence
is an abelian group.
Define the map η : HomR (R/I, M ) → M [I] by η(f ) = f (1 + I). (If this problem is to be nontrivial,
then 1 ∈
/ I, for otherwise I = R and HomR (R/I, M ) is just the zero map and either the action of R on M
will be trivial (r, m) 7→ 0, or M [I] = {0}.) Let i ∈ I and f ∈ HomR (R/I, M ) . Since iη(f ) = if (1 + I) =
f (i(1 + I)) = f (i + I) = f (I) = 0, η(f ) ∈ M [I], hence η is well defined. Let f, g ∈ HomR (R/I, M ) and
r ∈ R.
η(f + g) = (f + g)(1 + I) = f (1 + I) + g(1 + I) = η(f ) + η(g),
hence η is a group homomorphism. Now suppose that η(f ) = 0. Then η(f ) = f (1 + I) = 0 and f (r + I) =
f (r(1 + I)) = rf (1 + I) = r · 0 = 0, thus f is the zero map. Therefore η is injective. Let m ∈ M [I] and
define fm : R/I → M [I] by fm (1 + I) = m and extend it to all of R/I by fm (r + I) = rfm (1 + I). Since for
r + I, s + I ∈ R/I and a ∈ R we have:
fm ((r + I) + (s + I)) = fm ((r + s) + I) = (r + s)fm (1 + I) = (r + s)m = rs + rm = fm (r + I) + fm (s + I)
and
fm (a(r + I)) = fm (ar + I) = (ar)m = a(rm) = afm (r + I),
it is true that fm ∈ HomR (R/I, M ) . Then η(fm ) = fm (1 + I) = m and hence η is surjective.
∴ η is an isomorphism of abelian groups.
(iii) By definition R/I is an abelian group. Let r, s ∈ R and a + I, b + I ∈ R/I. Since I is a two-sided ideal
multiplication on the left and right are defined as follows:
r(a + I) = ra + I
(a + I)r = ar + I
since I absorbs multiplication on the left and right by R (i.e. RI ⊂ I ⊃ IR).
(left)
r[(a + I) + (b + I)] = r[(a + b) + I] = r(a + b) + I = (ra + rb) + I = (ra + I) + (rb + I) = r(a + I) + r(b + I)
(r + S)(a + I) = (r + s)a + I = (ra + sa) + I = (ra + I) + (sa + I) = r(a + I) + s(a + I)
(rs)(a + I) = (rs)a + I = r(sa) + I = r[sa + I] = r[s(a + I)]
Thus R/I is a left R-module.
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(right)
[(a + I) + (b + I)]r = [(a + b) + I]r = (a + b)r + I = (ra + rb) + I = (ar + I) + (br + I) = (a + I)r + (b + I)r
(a + I)(r + s) = a(r + s) + I = (ar + as) + I = (ar + I) + (as + I) = (a + I)r + (a + I)s
(a + I)(rs) = a(rs) + I = (ar)s + I = [ar + I]s = [(a + I)r]s
Thus R/I is a right R-module.
∴ R/I is an R-R-bimodule.
We have already shown that M [I] / M , so now we will show it is a submodule of M : Let r ∈ R and
m ∈ M [I]. Notice that for any i ∈ I, ri ∈ I since I is an ideal of R. Now the question is: is rm ∈ M [I]?
The answer is yes since for any i ∈ I:
i(rm) = (ir)m = 0
because ir ∈ I since I is a right ideal, and by definition of M [I], jm = 0 for any j ∈ I.
∴ M [I] is a submodule of M , and hence is a left R-module.
Recall the map η from part (ii):
η : HomR (R/I, M ) → M [I],
defined by η(f ) = f (1 + I). We showed that it is a group isomorphism.
HomR (R/I, M ) . Then
η(rf ) = (rf )(1 + I) = r[f (1 + I)] = rη(f ),
and hence η is a R-module isomorphism.
Now let r ∈ R and f ∈
(iv) Define the natural multiplication map R/I × M [I] → M [I] by (r + I)m = rm. The multiplication is well
defined since if r + I = r0 + I, then r = r0 + i for some i ∈ I and
rm = (r + I)m = [(r0 + i) + I]m = (r0 + i)m = r0 m + im = r0 m + 0 = r0 m.
This multiplication indeed lands in M [I] since for any i ∈ I
i[(r + I)m] = i(rm) = (ir)m = 0
because ir ∈ I since I is a right ideal and hence absorbs multiplication on the right. Now let’s check the
module properties. Let a + I, b + I ∈ R/I and m, n ∈ M [I]. Then:
(a + I)(m + n) = a(m + n) = am + an = (a + I)m + (a + I)n
[(a + I) + (b + I)]m = [(a + b) + I]m = (a + b)m = am + bm = (a + I)m + (b + I)m
and
[(a + I)(b + I)]m = (ab + I)m = (ab)m = a(bm) = (a + I)bm(a + I)[(b + I)m],
thus M [I] is a left R/I-module in the expected way.
(v) Assume that I 6= {0}. Recall from part (ii) that HomR (R/I, R) ∼
= R[I] as abelian groups. Since
(R/I)∗ := HomR (R/I, R) , this says that if R[I] is the trivial group, then so is (R/I)∗ , and hence (R/I)∗
can only be the zero module. So now we will proceed by showing that R[I] = 0. Clearly 0 ∈ R[I]. Let r ∈ R
be any nonzero element. If ir = 0 for some i ∈ I, then since R has no zero divisors, i = 0. Thus R[I] = {0},
and hence (R/I)∗ = 0.
The converse is not true. Let R = Z6 (which has unity) and let I = (1) = Z6 which is clearly a left ideal.
Then R/I = {0} and (R/I)∗ = HomR (R/I, R) = HomR (0, R) = {0}, however, Z6 has zero divisors since
2 · 3 = 6 ≡ 0 mod 6 for example.
Problem 4. Give an example of a non-free submodule of a free module.
Solution. Consider the ring Z6 . Since Z6 has identity, by the equivalence (iii) ⇐⇒ (iv) in Theorem IV.2.1 in Hungerford, Z6 is a free Z6 module. Now consider {0, 2, 4} ⊂ Z6 this is clearly a submodule of Z6 (and is isomorphic to Z3
which it will henceforth be refered to as) and is unitary since Z6 is.
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Claim. Z3 is not a free submodule of Z6 .
If it were a free submodule, it would have a nonempty basis (Theorem IV.2.1). Recall that since a basis B of any
R-module is required to be linearly independent, a consequence of that is that rxi = 0 (r ∈ R and xi ∈ B) implies
r = 0 for all xi ∈ B. Thus if B is a basis for Z3 clearly 0 ∈
/ B since r0 = 0 for any r ∈ Z6 . Since 3 · 2 = 6 ≡ 0
mod 6 and 3 · 4 = 12 ≡ 0 mod 6, we have that Z3 does not have a nonempty basis. Hence Z3 cannot be free.
Problem 5. Let n be a positive integer. Prove that nZ is a projective Z-module.
Proof. Since Z is a ring with identity, to show that nZ is projective, by Theorem IV.3.2 of Hungerford it is sufficient
to show that nZ is a free Z-module. By Theorem IV.2.1 of Hungerford, since nZ is a unitary Z-module, it is equivalent to show that nZ has a nonempty basis.
Let B = {n}. I claim that B is a basis of nZ. Since nZ ⊂ Z, if ab = 0 where a ∈ nZ and b ∈ Z then, since Z has
no zero divisors, either a = 0 or b = 0. This implies that B is linearly independent since an = 0 implies that a = 0
and n is the only element of B. By definition B spans nZ since nZ = {na | a ∈ Z}. Hence nZ has a nonempty basis
and by the above mentioned theorems, is projective.