Download 8 Solutions for Section 1

8
Solutions for Section 1
The amount of detail in the solutions varies: sometimes it’s just a sketch; sometimes all the details are there. You can always ask, at the examples class or at
some other time, if you need more details.
Examples 1.3 The first ring is Z4 ; the second is Z2 [X]/hX 2 i, that is Z2 [x :
x2 = 0]; the third is a field with four elements, obtained by taking the field Z2
and adding a root of the polynomial X 2 + X + 1 (note that α2 = α + 1 and
remember that +1 = −1 modulo 2).
Exercise 1.4 Let R be any ring. Show the following.
(1) There is just one element z with the property that z + r = r for every r ∈ R.
Proof: Suppose both z and w have this property. Then z + w = w (since z
has the property); also w + z = w (since w has the property). So w = z + w =
w + z = z.
(2) There is just one element e with the property that er = r for every r ∈ R.
Proof: Do the multiplicative version of the proof above.
(3) For each element r ∈ R there is just one element r′ such that r + r′ = 0.
Proof: Given r ∈ R suppose that r′ and r′′ are such that both r + r′ = 0 and
r + r′′ = 0. Then r′ = r′ + 0 = r′ + (r + r′′ ) = (r′ + r) + r′′ = 0 + r′′ = r′′ , so
there is only one such element.
Exercise 1.6 Show that if R is a domain then char(R) = 0 or char(R) is a
prime integer.
Proof: If char(R) = n > 0 and n is not prime then factorise: n = kl with
1 < k, l < n. Then 0 = n · 1 = (k · 1)(l · 1). Since R is a domain this means that
either k · 1 = 0 or l · 1 = 0, contradicting the minimality of n.
Exercise 1.17(4) Show that Q[ω] where ω is a primitive cube root of 1 is a
field.
Proof: Showing that this subset of C is a subring is straightforward: the interesting bit is showing that every non-zero element has an inverse.
Since ω 3 = 1 every element of Q[ω] can be written in the form a + bω + cω 2
with a, b, c ∈ Q. However this representation is deceptive and suggests that Q[ω]
is 3-dimensional as a vector space over Q, with basis {1, ω, ω 2 }. That is not so:
these three elements are not linearly independent and, in fact 1 + ω + ω 2 = 0.
Thus ω 2 = −ω − 1. So Q[ω] = {a + bω : a, b ∈ Q}. Here are two different
approaches to finding the inverse of a non-zero element a + bω.
(1) (charge ahead proof) An inverse has to have the form c + dω so solve
(a + bω)(c + dω) = 1. That is, 1 = ac + (bc + ad)ω + bd(−ω − 1) = (ac − bd) +
(bc + ad − bd)ω. Since {1, ω} are linearly independent over Q we can equate
coefficients and deduce: ac − bd = 1; bc + ad − bd = 0.
If a 6= 0 then the first equation gives c = (1+bd)/a and, feeding that into the
second equation and solving for d gives d = b/(ab−a2 −b2 ) (of course we have to
check that ab − a2 − b2 6= 0: this is true since a2 + b2 − ab = 12 ((a − b)2 + a2 + b2 )).
So, after feeding this expression for d back into (1 + bd)/a and simplifying,
b−a
b
(a + bω)−1 = ab−a
2 −b2 + ab−a2 −b2 ω if a 6= 0.
If a = 0, so b 6= 0, then the equations become bd = −1 and bc − bd = 0. So
d = −b−1 and c = d, hence (bω)−1 = −b−1 − b−1 ω.
38
(2) (lazier and more elegant) Multiplication by a + bω is a linear map from
Q[ω] to itself and a + bω will have a multiplicative inverse exactly if this map
is invertible. That will happen iff the determinant of this map with respect to
a chosen basis is non-zero.
Let’s choose the basis {1, ω}. Then the matrix of multiplication
bya + bω
a −b
. The
(acting on vectors from the left) is easily computed to be
b a−b
determinant of this matrix is a2 −ab+b2 which, being equal to 12 ((a−b)2 +a2 +b2 ),
is non-zero if a + bω 6= 0. And the expression for the inverse can be obtained
by writing down the inverse matrix and reading down the first column.
√
Exercise 1.19 Show that Z[ 2] and Z[ω] (ω a primitive cube root of 1) are
rings (check that they are subrings of C). Are they domains? In each case
identify the invertible elements.
√
√
consists
Proof: Checking that,
√
√ of
√ say, Z[ 2] = {a + b 2 : a, b ∈ Z} is a subring
and that (a + b√ 2) + (c + d√2) =
noting that 1 =√
1 + 0 2 and√0 have that form √
(a + c) + (b + d) 2, √
−(a + b 2) = (−a) + (−b) 2) and (a + b 2) × (c + d 2) =
(ac+2bd)+(ad+bc) 2) all have the correct form (i.e. that the coefficients all are
integers since a, b, c, d are). Similarly for Z[ω], bearing in mind (see solution to
1.17) that the standard form of elements is a+bω because ω 2 = −1−ω (so there’s
a little more computation involved in checking closure under multiplication).
And they are domains by 1.8 and 1.14.
For the last part, since they are subrings of C every non-zero element has
an inverse in C: the question is whether the inverse is in the subring
√ or not.
So, for instance,
you
can
easily
find
that
the
inverse
of
a
a
+
b
2 6= 0√is
√
a
b
2
2
−
2
(note
that
the
bottom
line,
a
−
2b
cannot
be
0
because
2
2
2
2
2
a −2b
a −2b
√
√
a
b
and
is irrational!). So the inverse of a + b 2 is in Z[ 2] iff both a2 −2b
a2 −2b2
√2
2
is
invertible
are integers
(rather
that
just
fractions).
So,
for
instance,
3
+
2
√
√
in Z[ 2] but 5 + 3 2 is not.
Example 1.20 The ring M2 (Z) of 2 × 2 matrices with integer entries is a ring
which is not commutative (exercise: show that this is not commutative). More
generally, if R is a ring, then the ring, Mn (R), of n×n matrices with entries in R
is not commutative if n > 1 (exercise show this), even if R itself is commutative.
(The ring operations in Mn (R) are matrix addition and multiplication; the 0
and 1 are the obvious candidates.)
Solution If you haven’t found two 2×2 matrices which don’t commute then you
haven’t looked for long enough. For the general case you can take (for instance)
one matrix to be that with 1 in the (1, 2)-entry and 0s elsewhere and take the
other to be the matrix with 1 in the (2, 1)-entry and 0s elsewhere.
Exercise 1.21 Find examples of non-zero nilpotent elements in M2 (Z). Find
examples of idempotent elements in M2 (Z) apart from 0 and 1.
2
1
1
0 0
0 1
,
,
,
Solution Here are some nilpotents:
−4
−1
−1
1
0
0
0
0 0
1 0
(if these were all you found,
,
And some idempotents:
0 0
0 1
5
10
3 −3
1 0
. There
,
,
see comment in solution of 1.20),
−2 −4
2 −2
1 0
39
1
−2
.
a b
, write
are many more: to find a solution, take a generic matrix A =
c d
down the four equations which are equivalent to the matrix equation A2 = A
and find solutions (dividing into cases, e.g. b = 0 and b 6= 0).
Exercise 1.22 Prove that if R is a domain then there are no nilpotent elements
other than 0 and no idempotent elements other than 0 and 1.
Proof: If r ∈ R is nilpotent, say rn = 0 for some n ≥ 1 and n is chosen minimal
such, then r · rn−1 = 0 so, since rn−1 6= 0, r is a zero-divisor hence, since R is a
domain, r = 0.
Similarly, if r2 = r, then r(r − 1) = 0 so, since R is a domain, either r = 0
or r = 1.
Exercise 1.23 Find nilpotent elements of M3 (Z) with indices of nilpotence 2
and 3. Is there a nilpotent element of order 4? What is the highest index of
nilpotence of an element of Mn (Z)? Can you prove it?

 

0 1 0
0 0 1
Solution For the first part,  0 0 0 ,  0 0 1  will do. The answer
0 0 0
0 0 0
to the second question is “no” and the answer to the third (as you probably
would guess from trying some examples) is n. To prove the last part, first
note that there is a matrix in Mn (Z) with index of nilpotence n - for instance
the matrix A with 1s just above the main diagonal and 0s elsewhere (you can
compute its powers inductively if you want a careful proof that An−1 6= 0 and
An = 0). It remains to show that no matrix A has index of nilpotence greater
than n. I give a proof of this below: there may be others (in any case, the
remaining part is not(!) an easy question).
So take A ∈ Mn (Z) and suppose that A is nilpotent. Regard A as a matrix
in Mn (Q) (OK since Mn (Z) is a subring of Mn (Q)). The point of doing this
is that we can now use linear algebra over the field Q: we can regard A as
the matrix of a linear transformation T : Qn → Qn from n-dimensional space
over Q to itself. By linear algebra the successive powers, A, A2 , A3 . . . , of A
correspond to repeating the linear transformation, T, T 2 , T 3 . . . (i.e. the matrix
of T k is Ak ). Let Vk = im(T k ) be the image of the k-th iterate, T k , of T . Note
that Vk+1 = T k+1 (Qn ) = T k (T (Qn )) ⊆ T k (Qn ) = Vk , so these images form a
sequence V1 ⊇ V2 · · · ⊇ Vk ⊇ Vk+1 ⊇ . . . of subspaces.
Since A is nilpotent it is non-invertible, so T is non-invertible, so the dimension of its image is at most n − 1: dim(V1 ) ≤ n − 1. Also, if at any stage
(before reaching the zero subspace) we had Vk = Vk+1 then we would also have
Vk+1 = Vk+2 = . . . (because T restricted to Vk would be acting as an isomorphism). So the sequence of subspaces is strictly decreasing until it reaches 0.
That means the dimension decreases by at least 1 each time we apply T . Since
dim(V1 ) ≤ n−1 that means that, by the time we get to Vn we must have reached
the zero(-dimensional) subspace. That is, T n = 0 and hence An = 0, as wanted.
Exercise 1.24 When does the ring, Zn , of integers modulo n have
nilpotent elements? (and what are they?)
Solution The answer, which you could reach by trying out various
and thinking about what happens, is that Zn has non-zero nilpotent
iff the square of some prime divides n, in which case the nilpotent
40
non-zero
numbers
elements
elements
are exactly those [a]n where every prime that divides n also divides a (for then
some power of a will be divisible by n). (A formal proof of this is not difficult
to write out.) For instance 2 × 3 × 5 × 72 gives a nilpotent element of Zn where
n = 2 × 34 × 5 × 72 but 34 × 5 × 72 does not.
Exercise 1.25 If R is a ring and r ∈ R the centraliser of r, denoted C(r), is
the set of elements which commute with r: C(r) = {s ∈ R : rs = sr}. Show that
C(r) is a subring of R. In the ring M2 (Z) find a description for the centraliser
of a general element. Is it always the case that the centraliser of an element is
a commutative ring?
Solution First, r0 = 0 = 0r and 1r = r = r1 so both 0, 1 ∈ C(r). If s, t ∈ C(r)
then sr = rs and tr = rt so (s + t)r = sr + tr = rs + rt = r(s + t), hence
s + t ∈ C(r) and similarly st,
Thus C(r) isa subring
of R.
−s ∈ C(r).
a b
e f
of M2 (Z). Then
∈ M2 (Z) is in
Take a general element
c d
g h
a b
e f
a b
e f
of M2 (Z), that
of M2 (Z)
=
C(r) iff
c d
g h
c d
g h
is, iff ... iff bg = cf, af + bh = be + df, ce + dg = ag + ch, cf = bg. If you choose
particular values of a, b, c, d then you can go on to solve the resulting system of
equations to find C(r) explictly.
The answer to the last part is ‘no’: just because two elements commute with
r doesn’t mean to say that they commute with each other. For example, in any
ring R, if we take r = 1 then C(1) = R but R might well be non-commutative.
Exercise 1.26 Find a pair of (non-trivial) orthogonal idempotents in the ring
M2 (Z). What are the idempotent elements of M2 (Z): are there just a few or
are they a-dime-a-dozen?
(If you think of the elements of M2 (R) as linear transformations of R2 to
itself, with respect to a fixed basis, then one may see that, geometrically, an
idempotent is a projection and that “orthogonal” means just that.)
Solution
this up to get pairs
idempotent matricesof the
Follow
of orthogonal
−sin(θ)cos(θ)
cos2 (θ)
sin2 (θ)
sin(θ)cos(θ)
form
,
. For
−sin(θ)cos(θ)
sin2 (θ)
sin(θ)cos(θ)
cos2 (θ)
√ !
3
3
1
1 0
0 0
4
4√
√4
,
, θ = π/3 gives
,
instance, θ = 0 gives
3
1
0 0
0 1
− 43
4
4
1 1 1
− 12
2
2
2
θ = π/4 gives
,
.
1
1
1
− 12
2
2
2
Exercise 1.27 Let e ∈ R be idempotent. Show that 1 − e is idempotent and
that e and 1 − e are orthogonal.
Proof: (1−e)2 = 1−e−e+e2 = 1−2e+e = 1−e and e(1−e) = e−e2 = e−e = 0,
(1 − e)e = e − e2 = e − e = 0
Exercise 1.29 Let R be the set of all functions f : [0, 1] −→ R from the
unit interval to the set of reals. Make this a ring by defining + and × to be
pointwise addition and multiplication respectively, for instance f + g is defined
to be the function which takes a ∈ [0, 1] to f (a) + g(a). Check that this gives
a commutative ring and identify the 0 and 1 of this ring. (This example also
works if we restrict to functions which are continuous, alternatively functions
which are differentiable: why?)
41
−
√
3
4
3
4
!
,
Part Solution Distributivity, for instance, is: take f, g, h ∈ R; we have to show
that f (g + h) = f g + f h, so take x ∈ [0, 1], then (f (g + h))(x) = f (x)(g(x) +
h(x)) = f (x)g(x) + f (x)h(x) = (f g)(x) + (f h)(x) = (f g + f h)(x). The 0
of this ring is the constant function which takes every r ∈ [0, 1] to 0 since,
if we write f0 for this function, then for any f ∈ R we have (f + f0 )(x) =
f (x) + f0 (x) = f (x) + 0 = f (x) so f + f0 = f . The 1 of R is the constant
function which takes every x ∈ [0, 1] to 1 (similar argument with × in place
of +). The parenthetical comment follows since, for instance, the sum and the
product of any two continous functions is continuous (a result from analysis).
Exercise 1.30 Let R be the set of all functions f : R −→ R with pointwise
addition for + and take the “multiplication” × to be composition of functions,
with the convention that f g means do g then do f. Show that R is not a ring
(although R satisfies some of the axioms for a ring it does not satisfy them all,
so you should find out what goes wrong).
Part Solution The problem comes with the distributive law f (g +h) = f g +f h
which is not satisfied for general f . For instance take both g and h to be the
identity function and f to be the squaring function. Then (f (g+h))(x) = f ((g+
h)(x)) = f (g(x) + h(x)) = f (x + x) = (2x)2 = 4x2 , whereas (f g)(x) + f h(x) =
f (g(x)) + f (h(x)) = f (x) + f (x) = x2 + x2 = 2x2 which is not the function
taking x to 4x2 .
Exercise 1.31 Could we do something like Exercise 1.29 but using functions
from [0, 1] to [0, 1]?
Solution No, because the sum of functions is defined by adding their values
and this will not be possible if we restrict the values to come from [0, 1].
Exercise 1.34 Is the binary operation on Z (the set of integers) which takes
(a, b) to a − b associative? Is there an identity element for this operation?
Solution No: a − (b − c) = a − b + c whereas (a − b) − c = a − b − c which is
not, in general (take, say, a = b = 0, c = 1 in particular), equal to a − b + c.
Exercise 1.35 Is the binary operation on P (the set of positive integers) given
by (a, b) 7→ ab associative?
2
Solution No: a(b ) 6= (ab )c in general, e.g. 2(1 ) = 2 whereas (21 )2 = 4. (By
c
c
the way, the usual convention is that ab means a(b ) .)
a b
Exercise 1.36 Let SL2 (Z) denote the set of all matrices of the form
c d
with a, b, c, d ∈ Z and ad − bc = 1. Is this set a ring under matrix addition and
multiplication?
Isthis set a group under matrix multiplication? What about
a b
: a, b, c, d ∈ Z, ad − bc = −1 : is this a group under matrix
the set
c d
multiplication?
Solution The set SL2 (Z)
is not
a ring under matrix addition. For
of matrices
−1 0
1 0
have determinant 1 but their sum,
and
example, both
0 −1
0 1
the zero matrix, has determinant 0, hence is not in SL2 (Z).
On the other hand, SL2 (Z) is a group under matrix multiplication: it has
an identity element (the 2 × 2 identity matrix), the operation is associative
c
42
(quoting the fact that this is so for matrices in general), the formula det(AB) =
det(A) det(B) shows that the product of two matrices with determinant 1 also
has determinant 1. Finally, (as you should know) the inverse of a matrix with
determinant 1 (more generally, with non-zero determinant) exists and itself has
determinant 1 (by that same formula, for example).
The set of matrices with determinant −1 is not even a group since the
product of two such matrices will not have determinant −1.
Exercise 1.37 Find some subrings of M2 (Z). This exercise becomes more
interesting if we don’t insist that subrings contain the identity element so, just
for this exercise, look for subsets containing {0} which are closed under addition,
subtraction and multiplication.
Solution There are lots. Here are some ways of producing such subrings (remember our convention is that all subrings are “unital”, i.e. contain the identity
of the ring).
a b
and consider the subring it generates,
Take a single 2 × 2 matrix
c d
a b
meaning the set of all elements you can produce from
by applying
c d
addition,
with
the matrix
subtraction and multiplication. For example, starting
0 0
0 0
with n ∈ Z
it’s easy to produce all matrices of the form
n 0
1 0
and, since it’s easy to check that this set of these matrices is a subring (in the
weaker
of this exercise), these matrices do form the subring generated
definition
0 0
. You can try other, less simple, “seed matrices” and see what you
by
1 0
get (the results will not, in general, be so easy to describe explicitly).
Or you
could start with more than one matrix. For example, taking the
0 0
matrix
above together with the identity matrix gives, on adding the
1 0
n 0
1 0
, and it’s easy to see that all matrices of the form
two,
m n
1 1
can be obtained. On the other hand, the set of such matrices
iseasily checked
0 0
.
to be a subring, hence equals the subring generated by
1 0
Another way of producing subrings is to come up with sets of conditions
which are preserved under addition and multiplication. For example, the condition
thatthe (2, 1) entry be 0 gives the ring of upper triangular matrices:
a b
: a, b, d ∈ Z . Or the condition that all entries be divisible by p
0 d
pa p2 b
2
and that the (1, 2) and (2, 1) entries be divisible by p gives
:
p2 c pd
a, b, c, d ∈ Z .
Or choose a subring S (still not insisting that 1 be in S) of the ring, Z, of
integers
and take
matrices with entries in S: for instance, taking S = pZ gives
pa pb
: a, b, c, d ∈ Z .
pc pd
Exercise 1.38 Suppose that R is a ring. How many solutions in R can there
be to the equation X 2 − 1 = 0?
43
Solution In other words how many square roots of 1 can there be in a ring?
Well, there’s always at least one, namely 1 itself. In the ring Z2 there is only
one. But there can be two roots (as in Z and in any Zn with n ≥ 3). Indeed, if
the ring is a domain then from 0 = X 2 − 1 = (X − 1)(X + 1) we deduce that
either X − 1 = 0, i.e. X = 1, or X + 1 = 0, i.e. X = −1. So a domain, in
particular any field, has at most two solutions.
But this argument is not valid in general rings and, indeed, there are rings
with more than two solutions. For instance, if R is any ring of characterisic 2
then we can add a new variable x and force the relation x2 = 0: then (1 + x)2 =
1 + 2x + x2 = 1 and we can repeat this process with new variables thus adding
as many square roots of 1 as we like. (The way to “force the relation” is to
form the factor ring R[X]/hX 2 − 1i. Factor rings will be dealt with later in the
course.)
Exercise 1.39 Suppose that R is a domain and let a, b ∈ R with a 6= 0. Prove
that the equation ax = b has at most one solution in R. What if, instead of
assuming a 6= 0, we assume b 6= 0: is the conclusion still correct?
Solution Suppose that c and c′ both are solutions, so ac = b = ac′ . Then
a(c − c′ ) = 0 so, since R is a domain and a 6= 0, c = c′ .
On the other hand if we know only b 6= 0 then this argument is not valid.
Which is not to say that the conclusion is incorrect (a proof of the “conclusion”
or a counterexample to it is required). Here’s one counterexample. Take R = Z4 ,
a = 2, b = 2, then we have both 2×1 = 2 and 2×3 = 2. Therefore the conclusion
when b 6= 0 is incorrect.
√ √
Exercise
1.40
√
√ Let Q[ 2, 3] denote the smallest subring of R which contains
Q, 2√and √
3. Show
√ that this consists exactly of the real numbers of the form
a + b 2 + c 3 + d 6 where a, b, c, d ∈ Q. Choose some non-zero elements
of √
this√ring and√find √
their inverses (this is, in fact, a field). Also show that
Q[ 2, 3] = Q[ 2 + 3] (hint: think about what you actually have to do here
- it’s not much).
Solution To show the first√point √
it will be
√ enough: (a) to show that the set of
elements of the form a + b 2 +√c 3 + d√ 6 does form a ring and (b) to show
that every√ring containing
√
√ Q, 2 and 3 has to contain all elements of the
form a + b 2 + c 3 + d 6. For (a) just note that 1 is of this form and that, if
you add, subtract or multiply two elements of that form then the result can be
rearranged to that form. That (b) is true follows directly from the definition of
a ring.
For the last part, you have to show that
each set√is contained
in√the √
other
√ √
√
and,
for
that,
it
is
enough
to
show
that
2,
3
∈
Q[
2
+
3]
and
2
+
3∈
p
√
√ √
3
Q[ 2, 3]. The second point
is
obvious.
For
the
first,
we
have
(
2
+
(3))
=
p
p √
√
√
√
√
"√
11 2 + 9 3 so − 12 ( 2 + (3))3 − 11( 2 + (3) = 3 and similarly 2 can
√
√
be expressed as
polynomial in 2 + 3 with rational coefficients, i.e. it is a
√ a√
member of Q[ 2, 3].
Exercise 1.41 Let R = Z[i] where i is a square root of −1. Prove that R is a
domain. Give another proof. And another.
Solution Since Z[i] is a subring of the field C and since any subring of a domain
(in particular of a field) is a domain, Z[i] is a domain.
Or, suppose that a + bi, c + di ∈ Z[i] and suppose that (a + bi)(c + di) = 0.
44
Take modulus squared of both sides to get (a2 + b2 )(c2 + d2 ) = 0 Since these
are integers and Z is a domain, either a2 + b2 = 0 or c2 + d2 = 0 so (since these
are all real numbers) either a = b = 0 or c = d = 0, i.e. either a + bi = 0 or
c + di = 0, as required.
Or, even more direct, suppose that (a+bi)(c+di) = 0 and suppose, say, that
a + bi 6= 0 (so we must prove that c + di = 0). Multiplying out and equating
real and imaginary parts gives ac − bd = 0 and ad + bc = 0, multiply the second
equation by a to get a2 d + acb = 0, now use the first equation to replace ac by
bd and obtained (a2 + b2 )d = 0 hence, since a2 + b2 6= 0, d = 0. Feed d = 0
into the two equations to get ac = 0 and bc = 0 so, since at least one of a, b is
non-zero, also c = 0, so c + di = 0, as required.
Exercise 1.42 If R1 is a ring with characteristic m and R2 is a ring with
characteristic n what is the characteristic of R1 × R2 ?
Solution The identity element of the product ring R1 × R2 is (1, 1) (that
is, (11 , 12 ) where 1i is the identity element of Ri ). Clearly mn(11 , 12 ) =
(n(m11 ), nm(n12 )) = (0, 0) since m11 = 0 and n12 = 0. But mn might not
be the smallest number, l which makes that happen: what is needed is that
both m and n divide l, so clearly the smallest such l is the lowest common multiple of m and n. (To see what’s happening here look at the example Z4 × Z6
and note why the characteristic is 12 not 24.) That is, the characteristic of
R1 × R2 is lcm(m, n).
Exercise 1.43 Show directly that any ring with exactly three elements must
be a domain.
Solution Two of the elements must be the identity, 1, and zero 0. Let’s denote
the remaining element as a. The only way in which the ring could fail to be
a domain is if a2 = 0 so let’s suppose that and aim for a contradiction. From
a2 = 0 we obtain (a + a)2 = 4a2 = 0. Since we cannot have a + a = a, either
a + a = 1 or a + a = 0. But 12 6= 0, so a + a = 0. Next, note that 1 + a = 0
(since neither 1 + a = 1 nor 1 + a = a is possible). So we have a + a = 1 + a,
hence a = 1, contradiction.
Exercise 1.44 Let R be a ring. Show that the identity a2 − b2 = (a + b)(a − b)
is true in R iff R is commutative
Solution If R is commutative then the usual computation gives the identity.
Conversely, (a + b)(a − b) = a2 − b2 + ba − ab so, if this is equal to a2 − b2 then
ba − ab = 0, that is ba = ab. So if the identity a2 − b2 = (a + b)(a − b) is true
in R then so is the identity ba = ab, that is, R is commutative.
Exercise 1.45 Find all idempotent and all nilpotent elements in the ring Z6 ×
Z12 .
Solution Since (a, b)2 = (a2 , b2 ), an element (a, b) ∈ Z6 × Z12 is idempotent,
that is satisfies (a, b)2 = (a, b) iff both a and b are idempotent (in Z6 , respectively, Z12 ). We can just check elements in turn, and obtain that the idempotents of Z6 are 16 and 36 , and the idempotents of Z12 are 112 and 412 . Therefore
the idempotents of Z6 × Z12 are (16 , 16 ), (16 , 412 ), (46 , 112 ), (46 , 412 ).
For nilpotence, note that (a, b) is nilpotent (i.e. some power is 0) iff both a
and b are nilpotent. The only nilpotent element of Z6 is (check) 06 ) and the
nilpotent elements of Z12 are 012 and 612 , so, apart from (0, 0) the only nilpotent
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element of Z12 is (0, 6) (with square equal to 0).
Exercise 1.46 Suppose that R is a ring such that a2 = a for every a ∈ R. Show
that a + a = 0 for every a ∈ R. Also show that R is commutative.
Solution Let a ∈ R. We have a + a = (a + a)2 = a2 + a2 + a2 + a2 = a + a + a + a
so, cancelling, 0 = a + a, proving the first part.
Let a, b ∈ R. We have (a + b)2 = a + b. Expanding the left-hand side and
using that a2 = a and b2 = b then cancelling we obtain ab + ba = 0. That is
ab = −ba. But, by the first part, −ba = ba, so ab = ba, as required.
Exercise 1.47 Let V be a finite-dimensional vector space over a field K and let
R be the ring of linear transformations from V to V (so addition is pointwise
addition and multiplication is composition of transformations). Prove that if
f : V −→ V is in R then f has a left inverse iff f has a right inverse. You may
quote basic results from linear algebra in your proof.
Solution Let’s use the convention that, if f, g are linear transformations, then
f g means the result of doing g then doing f . Let n = dim(V ) be the dimension
of V and recall that the rank of a linear transformation is the dimension of its
image.
Suppose that f has a left inverse, say hf = 1 (where 1 means the identity map
from V to V ). Then rank(hf ) = rank(1) = n. Notice that rank(f ) ≥ rank(hf )
(since the dimension of the image of f is at least as large as the dimension of the
image of hf , since the latter is the image of im(f ) under the map h). Therefore
rank(f ) = n and hence (by linear algebra results) f is invertible (right as well as
left) (more precisely, f is onto and its kernel must be 0, so it is also one-to-one).
The argument is not right/left symmetric (because there is an order of doing
maps) so suppose now that f has a right inverse, say f g = 1. That means that f
is onto. But then, by the dimension/rank formula, the kernel of f must therefore
be 0 so f is invertible, as required.
Exercise 1.48 Let R be the polynomial ring Z8 [X]. Show that the polynomial
1 + 2X is invertible.
Solution (Coming to this first solution requires a hunch, luck or rather more
insight/knowledge than you might be expected to have by this point.) Computing the successive powers of 1 + 2X, we have: (1 + 2X)2 = 1 + 4X + 4X 2 ;
(1+2X)3 = 1−2X +4X 2 (remember we’re computing modulo 8); (1+2X)4 = 1.
So (1 + 2X)−1 = (1 + 2X)3 = 1 − 2X + 4X 2 (= 1 + 6X + 4X 2 if you prefer).
If there is an inverse then it has to have the form f (X) = 1 + aX + bX 2 +
3
cX +. . . (eventually the coefficients would have to be 0 since we’re dealing with
polynomials, not power series). The necessary conditions on the coefficients are
obtained from the equation (1 + 2X)f (X) = 1 by multiplying out the left-hand
side and equating coefficients. This gives: a = −2, b = −2a = 4, c = 2b = 0
and then you can stop (all 0 from there), giving f (X) = 1 − 2X + 4X 2 .
Exercise 1.49 Suppose that K is a field with q elements and that G is a group
with n elements. How many elements does the group ring KG have?
Now let K = Z3 be the field with 3 elements and let G be a cyclic group of
order 3, generated by an element a. Write down the general form of an element
of R = KG. Simplify the following elements of R (i.e. write them in “standard
form”):
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(i) (1e + 2a + a2 ) + (1e + 2a2 );
(ii) (1e + 2a + a2 ) × (1e + 2a2 );
(iii) (1e + 2a + a2 )3 .
Solution If G = {g1 , g2 , . . . , gn } then the general form of an element of KG is
P
n
n
i=1 αi gi with αi ∈ K. The αi can be chosen independently so there are q
possibilities, hence this number of elements of KG (since two elements of KG
are equal iff they have identical representations of the above form).
Note that 3 = 0 and a3 = e and use these to compute (I don’t give the
steps): (i) 2e + 2a; (ii) e + a; (iii) e.
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