Download Solution

MATH 205 HOMEWORK #7 OFFICIAL SOLUTION
D&F 7.2.13:
(a) We need to check that g −1 Kg = K for all g ∈ G. We have
g −1 Kg = g −1 (k1 + · · · + km )g = g −1 k1 g + · · · + g −1 km g = K,
since K is a conjugacy class and thus g permutes it.
(b) Suppose that α is in the center of RG, and let G = {g1 , . . . , gn }. We can write
α=
n
X
ai gi ,
i=1
and this representation will be unique. Suppose that the gi are ordered in such a way that
for 1 ≤ j ≤ r, gj is a representative of the class Kr . We claim that
α=
r
X
ai Ki .
i=1
Note that
α−
r
X
n
X
ai Ki =
i=1
(ai − a0i )gi .
i=r+1
a0i
where = aj if gi is in the conjugacy class Kj . We need to show that this is zero. Suppose
that for some i0 , ai0 − a0i0 6= 0; let j0 be such that gi0 ∈ Kj0 . Since Kj0 is a conjugacy class,
we know that there exists g ∈ G such that g −1 gi0 g = gj0 . Then
g
−1
(α −
r
X
i=1
n
X
ai Ki )g =
(ai − a0i )g −1 gi gj0 .
i=r+1
This is in the center of RG, so we know that this must not have changed the coefficients;
on the other hand, this takes gi0 to gj0 , so we must have had ai0 − a0i0 = 0, since that is the
coefficient of gj0 originally. Thus this remaining sum is 0, as desired.
D&F 7.3.12:
(a) If we add two multiples of D we get another multiple of D, so it suffices to check that such
matrices multiply correctly. We have
a b
c d
ac + Dbd
ad + bc
=
Db a
Dd c
D(ad + bc) ac + Dbd
which is a matrix in exactly the correct form.
(b) Note that ϕ is clearly injective and commutes properly with addition. To check that it is a
ring isomorphism we need to check that multiplication works correctly. Note that
√
√
√
(a + Db)(c + Dd) = (ac + Dbd) + (ad + bc) D,
which is exactly correct. Thus ϕ is an injective ring homomorphism.
1
2
MATH 205 HOMEWORK #7 OFFICIAL SOLUTION
√
a b
to a + b D. This is also
Db a
clearly injective and commutes properly with addition, and it is (set-theoretically) the inverse of ϕ. Thus in particular we see that ϕ was surjective, and is therefore a ring isomorphism.
(c) It’s easy to see directly from the definition that this subset is closed under addition; thus it
remains to show that it is closed under multiplication.
a
b
c
d
(D − 1)b/4 a + b
(D − 1)d/4 c + d
ac + (D − 1)bd/4
ad + bc + bd
=
(D − 1)(cb + ad + bd)/4 (D − 1)bd/4 + (a + b)(c + d)
Consider the map ψ defined by sending a matrix
which is of the desired
form. Note that the
homomorphism from this subring to O given by
√
a
b
sending the matrix
to a + (1 + D/2)b. We just need to check that
(D − 1)b/4 a + b
multiplication works correctly, which follows directly from the definition.
D&F 7.3.13: This is an application of the method of proof of the previous problem, where D = −1.
Note that none of the properties of Z were used above, so in fact the matrix construction will√work
over any ring; here, we just take the ring to be R, and note that C is the set of expressions a+ −1b
over R.
D&F 7.3.15: We have a set-theoretic inverse to the map A 7→ χA : we take a function χ : X → Z/2
to the set χ−1 (1). Thus it suffices to prove that the given function is actually a ring homomorphism.
To see this, we just need to check that functions add and multiply correctly. Consider χA + χB :
this is 1 on every element of X which is either in A or in B but not in both, which is exactly
what A + B in the Boolean ring is. Thus it is equal to χA+B , and we see that we have a group
homomorphism.
To check multiplication, note that χA χB is 1 on all elements that are in both A and B; thus it is
exactly equal to χA∩B . Since A ∩ B is exactly the multiplication in the Boolean ring, we are done.
D&F 7.3.20: We know that both S and I are closed under addition and multiplication, so S ∩ I
is a subring of S. To check that it is an ideal, we need to check that for all s ∈ S and i ∈ S ∩ I,
si ∈ S ∩ I. By definition, si ∈ S, and si ∈ I because I is an ideal of R. Thus S ∩ I is an ideal of
S, as desired.
Now let R = R[x] and S = Z[x], and consider the ideal (2x) in S. If it were equal to S ∩ I
for some ideal I of R then we’d know that (2x) ⊂ I. But (2x) = (x) in R[x], so we’d get that
(x) ∩ Z[x] = (2x), a contradiction. Thus (2x) is not equal to I ∩ S for some ideal I of R[x].
D&F 7.4.6: Suppose that R is a division ring, and let I be an ideal of R. If I 6= (0) then there is
a nonzero a ∈ I; since R is a division ring, there exists b ∈ R such that ba = 1; so 1 ∈ I, and thus
I = R. Thus if R is a division ring then its only ideals are (0) and R.
Now suppose that R only has ideals (0) and R, and consider a nonzero a ∈ R. The ideal (a) is
not equal to (0), so it must be all of R; in particular, 1 ∈ (a). Thus there exists b ∈ R such that
ba = 1, and we see that a has a left inverse. Thus R is a division ring.
D&F 7.4.7: (x) is a prime ideal if and only if R[x]/(x) is an integral domain; since R[x]/(x) ∼
= R,
the desired result follows. Similarly, (x) is a maximal ideal if and only if R[x]/(x) ∼
= R is a field.
D&F 7.4.8: Note that if a = b = 0 this is trivial, and (a) 6= (0) for an nonzero a. Thus we can
assume that a, b 6= 0. Suppose that (a) = (b); then there exist c, d such that ca = b and db = a.
MATH 205 HOMEWORK #7 OFFICIAL SOLUTION
3
Then
dca = db = a.
Thus (dc−1)a = 0. Since we know that R is an integral domain, we must have dc = 1. Analogously,
cd = 1, so both c and d are units, and the desired result follows.
D&F 7.4.11: Suppose that a ∈ I and a ∈
/ P . Let j ∈ J be any nonzero element. Then aj ∈ I
since a ∈ I and aj ∈ J since j ∈ J, so aj ∈ I ∩ J ⊆ P . Since P is prime, we know that either a or
j is in P . By assumption, a is not in P , so j ∈ P . Since j was arbitrary we see that J ⊆ P . Thus
if I 6⊆ P we must have J ⊆ P , as desired.
D&F 7.4.12: We know that IJ contains each
Pn ai bj , so it just remains to show that these span
all of P
IJ. An element of
IJ
is
a
finite
sum
k=1 αk βk with αk ∈ I and βk ∈ J. We can write
P
αk = xki ai and βk =
ykj bj , so that
XX
αk βk =
xki ykj ai bj .
Thus each αk βk is in the span of the ai bj , and thus so is all of IJ, as desired.
Problem 4: Let R be a ring. (R, +) is an abelian group, and thus a Z-module. Show that the
multiplication in R gives a homomorphism of abelian groups R ⊗Z R → R.
Solution: We define µ : R ⊗Z R → R by sending a ⊗ b to ab. We thus need to check that it
satisfies the axiom for a bilinear map of Z-modules. We check that
µ((a + a0 ) ⊗ b − a ⊗ b − a0 ⊗ b) = (a + a0 )b − ab − a0 b = ab + a0 b − ab − a0 b = 0,
since multiplication distributes over addition. The other properties follow analogously.
Problem 5: Let R be a ring. Rop is defined to have the same underlying additive structure as R,
but multiplication is done in the opposite direction. Thus the product of a · b in Rop is the element
ba. Show that Rop is a ring, and that a left ideal of Rop is a right ideal of R.
Solution: We need to check the ring axioms for Rop . Since additively Rop is equal to R those
axioms clearly hold. We will write the multiplication in Rop as · and omit the operation for
multiplication in R to clearly distinguish between them; thus a · b = ba. We check associativity and
distributivity axioms:
(a · b) · c = (ba) · c = c(ba) = (cb)a = a · (cb) = a · (b · c).
(a + b) · c = c(a + b) = ca + cb = a · c + b · c.
a · (b + c) = (b + c)a = ba + ca = a · b + a · c.
Thus Rop is a ring. Note that if R is commutative then Rop is isomorphic to R.
Problem 6: Let R be a commutative ring. A polynomial p(x) ∈ R[x] is called irreducible if
whenever we write p(x) = q(x)r(x) we must have either q(x) or r(x) constant.
(a) Show that if R is a field and p has degree 2 or 3 then it is irreducible if and only if it has
no roots in R.
(b) Give an example of a ring R and a polynomial p(x) such that
{a ∈ R | p(a) = 0} > deg p(x).
(c) Show that if R is a field and p(x) is irreducible then (p(x)) is a prime ideal.
(d) Let ϕ : R → S be an injective ring homomorphism. Show that if ϕ(p(x)) is irreducible in
S[x] then p(x) is irreducible in R[x]. Give a counterexample to this if ϕ is not injective.
Solution:
4
MATH 205 HOMEWORK #7 OFFICIAL SOLUTION
(a) Note that the original form of this problem (where R is not a field) is not correct at all. For
example, over Z/6 the polynomial 5x + 1 is not irreducible:
(5x + 1) = (2x + 1)(3x + 1).
Thus the situation over rings is much more complicated.
However, over a field degrees add when polynomials are multiplied, since there are no zero
divisors. Thus we know that if a polynomial has degree 0 or 1 then it must be irreducible,
and all polynomials of degree 1 have roots. A polynomial of degree 2 or 3 is reducible only
if one of the factors has degree 1, which always has a root. On the other hand, if p(x) has
a root a then x − a|p(x), which means that p is reducible.
(b) Let R = Z/8, and let p(x) = x2 − 1. Then if x = 1, 3, 5, 7 then x2 ≡ 1 (mod 8), so x2 − 1
has 4 roots in Z/8.
(c) It suffices to check that F [x]/(p(x)) is an integral domain. We will show the stronger
statement that it is a field.
Note that, as we showed on the first homework set, for any polynomials a(x) and b(x) we
can write a(x) = q(x)b(x) + r(x) with deg r(x) < deg b(x) or r(x) = 0. By the Euclidean
algorithm this means that there exist polynomials α(x) and β(x) such that
a(x)α(x) + b(x)β(x) = d(x),
where d(x) is the greatest common divisor of a(x) and b(x). If p(x) is irreducible then its
greatest common divisor with any a(x) with deg a(x) < deg p(x) is equal to 1. If we can
write
a(x)α(x) + p(x)β(x) = 1
then a(x)α(x) = 1 ∈ F [x]/p(x). Thus F [x]/p(x) is a field, as desired.
(d) The desired statement is equivalent to showing that if p(x) is not irreducible in R[x] then it
is not irreducible in S[x]; in other words, if p(x) = a(x)b(x) in R[x] then ϕ(p(x)) is reducible
in S[x]. But since ϕ is a homomorphism we know that
ϕ(p(x)) = ϕ(a(x))ϕ(b(x));
thus if both ϕ(a(x)) and ϕ(b(x)) have degree less than deg ϕ(p(x)) we are done. For any
homomorphism ϕ, deg ϕ(p(x)) ≤ deg p(x), so we just need to check that they are equal.
However, the only way that ϕ can lower the degree of p(x) is if it takes a nonzero coefficient
to zero; since it is injective this never happens.
Let p(x) = 4x2 + 4x + 1 ∈ Z[x], and let ϕ : Z → Z/2 be the projection. Then ϕ(p(x)) = 1,
which is irreducible; however, p(x) = (2x + 1)2 in Z[x], which is reducible.