Download Solutions to Homework 1 - Math 4400 1. (# 24 p. 175) Describe all

Solutions to Homework 1 - Math 4400
1. (# 24 p. 175) Describe all ring homomorphisms from Z into Z × Z.
Solution. Let φ be such a ring homomorphism. Suppose that
φ(1) = (a, b)
with a, b ∈ Z. Since φ is a ring homomorphism it follows that
φ(x1 + · · · + xn ) = φ(x1 ) + · · · φ(xn )
(this follows from additivity of φ). In particular,
φ(m) = φ(m · 1) = m · φ(1) = m(a, b).
(1)
On the other hand, φ preserves multiplication. That is, φ(mn) = φ(m)φ(n). By (1) we
have
mn(a, b) = m(a, b) · n(a, b) ⇐⇒
mn(a, b) = mn(a2 , b2 ) ⇐⇒
(a, b) = (a2 , b2 ) if mn 6= 0.
This last inequality only holds if a = 0, 1 and b = 0, 1. It follows that there are four
ring homomorphisms which are given by
φ1 (1) = (0, 0),
φ2 (1) = (1, 0),
φ3 (1) = (0, 1),
φ4 (1) = (1, 1).
More explicitly, these are
φ1 (m) = (0, 0),
φ2 (m) = (m, 0),
φ3 (m) = (0, m),
φ4 (m) = (m, m).
It is straightforward to verify that each of these maps defines a ring homomorphism.
2. (# 46. p. 176) Show that if a and b are nilpotent elements of a commutative ring
then a + b is nilpotent.
Solution. We first show that
N
(a + b)
=
N X
N
j=0
j
aj bN −j .
This may be proven by induction. Instead we shall prove this by developing the lefthand side. Note that
(a + b)N = (a + b) · (a + b) · · · (a + b)
where there are N factors. If we multiply this out there will be 2N terms since in each
factor there are two choices (namely a or b). As R is commutative a typical term looks
like
a · · · a · b · · · b = aj bN −j
1
where there are j factors each of a and N − j factors each of b for some 0 ≤ j ≤ N . If
j is fixed and we would like to count the number of such terms with exactly j terms
of a then this is just Nj . We may think of having N boxes which will be filled with j
choices of the a’s. It follows that
N X
N j N −j
N
(a + b) =
a b
.
j
j=0
Now suppose that am = 0 and bn = 0. Observe that
au = 0 if u ≥ m
and
bv = 0 if v ≥ n.
By the binomial theorem
mn
(a + b)
=
mn X
mn
j=0
j
j mn−j
a b
=
m−1
X
j=0
mn mn j mn−j X mn j mn−j
a b
+
a b
.
j
m
j=m
Observe that aj = 0 for j ≥ m in the second sum. Thus the second sum is identically
zero. Hence
m−1
X mn
mn
aj bmn−j .
(a + b) =
j
j=0
On the other hand, if 0 ≤ j ≤ m − 1 then mn − j ≥ mn − (m − 1) = m(n − 1) + 1 ≥
n − 1 + 1 = n and thus bmn−j = 0 for 0 ≤ j ≤ m − 1. It follows that each term in the
first sum is identically zero. Hence (a + b)mn = 0 and a + b is nilpotent.
3. (# 47. p. 176) Show that a ring R has no nonzero nilpotent element if and only if
x = 0 is the only solution of x2 = 0 in R.
Solution. ⇐= Suppose that there exists a nonzero element a ∈ R such that a is
nilpotent. let n be the least integer such that an = 0. Note that n ≥ 2. If n = 2k is
even then
(ak )2 = a2k = an = 0
and ak 6= 0 since n is the least integer such that an = 0. This contradicts that x = 0 is
the only solution of x2 = 0 in R. If n = 2k + 1 is odd then a2k+1 = 0. Multiplying this
by a yields a2k = 0. As before,
(ak )2 = a2k = 0.
Since k < n it follows that ak 6= 0. Once again this contradicts that x = 0 is the only
solution of x2 = 0 in R.
=⇒. Let x be a solution of x2 = 0. If x 6= 0 then x is nilpotent. This contradicts that
R has no nonzero nilpotent elements.
4. (# 55. p. 177) A ring R is a Boolean ring if a2 = a for a ∈ R, so that every element
is idempotent. Show that every Boolean ring is commutative.
Solution. We begin by observing that
a = −a
for all a ∈ R in a Boolean ring. Note that a + (−a) = 0. Multiplying this by a on the
left we see that
a(a + (−a)) = 0 ⇐⇒ a2 + a(−a) = 0.
(2)
2
Also multiplying a + (−a) = 0 by −a on the left we see that
(−a)(a + (−a)) = 0 ⇐⇒ (−a)a + (−a)2 = 0.
(3)
Equating equation (2) and (3) it follows that
a2 + a(−a) = (−a)a + (−a)2
Recall that (−a)a = a(−a) = −a2 (This was proven in class). Thus by adding a2 to
both sides we obtain
a2 + (a(−a) + a2 ) = ((−a)a + a2 ) + (−a)2 ⇐⇒
a2 + 0 = 0 + (−a)2 ⇐⇒
a2 = (−a)2 ⇐⇒
a = −a since R is idempotent.
This establishes our initial claim.
Now consider x, y ∈ R. We expand (x + y)2 :
(x + y)2 = (x + y)(x + y) = x(x + y) + y(x + y) = x2 + xy + yx + y 2 .
On the other hand (x + y)2 = x + y as R is idempotent. Equating both expressions
yields
x2 + xy + yx + y 2 = x + y ⇐⇒
x + xy + yx + y = x + y ⇐⇒
x + xy + yx + y + (−(x + y)) = x + y + (−(x + y)) ⇐⇒
xy + yx = 0 ⇐⇒
xy = −yx ⇐⇒
xy = yx by our first observation.
Thus R is commutative.
3