Download Problem Set 1. Solutions.

Problem Set 1. Solutions.
Math 323, Fall 2017
(1) Let p ∈ Z, p 6= 0, ±1. Let us prove :
p is prime ⇔ for all a, b ∈ Z, if p divides ab then either p divides a or p divides b.
(⇒) : Let us prove the contrapositive. Assume p = mn with m, n ∈ Z, m, n 6= 1.
Then p divides mn = p but p does not divide either m or n.
(⇐) : “Normal proof ” : Assume p is prime. Let a, b ∈ Z such that p divides
ab. Assume that p does not divide a, let us show that p divides b. Since p is prime,
its only divisors are 0 and itself, so gcd(p, a) = 1. By Bezout’s Theorem, there are
m, n ∈ Z such that mp + na = 1. Multiplying by b, we get bmp + nab = b. Since
p divides bmp and nab, it divides their sum, so p divides b.
We proved that if p does not divide a, it must divide b, hence it must divide
either a or b.
Overly complicated proof : Assume p is prime. Let a, b ∈ Z such that p
divides ab. Let G = Ca × Cb the product of the cyclic groups. Since p divides
ab = |G|, by Cauchy theorem, there is (α, β) ∈ G of order p. Hence α and β are
either of order 1 (the identity) or p, but cannot be both of order 1. So Ca or Cb
contains an element of order p. By Lagrange theorem, either p divides |Ca | = a or
p divides |Cb | = b. (yay MATH 322 !)
Remark Note that this is not true if p = 0. Indeed 0 is not prime, and does not
divide anything, hence verifies the second statement. The numbers ±1 are also
not prime but they divide every other number, so they satisfy the condition as
well.
(2) Assume that d divides n. If d = 1 then ϕ(d) = 1 so there is nothing to prove.
Q
If d 6= 1 then use the fundamental theorem of arithmetic to write d = ki=1 pδi i
where p1 , . . . , pk are
numbers, and δ1 , . . . , δk ≥Q1. Since d divides n, we
Qkprime
ηi
0
can write n = n i=1 pi where ηi ≥ δi and gcd(n0 , ki=1 pηi i ) = 1. Since ϕ is
Q
Q
multiplicative, ϕ(d) = ki=1 ϕ(pδi i ) and ϕ(n) = ϕ(n0 ) ki=1 ϕ(pηi i ). Hence it is
enough to show that ϕ(pδi i ) divides ϕ(pηi i ) for all i ∈ {1, . . . , n}.
Let i ∈ {1, . . . , n}. We have ϕ(pηi i ) = pηi i −1 (pi − 1) = pηi i −δi pδi i −1 (pi − 1) =
pηi i −δi ϕ(pδi i ), hence ϕ(pδi i ) divides ϕ(pηi i ) as desired.
(3) Let d = gcd(a, b) and let (x0 , y0 ) be a solution of the equation ax + by = N .
Let t be an integer and x = x0 + db t, y = y0 − ad t. Then
b
a
ab
ab
ax + by = a(x0 + t) + b(y0 − t) = ax0 + by0 + t − t = N + 0 = N,
d
d
d
d
as desired.
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Remark Why does that give all solutions ? Let (x, y) be another solution of
the equation. Then a(x − x0 ) + b(y − y0 ) = 0 hence a(x − x0 ) = b(y0 − y), do
a
(x − x0 ) = db (y − y0 ). Since ad divides db (y0 − y) and is coprime with db then ad
d
divides (y0 − y). So there is t such that y0 − y = ad t, hence y = y0 − ad t. Plugging
that into ad (x − x0 ) = db (y0 − y), we get x = x0 + db t, as desired.
(4) We use the Euclidean algorithm.
89 = 1 × 69 + 20
20 = n − a
69 = 3 × 20 + 9
9 = a − 3(n − a) = −3n + 4a
20 = 2 × 9 + 2
2 = (n − a) − 2(−3n + 4a) = 7n − 9a
9=4×2+1
2 = 2 × 1 + 0.
1 = (−3n + 4a) − 4(7n − 9a) = −31n + 40a
Hence 1 is the greatest common divisor of 89 and 69, this can also be seen if one
checks that 89 is prime. The algorithm gave us the Bezout coefficients, so we know
that 40a − 31n = 1 hence 40 is the multiplicative inverse of a in Z/nZ.
(5) (a),(b),(e),(f) are subrings, not the others. You are encouraged to check it for
yourself.
0 if x = 0
For (c), it is not closed under addition, just take the function x 7→
−1 else
1 if x = 0
and add the function identically 1, the sum is the function x 7→
0 else
which has infinitely many zeroes but is not the zero function.
1 if x < 12
0 if x ≤ 12
For (d) add the functions x 7→
, their sum
1 and x 7→
0 if x ≥ 2
1 if x > 12
0 if x = 21
is the map x 7→
which has only one zero.
1 else
(6) Let R be a ring containing an identity 1 and let Z be its center. Note that 1 ∈ Z
since for all r ∈ R we have 1r = r1 = r, so Z is not empty. Likewise, Z also
contains −1.
Z is closed under additive inverses, indeed if a ∈ Z then for all r ∈ R we have
(−a)r = (−1)(ar) = (−1)(ra) = (−r)a = r(−1)a = r(−a) as desired. Let a, b be
elements of the center of a ring R. For every r ∈ R we have (a + b)r = ar + br =
ra + rb = r(a + b) hence a + b is in the center of R.
We proved that Z is an additive subgroup of R.
Also, we have (ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) hence ab is also in
the center of R. So Z is a subring of R.
By definition of the center, it is always a commutative ring.
Suppose now that R is a division algebra. We know Z is a commutative ring,
we only need to check it is closed under inverses. Let a ∈ Z and r ∈ R. We have
a−1 r = (r−1 a)−1 = (ar−1 )−1 = ra−1 ,
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so a−1 ∈ Z, so Z is a field.
(7) We want to prove that the center of H is R. The fact that elements of R are in
the center is immediate. Conversely, assume x = a + bi + cj + dk is in the center,
with a, b, c, d ∈ R.
Then, we have
xi = ix
ai − b − ck + dj = ai − b + ck − dj
Since {1, i, j, k} is an R−basis of H, we have that c = d = 0. Also,
xj = jx
aj + bk = a − bk,
so b = 0 too. We proved x = a ∈ R as desired.
(8) Let R be a boolean ring. First note that for all a ∈ R we have 2a = (2a)2 =
4a2 = 4a, removing 2a from both sides yields 2a = 0. Hence for all a ∈ R we have
a = −a. Now let a, b ∈ R.
We compute a + b = (a + b)2 = a2 + b2 + ab + ba, simplifying both sides by a + b
we get ab = −ba = ba as desired.
(9) (a) Assume that n = ak b for some positive integer k. Then (ab)k = ak bk =
k
k−1
ak bbk−1 = nbk−1 since Z is commutative. And so ab = nb
= 0bk−1Q= 0.
(b) (⇒) : Assume that all prime factors of n also divide a. Write n = ki=1 pηi i
Q
Q
with each pi prime and ηi ≥ 1. Then ki=1 pi divides a, hence ( ki=1 pi )max(ηj )
Q
max(ηj )
divides amax(ηj ) . Since max(ηj ) ≥ ηi for all i, we have that n divides ki=1 pi
=
Qk
max(ηj )
max(ηj )
max(ηj )
and therefore n divides a
, so a
= 0 in Z/nZ as
( i=1 pi )
desired.
(⇐) : Assume that a is nilpotent in Z/nZ. Let k ∈ Z such that ak = 0 in
Z/nZ. If p is a prime factor of n, since n divides ak , we know that p divides
ak . But since p is prime, this means that p divides a (one can use the first
question of the assigment for that). We proved the desired result.
We have 72 = 23 32 so nilpotents in Z/72Z correspond to integers in {0, . . . , 71}
divisible by 2 and 3 hence divisible by 6. There are 72/6 = 12 such elements
(counting 0).
(c) Here the product of two functions is given by the map whose values are the
product of the two functions. Suppose f : X → F is a nilpotent function
where X is a nonempty set, and F is a field. Then there is some positive
integer k such that for every x ∈ X we have (f (x))k = 0. But F is a field,
in particular it is an integral domain, so if f (x)k = 0 then we must have
f (x) P
= 0. Since this is true for all x ∈ X, we have that f is the zero function.
(10) Let N = ni=1 gi where G = {g1 , . . . , gN } is a finite group. Note that it suffices
to prove that xgi = gi x for all i ∈ {1, . . . , n} since every element
Pn of RG is obtained as an R−linear combination. Let g ∈ G. Then N g = i=1 gi g but the
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multiplication on the right or left by g only permutes the elements
Pn of the group,
hence {gg1 , . . . , ggn } = {g1 , . . . , gn } = {g1 g, . . . , gn g}, so N g = i=1 ggi = gN , as
desired.