Math 113 Homework 6
Solutions
due July 31, 2011
1. Let R be a ring of characteristic p. Prove that if a is nilpotent then 1 + a
is unipotent, that is, some power of 1 + a is equal to 1.
Solution: If a is nilpotent, there is some n > 0 such that an = 0. Let
k
k be the smallest positive integer such that pk > n. Then (1 + a)p =
k
k
k
1p + ap = 1 + (an )(ap −n ) = 1, i.e. 1 + a is nilpotent.
2. Find an ideal in the polynomial ring F [x, y] in two variables which is not
principal.
Solution: (x, y) is such an ideal, as only units divide both x and y.
3. (a) Let I, J be ideals of a ring R. Prove that I ∩ J is an ideal.
Solution: I ∩ J is an additive subgroup because intersections of subgroups are subgroups. If f ∈ I ∩ J and x ∈ R, then xf ∈ I because
f ∈ I and I is an ideal, and also xf ∈ J for the same reason, so
xf ∈ I ∩ J. Thus I ∩ J is an ideal.
(b) Show by example that the set of products {xyP: x ∈ I, y ∈ J} need
not be an ideal, but that the set of finite sums
xν yν of products of
elements of I and J is an ideal. This ideal is called the product ideal .
Solution: Let R = F [x, y, z, w], let I = (x, z), and J = (y, w). Then
xy and zw are both elements of the set of products {xy : x ∈ I, y ∈ J},
but their sum xy + zw is not, showing that this set cannot be an ideal.
The sum of two finite sums of products of elements of I and J is clearly
itself a finite sum of such products, while a product of two such sums
may be distributed over one of the sums to yield another, as per
X
X
X
X
(
xν yν )(
xµ yµ ) =
(xµ
xν yν )yµ
so the set of finite sums of products is indeed an ideal.
(c) Prove that IJ ⊂ I ∩ J.
Solution: Every element of the product set {xy : x ∈ I, y ∈ J} lies
in the intersection ideal, and since the latter is an ideal finite sums
of products also lie in the intersection ideal. This exactly says that
IJ ⊂ I ∩ J.
(d) Show by example that IJ and I ∩ J need not be equal.
Solution: Let I = J be a nontrivial ideal, say (x) ⊂ F [x]. Then
IJ = I 2 = (x2 ), which does not equal (x) = I = I ∩ J.
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4. An element e of a ring S is called idempotent if e2 = e. Note that in a
product R × R0 of rings, the element e = (1, 0) is idempotent. The object
of this exercise is to prove a converse.
(a) Prove that if e is idempotent, then e0 = 1 − e is also idempotent.
Solution: e02 = (1 − e)2 = 1 − 2e + e2 = 1 − e = e0 .
(b) Let e be an idempotent element of a ring S. Prove that the principal
ideal eS is a ring, with identity element e. It will probably not be a
subring of S because it will not contain 1 unless e = 1.
Solution: Let y = ex ∈ eS. Then ey = e · ex = e2 x = ex = y, so e
acts as the multiplicative identity on eS. eS is closed under addition
and multiplication as it is an ideal. Hence eS is a ring with identity
e.
(c) Let e be idempotent, and let e0 = 1 − e. Prove that S is isomorphic
to the product ring (eS) × (e0 S).
Solution: First observe that ee0 = e0 e = e(1 − e) = e − e2 = 0.
Now let φ : S → (eS) × (e0 S) be given by φ(x) = (ex, e0 x), and let
ψ : (eS) × (e0 S) → S be given by ψ(ey, e0 z) = ey + e0 z. These are
both homomorphisms, and are mutually inverse: for all x ∈ S,
ψ(φ(x)) = ψ(ex, e0 x) = ex + e0 x = (e + (1 − e))x = x
while for all (ey, e0 z) ∈ (eS) × (e0 S) we have φ(ψ(ey, e0 z)) = φ(ey +
e0 z) = (e2 y + ee0 z, e0 ey + e02 z) = (ey, e0 z). Hence these maps are in
fact isomorphisms.
5. Let I, J be ideals of a ring R such that I + J = R.
(a) Prove that IJ = I ∩ J.
Solution: IJ ⊂ I and also IJ ⊂ J, so IJ ⊂ I ∩ J. For the converse,
as R = I + J there exist r ∈ I, s ∈ J such that r + s = 1. If
x ∈ I ∩ J, since x ∈ J, rx ∈ IJ, while since x ∈ I, sx ∈ IJ also. Then
x = 1 · x = rx + sx ∈ IJ, so I ∩ J ⊂ IJ. Thus IJ = I ∩ J.
(b) Prove the Chinese Remainder Theorem: For any pair a, b of elements
of R, there is an element x such that x = a (modulo I) and x = b
(modulo J).
Solution: Let r, s be as above (i.e., r ∈ I, y ∈ J, r + s = 1). Observe
that s − 1 = −r ∈ I, while r − 1 = −s ∈ J. Let x = as + br.
Then x − a = a(s − 1) + br ∈ I, so x ∼
= a (modulo I). Similarly,
x − b = as + b(r − 1) ∈ J, so x ∼
= b (modulo J).
6. Prove that the maximal ideals of the ring of integers are the principal
ideals generated by prime integers.
Solution: All ideals of Z are principal. If (n) and (m) are two ideals
of Z, (n) ⊂ (m) exactly when m divides n. So the ideals containing a
given ideal (n) are the ideals generated by the divisors of n; there are no
nontrivial such ideals precisely when n is prime.
7. Let R be a ring, with M an ideal of R. Suppose that every element of R
which is not in M is a unit of R. Prove that M is a maximal ideal and
that moreover it is the only maximal ideal of R.
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Solution: If N were a maximal ideal of R not identical with M , it could
not be contained in M lest it fail to be maximal. The same is true of
any proper ideal properly containing M . Thus in either case there would
be some x ∈ N \M . However, as x ∈
/ M , x is by assumption a unit, so
(x) = R, contradicting the fact that N was assumed to be a proper ideal
(as x ∈ N implies that (x) ⊂ N ). Therefore there can be no such N , i.e.
M is the unique maximal ideal of R.
8. Prove or disprove the following.
(a) The polynomial ring R[x, y] in two variables is a Euclidean domain.
Solution: False. Euclidean domains are principal ideal domains, and
(x, y) is a non-principal ideal of R[x, y].
(b) The ring Z[x] is a principal ideal domain.
Solution: False. The ideal (p, x), where p ∈ Z is any prime, is a
non-principal ideal (the only divisor of both p and x is 1, but this
ideal is proper).
9. Prove that every prime element of an integral domain is irreducible.
Solution: Suppose R is an integral domain, and suppose p is an element
that is reducible as p = xy, with neither x nor y a unit. Then p certainly
divides itself, the product of x and y, but does not divide either x or
y (if it divided one, the other would have to be a unit as multiplicative
cancellation holds in an integral domain). Therefore p cannot be prime.
10. Let R be a principal ideal domain.
(a) Prove that there is a least common multiple [a, b] = m of two elements
which are not both zero such that a and b divide m, and that if a, b
divide an element r ∈ R, then m divides r. Prove that m is unique
up to unit factor.
Solution: Let a, b ∈ R, not both zero. Observe that an element
r of R is divisible by both a and b iff (r) ⊂ (a) and (r) ⊂ (b), i.e.
(r) ⊂ (a) ∩ (b). Further, as R is a PID, the intersection ideal (a) ∩ (b)
is principal, say generated by some element m. We now have that m
divides r iff both a and b divide r, and any such m must generate the
ideal (a) ∩ (b); m is therefore unique up to unit factor.
(b) Denote the greatest common divisor of a and b by (a, b). Prove that
(a, b)[a, b] is an associate of ab.
Solution: A greatest common divisor d of a and b is a generator of
the ideal generated by both a and b. The product dm will generate
the principal ideal (dm) = (d)(m) = ((a) + (b))((a) ∩ (b)). An element
r ∈ R is in this ideal iff it is the product of a linear combination
of a and b with something divisible by both a and b, i.e. if it is of
the form r = (sa + tb)c = sac + tbc, where c = va = wb for some
s, t, v, w ∈ R. That is, r = sw(ab) + tv(ab) = (sw + tv)(ab) for some
s, t, v, w ∈ R, which holds iff r ∈ (ab). Therefore (dm) = (ab), so these
two generators must be associates.
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11. Factor x3 + x + 1 in Fp [x], when p = 2, 3, 5.
Solution: Let f (x) = x3 + x + 1.
f is irreducible in F2 [x] as f (0) = f (1) = 1.
f factors as (x−1)(x2 +x−1) over F3 [x]; the quadratic factor is irreducible
as it has no zeros.
f is irreducible over F5 [x] as it has no zeros.
12. Factor 30 into primes in Z[i].
Solution: 30 = 2 · 3 · 5. As 3 = 3 modulo 4, 3 remains prime in Z[i].
2 and 5 factor as 2 = (1 + i)(1 − i) and 5 = (2 + i)(2 − i) respectively;
these factors are all prime as they have prime norms. Thus one prime
factorization of 30 in Z[i] is given by
30 = (1 + i) · (1 − i) · 3 · (2 + i) · (2 − i)
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