Download Math 230 – 2003-04 – Assignment 2 Due

Math 230 – 2003-04 – Assignment 2
Due: November 10, 2003
Instructions: Do two of the following questions. You may do more, but only your two best answers will be counted,
except for bonus parts, all of which will be counted. All questions have the same value (bonus parts not included).
All work must have adequate explanation. You must do your own work; answers showing evidence of collaboration
will receive reduced grades or other penalties as appropriate.
Question 1. An element a of a ring R is said to be nilpotent if there is some positive integer n with an = 0, where an
denotes the product of n copies of a. (Note that n may depend on a.)
(a) List the nilpotent elements in each of the following rings: Z4 , Z7 , Z16 , Z72 .
Solution:
Z4 : 0, 2, since 01 = 0, and 22 ≡ 0 mod 4; however, 1n ≡ 1 mod 4 for all positive n, and 3n ≡ 1 or 3 for all n.
Z7 : 0 is the only nilpotent element, since each non-zero element has an inverse and so cannot be a zero-divisor.
Z16 : 0, 2, 4, 6, 8, 10, 12, 14 are all nilpotent. If a is even, then eventually some power an will contain 24 = 16 as
a factor, and so an ≡ 0 mod 16. The powers of the odd numbers in Z16 will always be odd, so they will
not be multiples of 16.
Z72 : 72 = 8 × 9 = 23 × 32 . Any number a which has both 2 and 3 as factors will have some power an which has
both 23 and 32 as factors, i.e. an is a multiple of 72 and so a is nilpotent. These conditions are necessary,
too: if a has no factor 2, then an will never be a multiple of 72 and so a is not nilpotent. Similarly,
if a has no factor of 3, then a is not nilpotent. So the list of nilpotent elements is the multiples of 6:
0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.
(b) Given that a and b are nilpotent elements of a commutative ring R, prove that a + b is also nilpotent in R.
Solution: We are given that there are m and n positive integers so that am = 0 and bn = 0 in R. Then, since R is
commutative, we may use the binomial theorem:
m+n
(a + b)
=
m+n
X
i=0
m+n
= a
m + n m+n−i i
a
b
i
m+n−1
+ (m + n)a
m + n m+n−n n
b + ··· +
a
b + · · · + bm+n
n
In each term of the sum, the sum of the exponents on a and b is m + n. If the exponent on a is less than m,
then the exponent on b will be greater than n; ifthe exponent of b is less than n, then the exponent on a will be
greater than m. In any case, in each term m+n
am+n−i bi at least one of am+n−i and bi will be zero. Therefore
i
the whole sum is zero, (a + b)m+n = 0, and so a + b is nilpotent.
(c) Give an example of a ring R and nilpotent elements a and b of R such that a + b is not nilpotent.
Solution: Of course, in light of (b) we must use a non-commutative ring for such an example, so we turn to the
most familiar such rings, i.e. rings of matrices. Here is an example in M2 (Z):
0 1
0 0
Let A =
and B =
. Then A2 = 0 and B 2 = 0, so A and B are nilpotent. However
0 0
1 0
0 1
A+B =
and the powers of this matrix C are cyclic: C 2 = I2 , C 3 = C, C 4 = I2 , ...; we never have a
1 0
0 0
positive n with C n =
.
0 0
Bonus. Given that R is a commutative ring with identity, a is an element of R with an inverse in R, and b is a nilpotent
element of R, prove that a + b has an inverse in R.
Question 2.
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(a) Suppose that θ : R → S is a ring homomorphism, where R has identity 1R . Prove that θ(1R ) is an identity for
Im(θ).
Solution: We show that for all s ∈ Im(θ), θ(1R ) · s = s · θ(1R ) = s. So let s ∈ Im(θ) be given. Then there is some
r ∈ R with θ(r) = s; and since θ is a homomorphism,
θ(1R ) · s = θ(1R ) · θ(r) = θ(1R · r) = θ(r) = s,
and similarly for s · θ(1R ).
(b) Give an example of a ring homomorphism θ : R → S, where R has identity 1R , such that θ(1R ) is not an identity
for S.
Solution: From part (a), we see that if θ is surjective (i.e. Im(θ) = S), then θ(1R ) will be an identity for Im(θ) = S.
Thus our only hope is to use a homomorphism which is not surjective. We give two examples, one using numbers,
the other using sets.
(i) Let θ : Z2 → Z6 be given by 0θ = 0 and 1θ = 3.
Then θ is a homomorphism (in fact, an injective homomorphism, though we don’t really need this fact):
(0 + 0)θ = 0θ = 0 = 0θ + 0θ; (0 + 1)θ = 1θ = 3 = 0 + 3 = 0θ + 1θ, and similarly for 1 + 0; (1 +
1)θ = 0θ = 0 = 3 + 3 = 1θ + 1θ. For multiplication, (0 · z)θ = 0θ = 0 = 0 · zθ for both z ∈ Z2 ;
(1 · 1)θ = 1θ = 3 = 3 · 3 = (1θ) · (1θ).
However, 1θ = 3 is not an identity in Z6 , since, for example, 2 · 3 = 0 6= 2.
(ii) Let R = P({0}), i.e. R = {∅, {0}} with the usual operations of symmetric difference and intersection. Let
S = P({0, 1}), i.e. S = {∅, {0}, {1}, {0, 1}}; and let θ be the natural inclusion map from R to S, i.e. for
all X ∈ R, Xθ = X. It is simple to check that θ is a homomorphism, since the operations of R are the
same as those of S.
Then {0} is the identity for R, and {0}θ = {0} in S, but the identity of S is {0, 1}.
(c) Suppose that θ : Z60 → Z6 is a homomorphism. What are the possible values of |θ−1 (1)|? (Recall that
θ−1 (1) = {x ∈ Z60 : θ(x) = 1}.)
Solution: Since every element of Z60 can be expressed as a sum of 1s, any homomorphism θ : Z60 → Z6 is completely
determined by 1θ. We consider all possible cases:
1θ = 0. In this case xθ = 0 for all x ∈ Z60 , and |θ−1 (1)| = 0. (There are no elements of Z60 with image 1.)
1θ = 1. In this case, 2θ = (1+1)θ = 1θ+1θ = 1+1 = 2, and so on until we get to 6θ = (1+5)θ = 1θ+5θ = 1+5 = 0
in Z6 . Then 7θ = (1 + 6)θ = 1θ + 6θ = 1 + 0 = 1, and in general for x ∈ Z60 , xθ = x mod6 , i.e. xθ is the
remainder when x is divided by 6.
The trickiest part here is to realize that we must show θ is well-defined, i.e. that θ really is a function. It’s
important to realize that the elements of Z60 are really equivalence classes, cosets of 60Z in Z. So we must
show that if we take two different representatives x and y for some coset, we don’t get conflicting answers
for θ. If x and y are both representatives for some coset, then x − y ∈ 60Z, i.e. x = y + 60k for some
k ∈ Z. If y = 6q + r where q, r ∈ Z and 0 ≤ r < 6, we have x = y + 60k = (6q + r) + 60 = 6(q + 10) + r.
Thus r is the remainder when each of x, y is divided by 6, and so θ is properly defined.
We next check that θ is a homomorphism from Z60 → Z6 , i.e. that for all x, y ∈ Z60 , (x + y)θ = xθ + yθ
and (xy)θ = (xθ)(yθ).
So let x, y ∈ Z60 be given. We begin by integer division: x = 6x0 + r, y = 6y 0 + s where x0 , y 0 , r, s ∈ Z with
0 ≤ r < 6 and 0 ≤ s < 6. Furthermore we will have r + s = 6t + u for some t, u ∈ Z with 0 ≤ u < 6, and
rs = 6v + w for some v, w ∈ Z with 0 ≤ w < 6.
Then (x + y)θ = (6x0 + r + 6y 0 + s)θ = (6(x0 + y 0 + t) + u)θ = u =6 r + s = xθ + yθ, (where “=6 ” denotes
equality in Z6 ), so addition works properly.
For multiplication, (xy)θ = ((6x0 + r)(6y 0 + s))θ = (6(6x0 y 0 + x0 s + ry 0 ) + rs)θ = (6(. . .) + 6v + w)θ = w;
also (xθ)(yθ) = ((6x0 + r)θ)((6y 0 + s)θ) = (r)(s) =6 w.
At this point we know we have a homomorphism properly defined from Z60 to Z6 , and for this homomorphism we find |θ−1 (1)| = |{1, 7, 13, 19, 25, 31, 37, 43, 49, 55}| = 10.
2
θ(1) = 2. If θ is a homomorphism, then θ(1) = θ(1 · 1) = θ(1) · θ(1), so in this case we would have 2 = 2 · 2 in Z6 .
This is not true, so there is no such homomorphism.
θ(1) = 3. Checking as in the previous case, we don’t find the same problem: 3 = 3 · 3 is true in Z6 . Furthermore,
when we continue the checking, we find that the map is well-defined and is a homomorphism. It can be
briefly described by: xθ = 0 for even x, 3 for odd x. In this case |θ−1 (1)| = 0.
θ(1) = 4. Then θ(2) = θ(1 + 1) = θ(1) + θ(1) = 4 + 4 =6 2, θ(3) = 4 + 2 = 0, and the cycle repeats. There is no
x ∈ Z60 with xθ = 1, so |θ−1 (1)| = 0.
θ(1) = 5. In this case we would have 5 = θ(1) = θ(1 · 1) = θ(1) · θ(1) = 5 · 5 = 1, but 5 6= 1 in Z6 , so there is no such
homomorphism.
Summary: the two possible values for |θ−1 (1)| are 0 and 10.
Bonus. Suppose that θ : R → S is a surjective ring homomorphism, that S has more than one element, and that S has
an identity. Is it true that R must have an identity? To answer this question, you should either prove that R
must have identity, or give an example of such R, S and θ where R has no identity.
√
√
Question 3. Notation: For d a given integer, Z[ d] denotes the set {a + b d : a, b ∈ Z} with the usual
√ operations of
addition and multiplication for real (or complex, as appropriate) numbers. You may assume that Z[ d] is a ring.
√
√
List all homomorphisms from Z[ 2] to Z[ √
3]. You must prove that there are no other homomorphisms. [Hint: if θ
is such a homomorphism, and θ(1) = x + y 3, what are the possible values for x and y?]
√
√
Solution: The only homomorphism
from Z[ 2] to Z[ 3] is the zero homomorphism, i.e. the homomorphism where
√
xθ = 0 for all x ∈ Z[ 2].
√
√
√
√
Proof: following the hint, suppose that θ : Z[ 2] → Z[ 3] is a homomorphism. Then θ(1) ∈ Z[ √
3], so θ(1) = x+y
√ 2 3
for some x, y ∈√Z. Since θ is a homomorphism, θ(1) = θ(1 · 1) = θ(1) · θ(1). That is, x + y 3 = (x + y 3) =
x2 + 3y 2 + 2xy 3. At this point, we prove a lemma:
√
√
Lemma. If a, b, a0 , b0 ∈ Z with a + b 3 = a0 + b0 3, then a = a0 and b = b0 .
Proof: suppose b 6= b0 ; then b−b0 6= 0, and we can solve for
√
Therefore b = b0 , and so a = a0 by cancellation.
3=
√
a0 − a
. But 3 is not rational, so this is impossible.
b − b0
Returning to the problem, we find x = x2 + 3y 2 and y = 2xy. Then y(1 − 2x) = 0; since x, y are integers, we have
y = 0. Then x = x2 , so x = 0 or 1. The case x = 0 gives the zero homomorphism, so we are left with the case x = 1,
y = 0, i.e. θ(1) = 1. For this case, it follows that θ(2) = 2.
√
√
√ √
√
√
√
We will also have
θ( 2) = x + y 3 for some x, y ∈ Z. Then θ(2) = θ( 2 · 2) = θ( 2) · θ( 2), so 2 = (x + y 3)2 =
√
x2 + 3y 2 + 2xy 3. It then follows that x2 + 3y 2 = 2 and 2xy = 0. From this we obtain x = 0 or y = 0. If x = 0,
then 3y 2 = 2; if y = 0, then x2 = 2. Both of these are impossible since x and y are integers, so the case θ(1) = 1
leads to contradiction.
Question 4. Let R be the ring P(Z) with the usual operations of symmetric difference (∆) and intersection (∩).
(a) Give an example of a subring of R which is not an ideal of R.
Solution: Here is a simple example (there are many others!): we know that any subring must contain the zero for
∆, i.e. must contain ∅. However, the subring S = {∅} will not do for our example, since it is an ideal. So we must
put something else into S. If we put a singleton X = a into S, so that S = {∅, {a}}, then S is nonempty, closed
under subtraction and product, so S is a subring; however, S is still an ideal since the product (intersection) of
any ring element Y (i.e. Y ⊆ Z) will be either ∅ or {a}. The next simplest thing to try is to use a doubleton,
for example S = {∅, {0, 1}}. We need to know that this S is a subring; we use the second subring test.
(i) S is non-empty;
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(ii) Since the addition operation is ∆, and for this operation A∆A = ∅ for all subsets A of Z, we know that the
additive inverse of any such A is A itself. So for any A and B in R, A − B = A + (−B) = A + B = A∆B.
For the elements of S, ∅ − ∅ = ∅∆∅ = ∅ ∈ S, ∅ − {0, 1} = ∅∆{0, 1} = {0, 1} ∈ S, {0, 1} − ∅ is similarly
{0, 1}, and finally {0, 1} − {0, 1} = {0, 1}∆{0, 1} = ∅ ∈ S.
(iii) The multiplication operation is ∩. For the elements of S, ∅∩∅ = ∅ ∈ S, and also ∅∩{0, 1} = ∅, {0, 1}∩∅ = ∅.
Finally {0, 1} ∩ {0, 1} = {0, 1} ∈ S.
To see that S is not an ideal, we need some s ∈ S and r ∈ R with sr ∈
/ S. Here is a simple choice: {0, 1} ∈ S,
and {0} ∈ R, but {0, 1} ∩ {0} = {0} ∈
/ S.
(b) Let I be the collection of all subsets of 2Z. Prove that I is an ideal of R.
Solution: We will use the ideal test.
(i) I is non-empty since ∅ ∈ I.
(ii) Suppose that A and B are in I, that is, A is a set of even numbers and B is a set of even numbers. Then
A − B = A∆B as in part (a). We must show that A∆B is a set of even numbers. A∆B = (A \ B) ∪ (B \ A).
A \ B is a subset of A and so contains only even numbers; similarly B \ A contains only even numbers, so
the union contains only even numbers.
(iii) Now suppose that A ∈ I and B ∈ R, so that A is a set of even numbers, while B is an arbitrary set of
integers. Since the multiplication in R is intersection, AB = A ∩ B. Since A ∩ B ⊆ A, AB contains only
even numbers and so AB ∈ I as required.
(c) Let S be the collection of all sets of odd integers. Prove that S is a subring of R.
Solution: Use the second subring test:
(i) S is non-empty since ∅ ∈ S.
(ii) If A, B, are two sets of odd integers, then A − B = A∆B is also a set of odd integers.
(iii) If A, B are two sets of odd integers, then AB = A ∩ B is also a set of odd integers.
(d) Prove that R/I ∼
= S. Give a specific isomorphism θ from R/I to S. That is, you should define (I∆X)θ for all
X ⊆ R.
Solution: We’ll give two solutions: the first will just state what the function is, and prove it’s an isomorphism, the
second will rely on the first isomorphism theorem.
(i) For any element I∆X of R/I, let (I∆X)θ be the set of odd numbers in X, i.e. X ∩ {2n + 1 : n ∈ Z}.
First we show θ is well-defined, that is, two different representatives X and Y for the same coset do not
give different answers. So suppose I∆X = I∆Y , that is, X − Y ∈ I. Since X − Y = X∆Y in this ring,
this means that X \ Y ⊆ 2Z and Y \ X ⊆ 2Z. In other words, X and Y only differ on even numbers; they
contain the same odd numbers, so X ∩ {2n + 1 : n ∈ Z} = Y ∩ {2n + 1 : n ∈ Z}.
Next, we see that θ is a homomorphism: suppose I∆X, I∆Y ∈ R/I. Then
((I∆X) + (I∆Y ))θ
= (I∆(X + Y ))θ (definition of addition of cosets)
= (X + Y ) ∩ {2n + 1 : n ∈ Z} (definition of θ)
= (X∆Y ) ∩ {2n + 1 : n ∈ Z} (definition of + in R)
= (X ∩ {2n + 1 : n ∈ Z})∆(Y ∩ {2n + 1 : n ∈ Z}
(distributivity property of ∩ over ∆)
= (I∆X)θ + (I∆Y )θ,
and
((I∆X) · (I∆Y ))θ
=
=
=
(I∆(X · Y ))θ (definition of multiplication of cosets)
(I∆(X ∩ Y ))θ (definition of multiplication in R)
(X ∩ Y ) ∩ {2n + 1 : n ∈ Z} (definition of θ)
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=
=
(X ∩ {2n + 1 : n ∈ Z}) ∩ (Y ∩ {2n + 1 : n ∈ Z})
(property of intersection)
(I∆X)θ · (I∆Y )θ.
Next, we show θ is injective. Suppose (I∆X)θ = (I∆Y )θ. Then X ∩ {2n + 1 : n ∈ Z} = Y ∩ {2n + 1 :
n ∈ Z}, that is, X and Y have the same set of odd numbers. It follows that X − Y = X∆Y consists
entirely of even numbers, i.e. X − Y ∈ I, so I∆X = I∆Y .
Finally, θ is surjective. For suppose O ∈ S, i.e. O is a set of odd numbers. Then I∆O ∈ R/I and
(I∆O)θ = O ∩ {2n + 1 : n ∈ Z} = O.
(ii) Now for the proof using theorems. The simplest way to show that R/I ∼
= S is to give a homomorphism
φ : R → S such that S = Im(φ) and I = Ker(φ).
We define such a homomorphism by Xφ = X ∩ {2n + 1 : n ∈ Z} for all X ∈ R, i.e. all X ⊆ Z. We check
that φ is a homomorphism: let X, Y be elements of R; then
(X + Y )φ = (X + Y ) ∩ {2n + 1 : n ∈ Z}
= (X∆Y ) ∩ {2n + 1 : n ∈ Z}
= (X ∩ {2n + 1 : n ∈ Z})∆(Y ∩ {2n + 1 : n ∈ Z})
= Xφ + Y φ
and
(X · Y )φ = (X · Y ) ∩ {2n + 1 : n ∈ Z}
= (X ∩ Y ) ∩ {2n + 1 : n ∈ Z}
= (X ∩ {2n + 1 : n ∈ Z}) ∩ (Y ∩ {2n + 1 : n ∈ Z})
= (Xφ)(Y φ)
as required.
Next, we show that S = Im(φ). Let O ∈ S, i.e. O is a set of odd numbers. Then O is also an element of
R, and Oφ = O ∩ {2n + 1 : n ∈ Z} = O, so O ∈ Imφ.
It remains to show that I = Ker(φ). For X ∈ R,
X ∈ Ker(φ) ⇔
⇔
⇔
⇔
Xφ = 0S
X ∩ {2n + 1 : n ∈ Z} = ∅
X ⊆ 2Z
X ∈ I.
It now follows from the first isomorphism theorem that R/I ∼
= S. We are required to give a specific
isomorphism θ, and the one obtained in the proof of the first isomorphism theorem is the natural choice.
Note that the homomorphism names θ and φ have been interchanged here:
(I∆X)θ = Xφ = X ∩ {2n + 1 : n ∈ Z}.
End
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