22. Suppose x ∈ R so that ux = 0. As u is a unit, there is some v ∈ R with uv = 1, and
0 = v · 0 = v(ux) = (vu)x = 1 · x = x.
So if ux = 0, then x = 0; this shows u is not a zero divisor.
23. Suppose first that (a) = (b). So a ∈ (a) = (b), hence a = bx for some x ∈ R. Also,
b ∈ (b) = (a), hence b = ay for some y ∈ R. Thus a · 1 = a = bx = ayx. Then, since R is an
integral domain, this implies 1 = yx, which means x, y are units.
On the other hand, suppose a = bu for u a unit. Thus
(a) = {as : s ∈ R } = {bus : s ∈ R } ⊆ (b).
Since u is a unit, there is some v ∈ R so that uv = 1. Hence av = buv = b · 1 = b, and hence
(b) = {bs : s ∈ R } = {avs : s ∈ R } ⊆ (a).
Thus (a) = (b).
Finally, note that R = (1). So by the first part of the exercise, (a) = R if and only if
a = 1 · u = u for some unit u ∈ R.
24. (1), (2) Done in lectures; observe that non-zero constants C \ {0} are precisely the units
of R. (3), (4) Say I = (f ) = f R. If f is constant then I = {0} or I = C, in both cases not a
maximal ideal. If deg f ≥ 2, then f is reducible (Fundamental Theorem of Algebra) and I is
therefore not a prime ideal, hence not maximal. If deg f = 1, say f = c(x − a) with c 6= 0, then
R C, g 7→ g(a) has kernel I. As C is a field, I is maximal.
26. Note that S −1 R is given as a subset of a ring, the field of fractions of R. So we just need
to check that it is a subring: it must contain 1 and be closed under +,− and ×. For 1, take
r = s = 1. For +,− and ×, by definition of these operations in the field of fractions,
rs0 + s0 s
r
−r
r r0
rr0
r r0
+ 0 =
,
−
=
,
·
=
.
s s
ss0
s
s
s s0
ss0
These are all in S −1 R because S is a multiplicative set (s, s0 ∈ S implies ss0 ∈ S). Finally,
S = {1}
⇒ S −1 Z = Z
S = Z \ {0} ⇒ S −1 Z = Q
S = {2n |n ≥ 0}
⇒ S −1 Z = { 2an | a ∈ Z, n ≥ 0}
S = {odd integers} ⇒ S −1 Z = { ab | a, b ∈ Z, b odd}.
27. Step 1. It is tempting to say that by the 3rd isomorphism theorem, In ∩ A are ideals of
A × {0} ⊂ A × B, but this inclusion does not make A × {0} into a subring of A × B, because
A × {0} does not contain 1 = (1, 1) ∈ A × B. Nevertheless A × {0} is a ring, it is easy to check
directly that In ∩ A are ideals. They clearly form an ascending chain, that is In ∩ A ⊂ In+1 ∩ A
(though these inclusions may be equalities, even if In ( In+1 ). As A is Noetherian, this chain is
finite, so IN ∩ A = IN +1 ∩ A = IN +2 ∩ A = . . . for some N large enough.
Step 2. π is a ring homomorphism, so it takes ideals to ideals, and again this preserves
inclusions. As B is Noetherian, π(IM ) = π(IM +1 ) = ...;
Step 3. By taking N larger if necessary (so that N ≥ M ), we may assume that IN ∩ A =
IN +1 ∩A = . . . and π(IN ) = π(IN +1 ) = ... as well. It remains to show that, in fact, IN = IN +1 =
IN +2 = . . ., so the original chain of ideals terminates as well. For that suppose that I ⊂ J are
ideals of A × B, with I ∩ A = J ∩ A and π(I) = π(J), like all the IN , IN +1 , ... above.
Pick any element (a, b) ∈ J. Because π(I) = π(J) and π maps (a, b) to b, there is some
element of the form (a0 , b) which is in I. The difference (a0 , b) − (a, b) = (a − a0 , 0) lies in J,
so it actually lies in J ∩ A which is assumed to be the same as I ∩ A. So (a − a0 , 0) ∈ I, and
(a0 , b) ∈ I, hence their sum (a, b) ∈ I. This proves that I = J.