Download Math 312 Assignment 3 Answers October 2015 0. What did you do

Math 312
Assignment 3 Answers
October 2015
0. What did you do to celebrate Galois’s 204th birthday on 25 October?
[0]
1. She is called the mother of modern ring theory. A certain class of rings is named after her.
Who is she? What class? Write down the definition.
[2]
ANSWER. Emmy Noether, Noetherian Rings.
2. Identify all units in Zn , in Z[i], in Z[x], in R[x]. Short answers.
[4]
ANSWER. U (n), {1, −1, i, −i}, ±1, all nonzero constant polynomials.
[
]
3. Prove that { ab 2b
a√ : a, b ∈ Z} is a ring under the usual operations and is isomorphic to
√
Z[ 2] := {m + n 2 : m, n ∈ Z}.
ANSWER. (a) Show that the set is nonempty, closed under subtraction and multiplication.
Details omitted. You don’t have to prove associativity, distributivity, etc. The usual matrix [2]
operations satisfy all you need.
[
]
√
√
[
]
[ a 2b ] [ c 2d ]
a+c 2(b+d)
(b) Let Φ( ab 2b
)
=
a
+
b
2.
Then
Φ(
+
)
=
Φ(
= a + c + (b + d) 2 = [4]
a
b a
d c
b+d a+c
√
√
[
]
[ c 2d ]
(a + b 2) + (c + d 2) = Φ( ab 2b
a ) + Φ( d c ),
[
]
√
√
√
[
] [ c 2d ]
ac+2bd 2(ad+bc)
Φ( ab 2b
)
=
Φ(
= ac + 2bd + (ad + bc) 2 = (a + b 2)(c + d 2) =
a
d c
ad+bc ac+2bd
[
]
[ c 2d ]
Φ( ab 2b
a ) Φ( d c ). So Φ is a ring-homomorphism. It is clear that it is surjective.
[
]
To prove injectivity, it suffices to show that the kernel is trivial. The matrix ( ab 2b
belongs to
√
√ a
the kernel if and only
if
a
+
b
2
=
0.
First
b
must
be
zero,
since
otherwise
2
=
−a/b
which
√
is impossible since 2 is irrational. So b = 0 and then a = 0 follows. So the kernel is just the
singleton zero matrix.
4. Let R be a finite commutative ring with unity. Prove that every nonzero element a of R is [3]
either a unit or a zero divisor. Hint: Consider the product of a with members of R and use
the pigeon-hole principle.
Note: The word “nonzero” was missing from the original version. You won’t lose any marks
for errors caused solely by this omission.
ANSWER. Let |R| = n, R = {r1 , r2 , · · · rn }, a ∈ R, a ̸= 0. Consider the list {ar1 , ar2 , · · · arn }.
If these elements are are distinct n elements then 1 is among them, so ark = 1 for some k and
therefore a is a unit. If they are not distinct, then ari = arj for some i ̸= j, hence a(ri − rj ) = 0
and a is a zero divisor since ri − rj ̸= 0 and aj ̸= 0. Here you want a ̸= 0 since 0 is not a
zero-divisor.
5. (a) Let R be a ring and a ∈ R. Prove that each of the following is a left ideal
(i) Ra, (ii) {r ∈ R : ra = 0}.
ANSWER (i) Three things to verify: (i1) 0 ∈ R, so 0 = 0a ∈ Ra.
(i2) If r1 a, r2 a ∈ Ra, then r1 a − r2 a = (r1 − r2 )a ∈ Ra.
(i3) If ra ∈ Ra and s ∈ R, then s(ra) = (sr)a ∈ Ra since sr ∈ R.
(ii) Let L= {r ∈ R : ra = 0}. Three things to verify. (ii1) 0 ∈ L since 0a = 0.
(ii2) If b, c ∈ L, then (b − c)a = ba − ca = 0 − 0 = 0, so b − c ∈ L
(ii3) If b ∈ L and r ∈ R, then (rb)a = r(ba) = r0 = 0, so rb ∈ L.
1
[3]
(b) What are the ideals in part (a) for the ring M2 (Z) and a = [ 10 00 ]?
[2]
ANSWER. (i) Matrices whose 2nd column is zero. (ii) Matrices whose 1st column is zero.
6. Let R be a ring with unity and assume that the only left ideals of R are {0} and R. Prove that [3]
every nonzero element a of R has an inverse.
Hint: Use Q5 to first show that a has a left inverse b, then consider the left inverse of b.
What does this say about commutative rings with unity with no nonzero proper ideals?
ANSWER. Consider the left ideal Ra. It is not {0} since it contains a = 1.a. Therefore
Ra = R. So 1 ∈ Ra, i.e, a has a left-inverse b. The element b is also nonzero, so it has a left
inverse c. Now c = c1 = c(ba) = (cb)a = 1a = a. Therefore ab = cb = 1 and b is a two-sided
inverse of a.
Corollary: If R is a commutative ring with unity and with no nonzero proper ideals, then it is
a field.
7. Prove that 2Z/8Z is not isomorphic to Z4 as rings.
[2]
ANSWER: Claim Z4 has a unity but 2Z/8Z does not.
8. Let J be the ideal of all polynomials in Z[x] with even coefficients. Prove that x + J has no
inverse in Z[x]/J.
[3]
ANSWER: If x + J has an inverse f (x) + J, then xf (x) + J = (x + J)(f (x) + J) = 1 + J.
Therefore xf (x) − 1 ∈ J. But this is absurd since the constant term of xf (x) − 1 is −1 which
is not even.
9. If A and B are ideals in a commutative ring R, prove that AB ⊆ A ∩ B ⊆ A + B. Note: The [4]
ideal AB is defined in exercise 14.12.
ANSWER: (i) If a ∈ A and b ∈ B, then ab ∈ AR ⊆ A and similarly ab ∈ RB ⊆ B. so
ab ∈ A∩B. But these products may not all of AB. Every element of AB is a sum a1 b1 +· · · an bn
with every aj ∈ A and every bj ∈ B. By the above every summand aj bj belongs to A ∩ B and
so their sum is also in A ∩ B.
(ii) A ⊆ A + {0} ⊆ A + B and similarly B ⊆ {0} + B ⊆ A + B. Therefore A ∩ B ⊆ A + B.
10. (a) Let p be a prime. Prove that xp−1 − 1 = (x − 1)(x − 2) · · · (x − (p − 1)) in Zp [x].
[4]
Hint: The set of units of a ring is a group. Use a corollary of Lagrange’s Theorem.
ANSWER: The units of the ring Zp is the set of nonzero elements. It is a group of order
p − 1. By Lagrange j p−1 = 1 for every nonzero j ∈ Zp .
Alternatively use Fermat’s little Theorem to get the same conclusion.
Now Zp is a field, so every x − j (for nonzero j) is a factor of xp−1 − 1. Then you get
xp−1 − 1 = g(x)(x − 1)(x − 2) · · · (x − (p − 1)). But then the degree of g(x) must be 0, i.e,
g(x) is a constant C. Comparing the coefficients of xp−1 on both sides yields C = 1.
(b) Use the above to prove that (p − 1)! ≡ −1 mod p.
ANSWER: Evaluate the polynomials in part a at 0 to get −1 = (−1)p−1 (p − 1)! in Zp . If
p is odd, then (−1)p−1 = 1 hence (p − 1)! ≡ −1 mod p. The case p = 2 can be checked
separately or observe that (−1)2−1 = −1 ≡ +1 mod 2.
11. (a) Let R be a commutative ring with characteristic p (a prime).
[NOT MARKED]
p
p
p
Prove that (a + b) = a + b for every a, b ∈ R. Hint: The binomial theorem.
ANSWER. The binomial theorem holds in commutative rings. So
2
(a + b)p =
p ( )
∑
p j n−j
a b . The main part of the proof is showing that p divides all of the
j
j=0
binomial coefficients above except
first and the last. This immediately implies the
(p) for the
p!
result. If 1 ≤ j ≤ p − 1, then j = j!(n−j)! and the denominator (call it d) must divide
the numerator p! since binomial coefficients are integers. The prime factorization (of) the
denominator does not include p since all factors are < p. so p̸ |d However, p|p! = d pj , so
()
p| pj .
(b) Give an example of a commutative ring with characteristic 4 and elements a, b such that
(a + b)4 ̸= a4 + b4 .
()
Answer: In Z4 , (1 + 3)4 = 0, but 14 + 34 = 2. The culprit is that 4 - 42 .
12. Let J be a two-sided ideal in a ring R. Prove that R/J is commutative if and only if rs−sr ∈ J
for every r, s ∈ R.
ANSWER: R/J is commutative ⇐⇒ (r + J)(s + J) = (s + J)(r + J) for every r, s ∈ R ⇐⇒
rs + J = sr + J ⇐⇒ rs − sr ∈ J.
13. True or False. Short anwers.
[NOT MARKED]
(a) Every noncommutative ring is infinite. FALSE. Consider M2 (Z2 )
(b) There is a subring of Z6 which is a field. TRUE. One is {0, 3}. Another is {0, 2, 4}
(c) For every positive integer n, the characteristic of nZ is n. FALSE.
(d) An integral domain with characteristic 0 must be infinite. TRUE. The additive subgroup
generated by 1 is infinite.
(e) An integral domain with nonzero characteristic must be finite. FALSE. Consider Zp [x]
for a prime p.
(f) R ⊕ R is a field. FALSE: (1, 0) has no inverse
(g) 2x + 1 is invertible in Z4 [x]. TRUE (2x + 1)(2x + 1) = 1
(h) 2x + 1 is invertible in Q[x]. FALSE: The degree of (2x + 1)f (x) is ≥ 1, so xf (x) is not 1
(i) x3 + 2x + 3 is irreducible in Q5 [x] FALSE It factors as (x + 3)(x2 + 2x + 1).
(j) x4 + x2 + 1 is irreducible in Q[x]. FALSE It factors as (x2 + x + 1)(x2 − x + 1)
(k) If f (x), g(x) are in Z[x] and deg(f (x)) = 2, deg(g(x)) = 1, then there exist a polynomial
q(x) and an integer r such that f (x) = q(x)g(x)+r. FALSE: Try f (x) = x2 and g(x) = 2x.
(l) There is a ring R and two polynomials of degree 2 in R[x] whose product has degree 2.
TRUE: Consider (2x2 + 1)(3x2 + 1) in Z6 [x]
[ a b]
[ a b]
(m) { −b
a : a, b ∈ R} is isomorphic to C. TRUE Φ( −b a ) = a + ib is an isomorphism.
3
[3]