Winter 2016
Ma5b HW 1
Problem 1.
Proof.
(a) Recall that a = 0 ∈ Z/nZ if and only if n|a in Z. Thus ab is nilpotent in Z/nZ if there exists
m ∈ Z+ such that n|(ab)m . Since n = ak b, any m ≥ k has this property, hence ab is indeed nilpotent in
Z/nZ.
(b) From the fundamental theorem of arithmetic, we can write (uniquely) n in the form n = pa1 1 · · · pakk
where pk are distinct prime numbers. Let m = max(a1 , . . . , ak ). It is clear that n = pa1 1 · · · pakk divides
(p1 · · · pk )m , which in turn divides am if every pi divides a. As seen above n|am means precisely that a is
nilpotent. Conversely, assume n|am . Assume p is a prime dividing n. Then p|am . By looking at the prime
factorization of a, we see that p has to divide a.
Nilpotents of Z/72Z are {0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66}.
(c) Assume that R contains a nonzero nilpotent element, call it f : X → F . By definition for some integer
m > 0, f m is the zero function on X, i.e. f (x)m = 0 for all x ∈ X. Since F is a field, f (x)m = 0 implies
f (x) = 0 for all x ∈ X, i.e. that f is, in fact, the zero function, contrary to our assumption.
Problem 2.
Proof.
(a) Since x is nilpotent, by definition there is an integer m > 0 such that xm = 0. Then x · xm−1 = 0
so x is a zero-divisor (or zero).
(b) If xm = 0, then (rx)m = rm xm = rm · 0 = 0. Note that we used commutativity for the equality
(rx)m = rm xm .
(c) If xm = 0, then 1 = 1 + xm = (1 + x)(1 − x + x2 − x3 + · · · ± xm−1 ). The second parenthesis is a an
element of R, and the inverse of 1 + x, hence 1 + x is a unit.
(d) Let x be a nilpotent and u a unit with inverse u−1 . We want to show that x + u is a unit. Indeed
x + u = u(1 + u−1 x). By (b), u−1 x is nilpotent, and by (c), 1 + u−1 x is a unit. Thus x + u = u · (1 + u−1 x)
is the product of two units, hence a unit itself.
Problem 3.
Proof. Of contains the identity 1 = 1 + 0 · f ω. Closure under addition is straightforward: (a + bf ω) + (c + df ω) =
(a + c) + (b + d)f ω ∈ Of . Closure under multiplication is easy to verify as well: (a + bf ω)(c + df ω) =
(ac + bdf 2 D) + (ad + bc)f ω ∈ Of .
We show that 0, ω, . . . , (f − 1)ω is a complete set of coset representatives for Of in O. Two such elements
iω 6= jω represent distinct cosets because the difference is (i − j)ω, which is not of the form a + bf ω because f
does not divide i − j which is a nonzero integer satisfying −f < i − j < f .
It is a complete set of representatives because given any α = a + bω ∈ O, if we write b = qf + r with
0 ≤ r < q (the usual division with remainder in Z), then (a + bω) − r · ω = a + qf ω ∈ Of , so α and rω represent
the same coset.
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Winter 2016
Ma5b HW 1
Conversely, assume A is a subring of O containing the identity and having finite index.
Note first that A contains 1 some number of the form bω. Otherwise, the elements Zω yield infinitely
many distinct cosets. Let f 0 be the smallest, strictly positive, such number b. We claim that f 0 = f , i.e.
A = Of 0 . Indeed, since 1, f 0 ω ∈ A, we have Of 0 ⊂ A. Moreover since Of 0 has index f 0 in O, it follows that
[O : A] ≤ [O : Of 0 ], i.e. f ≤ f 0 . On the other side, the elements 0, ω, . . . , (f 0 − 1)ω yield distinct cosets, hence
f ≥ f 0 . Thus f = f 0 , as wanted.
Problem 4.
Proof.
(a) Since φ is a nonzero homomorphism, φ(1R ) 6= 0S . Now, from φ(1R ) = φ(1R · 1R ) = φ(1R ) · φ(1R ),
it follows that φ(1R ) · (1 − φ(1R )) = 0. Thus either φ(1R ) = 1S or φ(1R ) is a zero-divisor. The only
zero-divisor in an integral domain is zero. If φ is nonzero, by the above φ(1R ) is either 1S or a non-zero
divisor. Since the second case can’t happen, φ(1R ) = 1S , as wanted.
(b) Assume φ(1R ) = 1S and let u be a unit of R. Then 1S = φ(u · u−1 ) = φ(u)φ(u−1 ), hence φ(u) is invertible
with inverse φ(u−1 ).
Problem 5.
Proof.
(a) Since I and J are subrings of R, they both contain 0, hence so does I ∩ J. If a, b ∈ I ∩ J, then
a + b, ab ∈ I and a + b, ab ∈ J, hence a + b, ab ∈ I ∩ J. Thus I ∩ J is a subring of R. If a ∈ I ∩ J and
r ∈ R, then ra ∈ I and ra ∈ J since I and J are ideals. Thus ra ∈ I ∩ J. Similarly ar ∈ I ∩ J. We have
verified all the conditions in the definition of an ideal for I ∩ J.
(b) Let now (Ij )j∈J be an arbitrary nonempty collection of ideals (indexed by J). Let A = ∩j Ij . We want to
show A is an ideal. Each Ij is a subring of R, hence contains 0. Thus so does A. If a, b ∈ A, then a, b ∈ Ij
for every j ∈ J. Since Ij is closed under addition and multiplication, a + b and ab belong to Ij for every j.
Thus a + b and ab belong to ∩j Ij = A. This shows that A is a subring of R. Finally, let a ∈ A and r ∈ R.
Since a ∈ Ij for every Ij and Ij is an ideal, we have that ra and ar are in Ij for every j. Thus ar and ra
belong to the intersection of all Ij ’s, which is A. Thus A satisfies all the conditions in the definition of an
ideal.
Problem 6.
Proof. (a) Let I = {x ∈ R| ax = 0}. Since a · 0 = 0, we have 0 ∈ I. If x, y ∈ I, then a(x + y) = ax + ay =
0 + 0 = 0, hence x + y ∈ I. Moreover, a(xy) = (ax)y = 0 · y = 0, hence xy ∈ I. Thus I is a subring of
R. Let now r ∈ R and x ∈ I. Then a(xr) = (ax)r = 0 · r = 0, thus rx ∈ I. We have verified, for I,
all the conditions of a right ideal. The proof for left ideals is almost identical.
(b) Let I = {x ∈ R| xa = 0 for all a ∈ L}. Given a ∈ L, let Ia = {x ∈ R| xa = 0}. Then Ia is the left
annihilator of a from part (a) and I = ∩a∈L Ia . We have verified in Problem 18 that ∩Ia is a left
ideal. It remains to check that it is also a right ideal, i.e. that x ∈ I and r ∈ R imply xr ∈ I.
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Winter 2016
Ma5b HW 1
Indeed, x ∈ I means, by definition, that xa = 0 for all a ∈ L. Since L is a left ideal rL ⊂ L, i.e.
x(ra) = (xr)a = 0 for all a ∈ L. This, by definition of I, implies xr ∈ I, hence I is also a right ideal.
Remark: Note that we did not apply Problem 18 directly. It is not, in general, true that Ia are
both left and right ideals. Instead we have applied part of the proof in Problem 18 to conclude that
I = ∩a∈L Ia is a left ideal and then proved that it is also a right ideal.
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