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) wehave

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 fourring 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 ringthen a + b is nilpotent.Solution. We first show that

(a + b)N =N∑

j=0

(N

j

)ajbN−j .

This may be proven by induction. Instead we shall prove this by developing the left-hand 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 eachfactor there are two choices (namely a or b). As R is commutative a typical term lookslike

a · · · a · b · · · b = ajbN−j

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where there are j factors each of a and N − j factors each of b for some 0 ≤ j ≤ N . Ifj is fixed and we would like to count the number of such terms with exactly j termsof 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

(a + b)N =N∑

j=0

(N

j

)ajbN−j .

Now suppose that am = 0 and bn = 0. Observe that

au = 0 if u ≥ m

andbv = 0 if v ≥ n.

By the binomial theorem

(a + b)mn =mn∑j=0

(mn

j

)ajbmn−j =

m−1∑j=0

(mn

j

)ajbmn−j +

mn∑j=m

(mn

m

)ajbmn−j .

Observe that aj = 0 for j ≥ m in the second sum. Thus the second sum is identicallyzero. Hence

(a + b)mn =m−1∑j=0

(mn

j

)ajbmn−j .

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 thefirst 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 ifx = 0 is the only solution of x2 = 0 in R.Solution. ⇐= Suppose that there exists a nonzero element a ∈ R such that a isnilpotent. let n be the least integer such that an = 0. Note that n ≥ 2. If n = 2k iseven 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 isthe only solution of x2 = 0 in R. If n = 2k + 1 is odd then a2k+1 = 0. Multiplying thisby 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 onlysolution of x2 = 0 in R.

=⇒. Let x be a solution of x2 = 0. If x 6= 0 then x is nilpotent. This contradicts thatR 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 elementis 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 theleft we see that

a(a + (−a)) = 0 ⇐⇒ a2 + a(−a) = 0. (2)

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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 toboth 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 + y2.

On the other hand (x + y)2 = x + y as R is idempotent. Equating both expressionsyields

x2 + xy + yx + y2 = 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.

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