Download 9 Solutions for Section 2

9
Solutions for Section 2
Exercise 2.1 Show that isomorphism is an equivalence relation on rings. (Of
course, first you’ll need to recall what is meant by an equivalence relation.)
Solution Most of this was done in class. After noting that, for any ring R, the
identity function on R is an isomorphism from R to itself (hence reflexivity of
isomorphism), we have to show that if R is isomorphic to S, say θ : R → S is an
isomorphism then there is an isomorphism from S to R (symmetry); also that
if R is isomorphic to S, say θ does the job, and S is isomorphic to the ring T ,
say ψ : S → T is an isomorphism, then R is isomorphic to T (transitvity). For
the symmetry, show that the inverse, θ−1 : S → R, of θ, defined by θ−1 (s) = r
iff s = θ(r) is an isomorphism. For transitivity, show that the composition,
ψθ : R → T , defined by (ψθ)(r) = ψ(θ(r)) is an isomorphism.
1 0
+
Exercise 2.5 Show that the ring of 2 × 2 matrices of the form a
0 1
0 i
i 0
0 −1
with a, b, c, d ∈ R (and i a square
+d
+c
b
i 0
0 −i
1 0
root of −1) is isomorphic to the ring of quaternions.
Proof: Let R denote the ring of matrices
of the above form.
Define the map
a + ci −b + di
. Showing that this
θ : H → R by θ(a + bi + cj + dk) =
b + di a − ci
is 1-1 and onto is just as in Example 2.4 (which was done in lectures). Clearly
θ(1) is the identity matrix.
Take a + bi + cj + dk, a′ + b′ i + c′ j + d′ k ∈ H. To check that θ preserves
addition we have to show that θ (a + bi + cj + dk) + (a′ + b′ i + c′ j + d′ k =
θ(a + bi + cj + dk) + θ(a′ + b′ i + c′ j + d′ k), which is, when you begin to write it
down, obvious.
To check
multiplication we have to show that θ (a + bi + cj + dk) × (a′ + b′ i +
c′ j+d′ k = θ(a+bi+cj+dk)×θ(a′ +b′ i+c′ j+d′ k). The LHS equals θ (aa′ −bb′ −
′
′
′
cc
+a′ b+cd′ −c′ d)i+(ac′ +a′ c+b′ d−bd′ )j +(ad′ +a′ d+bc′ −b′ c)k =
−dd′ )+(ab
′
(aa − bb − cc′ − dd′ ) + (ac′ + a′ c + b′ d − bd′ )i −(ab′ + a′ b + cd′ − c′ d) + (ad′ + a′ d + bc′ − b′ c)i
.
(ab′ + a′ b + cd′ − c′ d) + (ad′ + a′ d+ bc′ − b′ c)i (aa′ − bb′ − cc′ − dd′ ) − (ac′ + a′ c + b′ d − bd′ )i
a + ci −b + di
a + ci −b + di
which, if you mul×
The RHS equals
b + di a − ci
b + di a − ci
tiply it out, gives exactly the matrix above.
Exercise 2.7 Show that all rings with three elements are isomorphic.
Solution Suppose that R is a ring with three elements. Two of these elements
must be 0, 1; denote the third by a. It will be enough to show that there is no
choice in drawing up the addition and multiplication tables of R, for then, if S
is another ring with three elements, 0, 1, b say, then the map taking 0 to 0, 1 to
1 and a to b must preserve addition and multiplication (since there is no choice
over these), so must be an isomorphism.
Addition: what is 1 + 1? It can’t be 1 because from 1 + 1 = 1 we’d get the
contradiction 1 = 0. Could 1 + 1 = 0? No, because by the kind of cancelling
argument just used, a + 1 cannot equal a or 1 hence a + 1 = 0, but then we’d
have 1 + 1 = a + 1 so a = 1 contradiction. So we must have 1 + 1 = a, and we’ve
also noted that a + 1 = 0, that is 1 + 1 + 1 = 0 hence the ring is of characteristic
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3 (actually you could also have got to this point by using Lagrange’s Theorem).
It follows that a + a = 2 + 2 = 1. So now we have the complete addition table.
Multiplication: since 0 times anything is 0 and 1 times anything is that
thing, the only question to be answered is what is a × a? But a = 2 and the
ring has characteristic 3, so a × a = 1.
There was no choice anywhere, so we have proved that, up to isomorphism,
there is just one ring with three elements (familiar as Z3 , the integers modulo
3).
Example 2.8 There are rings with four elements which are not isomorphic. In
fact, the tables in Examples 1.3 do all give different rings (indeed, there is even
one more).
Comment The fourth ring is Z2 × Z2 , the product of Z2 with itself. You can
see from the 1s on the diagonal of the multiplication table of this ring that it is
not isomorphic to any of the other three.
Exercise 2.10 Give an example to show that if θ : R −→ S is a homomorphism
and if r ∈ R is such that θ(r) is invertible in S, it need not be the case that r
is invertible in R.
Solution For instance, take the canonical homomorphism Z −→ Z5 which takes
a to [a]5 . Then 2 ∈ Z is not invertible but its image, [2]5 is (its inverse is [3]5 ).
Exercise 2.12 Prove that there is no homomorphism from Zn to Zl if l is not
an integer factor of n. [Hint, make use of the notion of characteristic]
Proof: Suppose there is such a homomorphism θ. Since [1]n n = 0 we must
have θ([1]n n) = 0, that is, [1]l n = 0. But the characteristic of Zl is l so, see
Lemma 1.5(2), it must be that l divides n, as required.
Exercise 2.15 Show that if θ : R −→ S and β : S −→ T are homomorphisms
such that the composition βθ : R −→ T is an embedding then it need not be
the case that β is an embedding.
Solution There are many possible examples. For instance take θ : Q −→ Q[X]
to be the embedding which takes a rational r ∈ Q to that same rational, regarded
as a constant polynomial. And take β : Q[X] −→ Q to be evaluation (of a
polynomial) at, say 0. The composition simply returns r ∈ Q to r, so certainly
is an embedding. But β is not: it takes any polynomial, such as X, with zero
constant term to 0.
√
√
Exercise
√ 2.20 Show that the map from Z[ 2] to itself given by sending a+b 2
to a −
√ b 2 is an automorphism. Show that there are no more automorphisms
of Z[ 2] apart from the identity map.
Solution First we show
thismap, θ say, is a homomorphism.
Clearly
√
√ that √
θ(1) = 1. Also θ (a+b 2)(c+d 2) = θ (ac+2bd)+(ad+bc) 2) = (ac+2bd)−
√
√
√
√
√
(ad + bc) 2. On the other
√ hand θ(a + b 2)θ(c + d 2) = (a − b 2)(c − d 2) =
(ac + 2bd) − (ad + bc) 2. So θ preserves multiplication and it is even easier to
show that it preserves addition, hence is a homomorphism.
√
Certainly θ√is onto: given a typical element a + b 2 of the ring, this is the
image of a − b 2 under θ. Also θ is 1-1: directly or since θ2 is the identity map
(exercise: make sure you see why this observation is enough).
√
To see that there are no more automorphisms: suppose that β : Z[ 2] →
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√
√
Z[ 2] is an automorphism of Z[ 2]. Since β(1) = 1 it must also be that, for
any positive integer n = 1 + · · · + 1 (n 1s) we have β(n) = β(1 + · · · + 1) =
β(1) + · · · + β(1) = 1 + · · · + 1 = n. Since β(−n) = −β(n) and since β(0) = 0
we deduce that√β(n) = n for √
every n ∈ Z.
√
So β(a + b √
2) = a + bβ(
2) and it remains to
determine what β( 2) can
√
√
2
2
be. We have
√ β( 2) = β(( 2) ) = β(2) = 2 so β( 2) must be a square root of
2. Since Z[ 2] is contained in√the field of complex√numbers
√ and since the only
2
it
follows
that
β(
2)
=
2 - in which case β
square roots of 2 in C are ±
√
√
is the identity map - or β( 2) = − 2 - in which case√β is the map θ above.
Therefore there are just these two automorphisms of Z[ 2].
Exercise 2.24 Give an example of a ring R and a subset H of R which contains
0 and is closed under addition and multiplication but which is not an ideal of
R.
Solution For instance, take R = Z and H = N.
Exercise 2.27 Let
R = M2 (Z) be the ring of 2 × 2 matrices with integer
1 0
. Compute the right ideal generated by a, the left
entries. Let a =
0 0
ideal generated by a and the (two-sided) ideal generated by a.
a b
a b
1 0
so the right ideal generated by
=
Solution
0 0
c d
0 0
a b
1 0
.
is the set of all matrices of the form
a=
0 0
0 0
a 0
1 0
a b
so the left ideal generated by a
=
Also
0 0
c d
c 0 a 0
.
consists of all matrices of the form
c 0
The elements
of the
generated
by a will be sums of terms
ideal
(two-sided)
a′ b′
1 0
a b
of the form
. So it is necessary to do a bit of ex0 0
c d
c′ d′
0 1
0 0
1 0
=
perimentation to see what can be got in this way. You might find:
0 0
1 0
0 0
0 0
1 0
0 0
are in this ideal, so is
and
. So, since both
0 1
0 0
0 1
1 0
. But any ideal which contains the identity element must equal the
0 1
1 0
is
whole ring. So we conclude that the two-sided ideal generated by
0 0
all of M2 (Z).
Exercise 2.29 Show that if R is any ring then R is a division ring iff the only
right ideals of R are {0} and R (and equivalently for left ideals). It is the case
(and showing this is a substantial exercise) that the first Weyl algebra, defined at
Example 1.31, is a (non-commutative) ring which has no ideals except {0} and
R yet which is not a division ring. So the analogue of 2.28 for non-commutative
rings must refer to one-sided ideals: the condition on two-sided ideals is not
enough to give a division ring.
Proof: Suppose that R is a division ring and let I be any non-zero ideal of R.
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Choose a ∈ I with a 6= 0. Since every non-zero element of R is invertible we have
1 = aa−1 ∈ I and hence I = R (for then, if r ∈ R we have r = 1r = a(a−1 r ∈ I).
Conversely if there are no right ideals except {0} and R, take a ∈ R with
a 6= 0. Certainly the right ideal, aR, generated by a is non-zero (it contains a)
so, by assumption it must equal R. In particular 1 belongs to this ideal, so there
is r ∈ R with ar = 1. Hence a has a right inverse. Of course the same applies
to r (since certainly r 6= 0), say rs = 1. Then we have a = a1 = ars = 1s = s.
Thus r is also a left inverse for a and hence every non-zero element of R is right
and left invertible, that is, R is a division ring.
(For the comment, Exercise 2.12 on the “More Examples” sheet goes more
than half-way towards suggesting a proof of the fact that the first Weyl algebra
has no two-sided ideals apart from {0} and the whole ring.)
Exercise 2.30 Show that if A is a subset of a ring R then hAi is the smallest
ideal of R which contains A.
Proof: Any ideal containing A must be closed under right and left multiplication and under addition, hence every element of the form r1 a1 s1 + . . . rn an sn
with the ai ∈ A and the ri , si ∈ R must be in every ideal containing A. On the
other hand, it is pretty clear that the set of such elements is closed under right
and left multiplication and under addition, hence forms an ideal. Thus the set
of these elements is indeed the smallest ideal containing A.
Example/Exercise 2.31 Let R = Z[X] and let I = h2, Xi be the ideal of Z[X]
generated by {2, X}. Prove that I is not a principal ideal. [Hint: suppose, for a
contradiction, that it is, choose a generator, ...]
Proof: Suppose, for a contradiction, that this ideal were principal, say is generated by the polynomial p. Since 2 ∈ hpi it must be that 2 = pq for some
polynomial q. Comparing degrees it follows that p must have degree 0, that is,
p must be a constant polynomial, that is, an integer, k say. But also X must be
a multiple, X = kq ′ of p = k. Comparing leading coefficients, we deduce that k
divides 1, hence that k = ±1. Whichever, it follows that the ideal generated by
p contains 1 (hence is the whole of Z[X]). But, we claim, the ideal generated
by {2, X} does not contain 1 for, if it did, we would have 2q + Xq ′ = 1 for some
polynomials q, q ′ . Substitute X = 0 on both sides to deduce 2q(2) = 1, hence
that 1 is an integer multiple of 2, which it is not, the required contradiction.
Exercise 2.32 The ideal h4, 6i of Z generated by 4 and 6 together is principal
- find a generator for it. Generalise this.
Solution 2 is a generator: certainly 2 = 6 − 4 is in the ideal h4, 6i. Conversely,
every element of that ideal must be divisible by 2 (since each of the generators
is), that is, lies in the ideal generated by 2.
More generally, ha, bi = hgcd(a, b)i. For the gcd has the form ak +bl for some
integers k, l, hence is in the ideal generated by a and b, thus hgcd(a, b)i ⊆ ha, bi
and, conversely, each of a and b is a multiple of gcd(a, b), hence is in the ideal
generated by gcd(a, b), hence ha, bi ⊆ gcd(a, b) and so these ideals are equal.
Exercise 2.36 Let R = Z and let I = h3i. What is I 2 ? What is I n ? Let
J = h12i. What is IJ?
Solution I 2 = h9i; I n = h3n i; IJ = h36i.
Exercise 2.37 Let I, J be ideals of a ring R. Show that IJ ⊆ I ∩ J. Give an
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example to show that this inclusion may be proper.
Solution The typical element of IJ is a sum of terms of the form ab with a ∈ I
and b ∈ J. Note that any such element ab is both in I and in J, hence is in
I ∩ J. Hence any sum of such elements is in I ∩ J. That is, IJ ⊆ I ∩ J.
Look to the previous exercise for an example: take I = h3i and J = h12i,
then IJ = h36i, whereas I ∩ J = h12i (note that h12i ⊆ h3i).
Exercise 2.38 Let R = Z. Show that h2i + h5i = Z. Compute h2i ∩ h5i and the
product h2ih5i, writing each of these as a principal ideal, that is, in the form hni
for some integer n. Now replace 2 and 5 by arbitrary integers a and b. What
are hai + hbi, hai ∩ hbi and haihbi? [Hint: unless you’ve guessed the answers
this might be difficult to answer, let alone prove, so try with some other pairs
of integers in place of 2 and 5. In other words, explore until you see what’s
happening.] Compute (h2i : h5i) and (h6i : h8i). What about the general case
(hai : hbi)?
Solution Note that 1 = 5 − 2 × 2 ∈ h2i + h5i = h2, 5i. Both h2i ∩ h5i and h2ih5i
equal h10i.
More generally ha, bi = hgcd(a, b), hai ∩ hbi = hhcf(a, b)i and haihbi = habi.
These can be proved by, for example, considering the prime factorisations of a
and b.
Finally (h2i : h5i) = h2i and (h6i : h8i) = h3i and, in general, (hai : hbi) =
a
a
i (since gcd(a,b)
is the smallest integer which turns b into a multiple of
h gcd(a,b)
a).
Exercise 2.39 Find all ring homomorphisms from Z × Z to itself.
Solution Set e1 = (1, 0) and e2 = (0, 1). If θ is a homomorphism from Z × Z to
itself then, once we know the values of θ on these idempotents, we know θ (for
a typical element of Z × Z is (m, n) = me1 + ne2 so θ(m, n) = mθ(e1 ) + nθ(e2 )).
Any homomorphism preserves idempotents and the only idempotents of Z × Z
are 0, 1 = (1, 1), e1 and e2 , so each of θ(e1 ), θ(e2 ) has to be one of these four.
But since e1 + e2 = 1 we must have θ(e1 ) + θ(e2 ) = 1. So let’s look at the
possibilities.
If θ(e1 ) = e1 then θ(e2 ) = 1 − e1 = e2 so this is the identity map.
If θ(e1 ) = e2 then θ(e − 2) = 1 − e2 = e1 so this is the map which switches
the two copies of Z.
If θ(e1 ) = 1 then θ(e2 ) = 1 − 1 = 0 so this is the map which takes (m, n) to
(m, m) and it’s easy to check that this is indeed a homomorphism.
If θ(e1 ) = 0 then θ(e2 ) = 1 − 0 = 1 so this is the homomorphism (symmetric
to the previous one) which takes (m, n) to (n, n).
√
a 2b
Exercise 2.40 Let R = Z[ 2] and let S =
: a, b ∈ Z . Define the
b a
√
a 2b
. Prove that θ is an isomorphism.
map θ : R −→ S by θ(a + b 2) =
b a
√
Solution First we show that θ is a homomorphism.
c +
any
a + b 2,
So choose
√
√
√
c 2d
a 2b
d 2 ∈ R. Then θ(a + b 2) + θ(c + d 2) =
=
+
d c
b a
√
√
√
a + c 2b + 2d
and θ((a + b 2) + (c + d 2)) = θ((a + c) + (b + d) 2) =
b+d
a+c
52
√
√
√
√
a + c 2(b + d)
. Thus θ(a + b 2) + θ(c + d 2) = θ((a + b 2) + (c + d 2)),
b+d
a+c
√
√
so θ does preserve addition.
Similarly compute
√
√ θ((a + b 2)
√ × (c + d 2)) =
θ((ac + 2bd) +√(ad + bc) 2) = · · · = θ(a + b 2) × θ(c + d 2). Furthermore
θ(1) = θ(1 + 0 b) is, by definition of θ, the identity matrix. All this shows that
θ is a ring homomorphism.
Also θ is onto (=surjective) since
√the matrix in S with parameters a, b√is the
image under θ of the element a + b 2 in R.Andθ is 1-1 since
if θ(a + b 2) =
√
c 2d
a 2b
so (by definition of
=
θ(c+d 2) then, by definition of θ,
d c
b a
√
√
what it means for matrices to be equal) a = c and b = d, hence a+b 2 = c+d 2.
(You might also be careful and note that θ is well-defined
since each member of
√
R can be written in only one way in the form a + b 2.)
Exercise 2.41 Prove that Z6 ≃ Z2 × Z3 . Is Z10 ≃ Z2 × Z5 ? Is Z8 ≃ Z2 × Z4 ?
Justify your answers.
Solution To show that the rings are isomorphic we produce a map from one to
the other and show that it is an isomorphism. So you have a choice, define a
map from Z6 to Z2 × Z3 or define a map from Z2 × Z3 to Z6 . Let’s do the first
(at the end I give a map in the other direction).
Define the map θ from Z6 to Z2 × Z3 by θ(a) = (a2 , a3 ) for a ∈ Z6 : here
by an I mean the reduction of a modulo n. (Thus θ(0) = (0, 0), θ(1) = (1, 1),
θ(2) = (0, 2), θ(3) = (1, 0), θ(4) = (0, 1), θ(5) = (1, 2) is a complete definition
of θ.) Each of the reduction modulo n maps from Z6 to Zn (n = 2, 3) is
a homomorphism (by the notes) so it follows directly (I won’t write out the
details but you can) that this map θ is a homomorphism. It’s clear from the
above explicit list that θ is a bijection, hence an isomorphism. However, that
last bit is not a good proof because something general is going in and the fact
that we can see it’s a bijection directly (because the numbers involved are small)
hides this. So let’s see what the general point is, in the next paragraph.
Suppose we start with Zm and we have two factors, l, n of m. Then we can
define a map θ : Zm → Zl × Zn by θ(a) = (al , an ). Because both l and n divide
m, this map is well-defined and it’s a homomorphism for the same general reason
as in the previous paragraph. Let’s see if it’s injective. So suppose θ(a) = θ(b),
that is, (al , an ) = (bl , bn ), that is, a ≡ b mod l and a ≡ b mod n (here I’m
reading a and b as being in Z). Now, it’s a fact from basic number theory (not
so difficult to prove) that what follows from this is that a ≡ b mod lcm{l, n}.
So in order to conclude that a = b, and hence that θ is injective, we need to
have that m is the least common multiple of l and n. What about surjectivity?
In order that θ be surjective, for each pair (c, d) ∈ Zl × Zn there should be some
integer a (modulo m) such that a ≡ c mod n and a ≡ d mod n. A result from
number theory, called the Chinese Remainder Theorem says that this is possible
iff l and n are coprime (that is, have no common factor greater than 1). So the
conclusion, in the general case, is that the “natural” homomorphism from Zm
to Zl × Zn is an isomorphism if l and n are coprime and ln = m (note that,
since they are coprime, ln is the lowest common multiple of l and n).
[In the original case a map η from Z2 × Z3 to Z6 which does the job is given
by η(b, c) = 3b + 2c mod 6. You can check that this is a homomorphism and is
a bijection (alternatively you can check that it is the inverse of θ.]
By the general considerations a couple of paragraphs above, we do have
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Z10 ≃ Z2 ×Z5 but we don’t have Z8 ≃ Z2 ×Z4 (because 2 and 4 aren’t relatively
prime). Well, for the latter, that’s not really a proof because, although the
“natural” map is not an isomorphism, there could be some other map which is.
But note that the characteristic of Z8 is 8, whereas the characteristic of Z2 × Z4
is 4, so they can’t be isomorphic rings.
Exercise 2.42 Prove that if θ : R −→ S is a surjective homomorphism of rings
and R is commutative then S must be commutative. Give an example to show
that the conclusion may be false if θ is not surjective.
Solution Let s, s′ ∈ S (we must show that ss′ = s′ s). Since θ is surjective
there are r, r′ ∈ R such that θ(r) = s and θ(r′ ) = s′ . Since R is commutative
rr′ = r′ r, so θ(rr′ ) = θ(r′ r). Since θ is a homomorphism, the latter equation
gives θ(r)θ(r′ ) = θ(r′ )θ(r), that is, ss′ = s′ s, as required.
For an example, choose your favourite non-commutative ring for S and take
R = Z and let θ be the map which sends n ∈ Z to n ∈ S (meaning n 1S s
added together).
(For a specific example take S = M2 (Z) so then θ is a map
n 0
.
n 7→
0 n
Exercise 2.43 Let R be the ring of all infinitely differentiable functions from
the unit interval [0, 1] to R (with pointwise addition and multiplication as the
operations). Let D : R −→ R be the map which takes a function f to its
derivative f ′ . Is D a ring homomorphism?
Solution No, although it does preserve addition (D(f + g) = D(f ) + D(g)) it
fails to preserve multiplication (D(f g) equals D(f )g + f D(g), not D(f )D(g))
(and also does not take 1 to 1: the identity element of R is the constant function
which maps everything to 1 ∈ R and this has derivative the zero function).
Exercise 2.44 Suppose that R is a commutative ring of characteristic 3. Prove
that the map θ : R −→ R defined by θ(r) = r3 is a homomorphism. Suppose
also that R has no non-zero nilpotent elements: prove that θ is injective.
Solution Check: θ preserves 1 since 13 = 1; θ preserves multiplication since
θ(r)θ(s) = r3 s3 = (rs)3 (since R is commutative) = θ(rs); θ preserves addition since θ(r + s) = (r + s)3 = r3 + 3r2 s + 3rs2 + s3 (again using that R is
commutative), = r3 + s3 (since 3 = 0 in R), = θ(r) + θ(s).
An element r ∈ R is in the kernel of θ, i.e. θ(r) = 0, iff r3 = 0 so, if there
are no non-zero nilpotent elements, then this implies r = 0, hence ker(θ) = 0,
hence θ is injective.
Exercise 2.45 Show that if I is a right ideal of a ring R and if there is an
element a ∈ I with a right inverse then I = R.
Solution Say ar = 1. Since a ∈ I and r ∈ R we have, by definition of right
ideal, that 1 ∈ I. But then, if s ∈ R we have s = 1 · s ∈ I, so every element of
R is in I, as required.
Exercise 2.46 Let R be a commutative ring and set N (R) = {r ∈ R : ∃n, rn =
0} to be the set of all nilpotent elements of R. Prove that N (R) is an ideal of
R. Compute N (R) for:
(i) R = Z12 ;
(ii) R = Z24 ;
54
a b
: a, b, c ∈ Q .
(iii) R =
0 c
Solution First, 0 clearly is in N (R). Next, suppose r ∈ N (R), say rn = 0, and
let s ∈ R. Then, since R is commutative, (rs)n = rn sn = 0, so rs ∈ N (R).
Finally, if r, s ∈ N (R), say rn = 0 and sm = 0 then consider (r + s)n+m . If this
is expanded using the binomial theorem (valid since R is commutative) then
each term has the form ai ri sn+m−i and so either i ≥ n or n + m − i ≥ m:
either way the term equals 0. So since (r + s)n+m = 0 and hence r + s ∈ N (R).
Therefore N (R) is indeed an ideal of R.
(i) You can just take the elements in turn and see whether some power is 0
(they soon cycle so this doesn’t take long): obviously 0 ∈ N (R); clearly neither
1 nor −1 = 11 is in N (R); the powers of 2 go 2, 4, 8, 4, . . . so neither it nor
−2 = 10 has 0 as a power, et cetera. You will find that 6 is nilpotent but that’s
it, so N (R) = {0, 6}.
(ii) A similar check will give N (R) = {0, 6, 12, 18}. Alternatively, if you
think about what’s happening, you might realise that the condition for a ∈ Zn
to be nilpotent is that every prime factor of n is also a factor of a (then some
high enough power of a will contain enough powers of each prime to have n as a
factor, that is to be 0 modulo n and, on the other hand, if some prime is missing
then no power of a will ever be divisible by n).
a b
you will see
(iii) Computing successive powers of a generic matrix
0 c
(and can prove by induction) that the diagonal entries are powers of a and c so
the only way these can ever be 0 is if both a and c are 0. But
with
any matrix
0 b
: b ∈ Q}.
a = 0 = c has square 0 so is nilpotent. Therefore N (R) = {
0 0
Exercise 2.47 Prove that every non-zero ideal of the ring Z[i] contains a nonzero integer.
Solution Let a + bi ∈ I be a non-zero element of an ideal I. Since a − bi ∈ R
and I is an ideal (a + bi)(a − bi) = a2 + b2 ∈ I, so a2 + b2 is a non-zero integer
in I.
Exercise 2.48 Determine whether or not each of the following statements about
ideals in the polynomial ring Q[X, Y ] is true:
(i) hXY + Y 2 , X 2 + XY + Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i;
(ii) hX 2 + XY, XY − Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i.
Solution (i) Since each of XY + Y 2 , X 2 + XY + Y 2 and X 2 + Y 2 is a “linear
combination” of X 2 , XY and Y 2 certainly hXY +Y 2 , X 2 +XY +Y 2 , X 2 +Y 2 i ⊆
hX 2 , XY, Y 2 i. To show the converse we have to produce each of X 2 , XY and
Y 2 as a “linear combination” of XY + Y 2 , X 2 + XY + Y 2 and X 2 + Y 2 .
We have: XY = (X 2 + XY + Y 2 ) − (X 2 + Y 2 );
2
Y = (XY + Y 2 ) − XY = (XY + Y 2 ) − ((X 2 + XY + Y 2 ) − (X 2 + Y 2 ));
X 2 = (X 2 +Y 2 )−Y 2 = (X 2 +Y 2 )−((XY +Y 2 )−((X 2 +XY +Y 2 )−(X 2 +Y 2 ));
X 2 = (X 2 + Y 2 ) − Y 2 ), which you can simplify if you want, but already the
question is answered: hXY + Y 2 , X 2 + XY + Y 2 , X 2 + Y 2 i = hX 2 , XY, Y 2 i.
(ii) As in the first part, we have hX 2 +XY, XY −Y 2 , X 2 +Y 2 i ⊆ hX 2 , XY, Y 2 i.
Notice that X 2 + Y 2 = (X 2 + XY ) − (XY − Y 2 ) so the generator X 2 + Y 2 is
not actually needed and hence
hX 2 + XY, XY − Y 2 , X 2 + Y 2 i = hX 2 + XY, XY − Y 2 i.
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Seeing that, you might suspect that this ideal is really smaller than hX 2 , XY, Y 2 i:
how could these three “independent” homogeneous quadratic polynomials be
generated using just two homogeneous quadratic polynomials? To prove that
they can’t we can take the idea of “independence” seriously.
Consider the Q-vectorspace consisting of all homogeneous quadratic polynomials in X and Y : {rX 2 + sXY + tY 2 : r, s, t ∈ Q}. Observe that this is
indeed a vector space and {X 2 , XY, Y 2 } is one basis (in part (i) we showed
that {XY + Y 2 , X 2 + XY + Y 2 , X 2 + Y 2 } is another basis). So this vector
space is 3-dimensional. In particular it is impossible to generate all of X 2 ,
XY and Y 2 as linear combinations of the form u(X 2 + XY ) + v(XY − Y 2 )
with u, v ∈ Q. We’re not quite finished yet, however, since the question
is whether we can generate all of X 2 , XY and Y 2 as linear combinations
of the form p · (X 2 + XY ) + q · (XY − Y 2 ) with p, q polynomials in Q[X].
But note that the degree 2 homogeneous part of any polynomial of the form
p · (X 2 + XY ) + q · (XY − Y 2 ) is just the constant term of the polynomial p
times X 2 + XY plus the constant term of q times XY − Y 2 , and that is what
we’ve covered with the linear algebra. So the conclusion is that it is impossible
to obtain all of X 2 , XY and Y 2 in the form p · (X 2 + XY ) + q · (XY − Y 2 ),
hence hX 2 + XY, XY − Y 2 , X 2 + Y 2 i =
6 hX 2 , XY, Y 2 i.
Exercise 2.49 Suppose that I, J are ideals of the commutative ring R such
that I + J = R. Prove that IJ = I ∩ J.
Solution Since I and J are (two-sided) ideals IJ is a subset of both I and
J hence IJ ⊆ I ∩ J. We have to show the converse, so let r ∈ I ∩ J. Since
this converse is not true in general we must use the additional information that
I + J = R, in particular that there is a ∈ I and b ∈ J such that a + b = 1. Then
r = r1 = ra + rb. Now, ra ∈ JI and rb ∈ IJ. Since R is commutative JI = IJ
so ra ∈ IJ, hence r ∈ IJ + IJ = IJ (this equality directly from the definition
of IJ), as required.
Exercise 2.50 Let R = RhX, Y : Y X = XY + 1i be the first Weyl algebra (as
defined in the notes). Use the defining relation Y X = XY + 1 to write each of
the following elements of R as a sum of monomials of the form X i Y j (i, j ≥ 0):
Y X 2; Y 2X 2; Y X 3; Y X 2 − X 2Y ; Y X 3 − X 3Y .
Prove that, for every n ∈ N, the ideal generated by X n is equal to R.
Solution Y X 2 = (Y X)X = (XY + 1)X = XY X + X = X(XY + 1) + X =
X 2 Y + 2X;
Y 2 X 2 = Y (Y X 2 ) = Y (X 2 Y + 2X) = (XY + 1)XY + 2(XY + 1) = XY XY +
3XY + 2 = (X(XY + 1)Y + 3XY + 2 = X 2 Y 2 + 4XY + 2;
Y X 3 = (Y X 2 )X = (X 2 Y + 2X)X = X 2 (XY + 1) + 2X 2 = X 3 Y + 3X 2 ;
Y X 2 − X 2 Y = X 2 Y + 2X − X 2 Y = 2X;
Y X 3 − X 3 Y = X 3 Y + 3X 2 − X 3 Y = 3X 2 .
The second part depends on your noting that the last two equations can
be seen as Y X n − X n Y = nX n−1 for n = 2, 3. This is generally true and is
proved by induction (the induction step is: assuming Y X n − X n Y = nX n−1 ,
we have Y X n+1 − X n+1 Y = (Y X n − X n Y )X + X n Y X − X n+1 Y = ((n −
1)X n )X + X n (XY + 1) − X n+1 Y (using the induction assumption) = (n −
1)X n+1 + X n+1 Y + X n − X n+1 Y = nX n+1 ). Next, note that Y X n − X n Y is
in the ideal generated by X n , that is, nX n−1 is in the ideal hX n i. Since n1 ∈ R
this means that X n−1 ∈ hX n i. Therefore X n−2 ∈ hX n−1 i also is true and so
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X n−2 ∈ hX n i. And so on (i.e. by induction), to conclude that 1 = X 0 ∈ hX n i.
Exercise 2.51 Let R = R[X, Y ] and let S ⊆ R2 be any subset of the real plane.
Set I(S) = {p ∈ R : ∀(r, s) ∈ S, p(r, s) = 0}. Prove that I(S) is an ideal of
R. [Note that this is the kernel of the two-dimensional version of the evaluation
map for polynomials which is defined in the printed notes.]
Solution Suppose that p, q ∈ I(S). Let (r, s) ∈ S. Then (p + q)(r, s) = p(r, s) +
q(r, s) = 0. So p + q ∈ I(S). Let f ∈ R. Then (pf )(r, s) = p(r, s)f (r, s) =
0 · f (r, s) = 0 so pf (= f p) ∈ I(S). Also the zero polynomial is in I(S). Hence
I(S) is an ideal of R.
Exercise 2.52 Prove that if θ : R −→ S is a surjective homomorphism of rings
and I is a right ideal of R then θ(I) is a right ideal of S. Give an example to
show that the conclusion may be false if θ is not surjective.
Solution Since 0 ∈ I we have 0S = θ(0R ) ∈ θ(I). Also if s, s′ ∈ θ(I), say
s = θ(r) and s′ = θ(s′ ) then s + s′ = θ(r) + θ(r′ ) = θ(r + r′ ) ∈ θ(I) since
r + r′ ∈ I. Let s ∈ θ(I), say s = θ(r) with r ∈ I. Also let t ∈ S. Since θ is
surjective there is u ∈ R with θ(u) = t. Then st = θ(r)θ(u) = θ(ru) (since θ is a
homomorphism). Since r ∈ I and I is a right ideal of R, we have ru ∈ I, hence
θ(ru) ∈ θ(I). But θ(ru) = θ(r)θ(u) = st, so st ∈ I. Thus θ(I) contains the zero
element of S, is closed under addition and is closed under multiplication by all
elements of S, so is a right ideal of S, as required.
The part of the argument that seems to need θ to be surjective is where we
took t ∈ S and could say that there is u ∈ R with θ(r) = u. That observation
can help in trying to find an example (to show failure without the surjectivity
assumption). Here’s one: take R = Z, I = R, S = M2
(Z) and θ : R → S the
n 0
. The image
map which takes an integer n to the diagonal matrix
0 n
of R consists of all diagonal matrices with equal diagonal entries but this is
not a right ideal: even though it contains the zero matrix and is closed under
addition,it is not closed under multiplication by arbitrary matrices, for instance
1 0
.
take t =
0 2
Exercise 2.53 Let a be an element of a ring R. Show that the right annihilator of a, rann(a) = {r ∈ R : ar = 0} is a right ideal of R. If also b ∈ R show
that rann(a + b) ⊇ rann(a) ∩ rann(b). Show that
can be proper.
the inclusion
2 3
.
Compute rann(a) where R = M2 (Z6 ) and a =
0 0
Show that rann(a) need not be a left ideal of R.
Solution Certainly 0 ∈ rann(a). If r, s ∈ rann(a) then ar = 0 = as so a(r+s) =
0 hence r + s ∈ rann(a). Finally, if r ∈ rann(a), so ar = 0 and if t ∈ R then
a(rt) = (ar)t = 0t = 0 so rt ∈ rann(a). Thus rann(a) is a right ideal of R.
Next, suppose r ∈ rann(a) ∩ rann(b), so ar = 0 = br. Then (a + b)r = 0 so
r ∈ rann(a + b), showing that rann(a) ∩ rann(b) ⊆ rann(a + b). To see that the
inclusion can be proper take, say b = −a: certainly rann(−a) = rann(a) so the
intersection is just rann(a), but rann(a + b) = rann(0) = R which can be larger
than rann(a) (e.g. take R = Z, a = 1).
a b
2 3
2 3
=
. We have
Now we compute the right annihilator of
c d
0 0
0 0
57
2a 2b + 3d
so this is zero iff 2a = 0 and 2b = 3d (note −3 = 3 in Z6 ).
0
0
So a = 0 or 3 and there is no restriction on c. The values of (b,
d) satisfying
2 3 is
2b = 3d are (0, 0), (0, 2), (0, 4), (3, 0), (3, 2), (3, 4). Thus rann
0 0
a b
where c can take any value, a is either 0 or 3 and
the set of matrices
c d
the pair (b, d) takes one of the six values above.
Finally, to see that rann(a) need not be a left ideal,
take (for instance),
t u
r s
0 1
0 1
=
R = M2 (Z) and let a =
. We have
0 0
0 0 t u 0 0
r s
r s
: r, s ∈ Z .
∈ rann(a) iff t = 0 = u, that is, rann(a) =
so
0
0
t u
0 1
is in the set but
But this set is not a left ideal since, for instance
0 0
0 0
0 1
0 0
is not.
=
0 1
0 0
1 0
Exercise 2.54 Let p be a prime integer and set Z(p) = m
n ∈ Q : p ∤ n}. Prove
that Z(p) is a subring of Z. Show that the only proper non-trivial ideals of Z(p)
are those generated by powers of p.
k
Solution Certainly Z(p) contains 1 = 1/1. If r = m
n and s = l are in Z(p)
then (assuming that these fractions are written in simplest form, hence with
numerator and denominator having no non-trivial factor) p ∤ n and p ∤ l. Now
k
ml−nk
r−s = m
and, since p is prime, p ∤ nl, so r − s ∈ Z(p) . Moreover,
n − l =
nl
mk
rs = nl and p ∤ nl so rs ∈ Z(p) . Thus Z(p) is a subring of Q.
To find the ideals, take any non-zero r ∈ Z(p) . Write r = pu m
n where p is
n
not a factor of m or n. Note that m
∈ Z(p) , so the ideal generated by r contains
n
the ideal generated by pu = r m
, indeed the ideal generated by r is equal to that
u
generated by p since each of these is an invertible element times the other. It
follows that the cyclic ideals of Z(p) are Z(p) > hpi > hp2 i > · · · > . . . {0}. It
follows that every ideal is cyclic (since, given two elements, one is already in the
ideal generated by the other). Therefore there are no more ideals other than
those just listed.
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