Math 902
Solutions to Homework # 6
1. Let G be a finite group and ρ : G → GLC (V ) and ψ : G → GLC (W ) be finite
dimensional representations for G over C. Define ρ · ψ : G → GLC (V ⊗C W ) by
(ρ·ψ)(g) = ρ(g)⊗ψ(g). Prove that ρ·ψ is a group homomorphism and that χρ·ψ = χρ χψ .
(Hence, the product of two characters of G is a character of G.)
Solution: For all g, h ∈ G, note that
(ρ · ψ)(g)(ρ · ψ)(h) = (ρ(g) ⊗ ψ(g))(ρ(h) ⊗ ψ(h))
= ρ(g)ρ(h) ⊗ ψ(g)ψ(h)
= ρ(gh) ⊗ ψ(gh)
= (ρ · ψ)(gh).
Thus, ρ · ψ is a group homomorphism. For g ∈ G observe that:
χρ·ψ (g) = tr(ρ(g) ⊗ ψ(g)) = tr(ρ(g)) tr(ψ(g)) = χρ (g)χψ (g).
2. Let G be a finite group χ and φ be (complex) characters of G. Suppose χ is irreducible
and deg φ = 1. Prove that χφ is irreducible. (Hint: Show that hχφ, χφi = 1.)
Solution: Observe that as deg φ = 1, φ(g) = λg for some root of unity λg ∈ C. Hence,
φ(g) = λg = λ−1
g . Thus, φ(g)φ(g) = 1 for every g ∈ G. Hence
X
χ(g)φ(g)χ(g)φ(g)
hχφ, χφi =
g∈G
=
X
=
X
χ(g)χ(g)φ(g)φ(g)
g∈G
χ(g)χ(g)
g∈G
= hχ, χi = 1.
Thus χφ is an irreducible character of G.
3. Suppose a finite group G has the following two rows in its character table over C (where
ω is a primitive cube root of unity). Complete the rest of the table.
χ1
χ2
g1
1
2
g2
1
-2
g3
1
0
g4
ω2
-1
g5
ω
-1
g6
ω2
1
g7
ω
1
Solution: As there are seven conjugacy classes, there must be seven irreducible characters. Let χ0 be the trivial character. Using Problem #2, we have irreducible characters
χ3 = χ1 χ1 , χ4 = χ1 χ2 , and χ5 = χ3 χ2 . We obtain the last irreducible character χ6
using the orthogonality relations between columns g1 , g2 , and g3 , and then between
columns g3 with g5 , g6 , and g7 .
χ0
χ1
χ2
χ3
χ4
χ5
χ6
g1
1
1
2
1
2
2
3
g2
1
1
-2
1
-2
-2
3
g3
g4
1
1
1
ω2
0
-1
1
ω
0 −ω 2
0
−ω
−1
0
g5
1
ω
-1
ω2
−ω
−ω 2
0
g6
1
ω2
1
ω
ω2
ω
0
g7
1
ω
1
ω2
ω
ω2
0
4. Let G be a finite group and M and N finitely generated C[G]-modules. Let χ and φ
be the characters of G associated to M and N , respectively. Prove that
hχ, φi = dimC HomC[G] (M, N ).
(Hint: Let I1 , . . . , It denote the distinct (up to isomorphism) simple left ideals of C[G]
and let χ1 , . . . , χt be their associated characters. Then M ∼
= m1 I1 ⊕ · · · ⊕ mt It for some
integers mi ≥ 0 and χ = m1 χ1 + · · · + mt χt . Similarly for N .)
Solution: As in the Hint, let N ∼
= n1 I1 ⊕ · · · ⊕ nt It . Then φ = n1 χ1 + · · · + nt χt .
Using that hχi , χj i = δij , we get that hχ, φi = m1 n1 + · · · + mt nt . Now,
HomC[G] (M, N ) ∼
= ⊕ij mi nj HomC[G] (Ii , Ij ).
However, as Ii ∼
6= Ij for i 6= j, HomC[G] (Ii , Ij )) = 0 for i 6= j. Also, HomC[G] (Ii , Ii ) =
∼
EndC[G] (Ii ) = C, since Ii is simple (and thus the endomorphism ring is a division ring)
and C is algebraically closed (noting that the endomorphism ring is finitely generated
as C vector space). Thus, HomC[G] (M, N ) ∼
= m1 n1 C ⊕ · · · ⊕ mt nt C. Consequently,
dimC HomC[G] (M, N ) = m1 n1 + · · · + mt nt = hχ, φi.
5. A commutative ring R is said to be essentially of finite type over a field k if R is
the localization at some multiplicatively closed set of a finitely generated k-algebra.
Suppose R is a domain which is essentially of finite type over a field k. Prove that the
integral closure of R in its field of fractions is a finitely generated R-module. (Hint:
First note that R is the localization of an affine k-domain: We have R = TS where T is
a finitely generated k-algebra and S a multiplicatively closed set of T . Then T = A/I
where A is a polynomial ring over k and I an ideal of A. Lift S to a multiplicatively
closed set W of A. Then R = TS = (A/I)S = AW /IW . As R is a domain, IW is a prime
ideal of AW . Thus IW = PW for some prime P of A. Then R = AW /PW = (A/P )W .
Hence, R is the localization of A/P , which is an affine k-domain. Now use the result
we proved in class for affine k-domains.)
Solution: The Hint reduces the problem to the case that R is the localization of an
affine domain A, say R = AS , where S is some multiplicatively closed set of A. Let F
be the field of fractions of A. Then F is also the field of fractions of R = AS . Let T
be the integral closure of A in F . Then, by the result from class, we know that T is a
finitely generated A-module. Now, the integral closure of R = AS in F is TS . As T is
a finitely generated A-module, so TS is a finitely generated AS = R-module.
6. Let R be an integrally closed domain with field of fractions F . Let E/F be a finite
field extension and let S be the integral closure of R in E. Let Aut(S/R) denote the
group of ring automorphisms of S which fix R. Prove that Aut(E/F ) ∼
= Aut(S/R).
Solution: Let ψ ∈ Aut(E/F ). Note that if s ∈ S, then s is integral over R. Hence,
s is a root of a monic polynomial f (x) ∈ R[x]. Since ψ fixes F (and hence R),
ψ(s) is also a root of f (x). Hence, ψ(s) is integral over R and in E. Therefore,
ψ(s) ∈ S. As ψ −1 ∈ Aut(E/F ), ψ −1 (s) ∈ S. Hence, ψ(S) = S. Define a map
f : Aut(E/F ) → Aut(S/R) by f (ψ) = ψ|S , the restriction of ψ to S. Clearly, ψ is
a group homomorphism. Let W = R \ {0}. Then SW = F . (As SW is integral over
RW = F , SW is a field. Further, SW is the integral closure of RW = F in E. As E is
algebraic over F , we obtain that SW = E.) Suppose σ : S → S is an automorphism
fixing R. Then localizing at W , σ1 : SW → SW is an automorphism of E = SW fixing
F = RW ; i.e., σ1 ∈ Aut(E/F ). Note that f ( σ1 ) = σ. Hence, f is surjective. Similarly,
)
if τ ∈ Aut(E/F ) then f (τ
= τ . This implies that f is injective. Consequently, f is an
1
isomorphism.
7. With the notation as in the previous problem, let N = NFE be the norm function for
the extension E/F . Prove that for all s ∈ S, N (s) ∈ R. (Note: I am not assuming
E/F is separable.)
Solution: Let {σ1 , . . . , σt } be the set of distinct embeddings of E into E fixing F . Let
r = [E : F ]i . Then N = (σ1 · · · σt )r . By a result from class, N (α) ∈ F for every α ∈ E.
Let s ∈ S. As each σi fixes F , we see that σi (s) is integral over R. As the set of elements
of E which are integral over R forms a ring, we have that N (s) = (σ1 (s) · · · σt (s))r is
integral over R. But since N (s) ∈ F and R is integrally closed in F , we have that
N (s) ∈ R.