Math 503: Abstract Algebra

Homework 2

Lawrence Tyler Rush

<me@tylerlogic.com> In Collaboration With

Caitlin Beecham
Adam Freilich
Keaton Naff
Matt Weaver

February 6, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework02

1

Let R be a commutative ring and G be a finite group.

(a) Show that R ⊗_ℤℤ[G] has a structure as a ring

The tensor product R ⊗_ℤℤ[G] is already an abelian group, so we need to find a multiplication operation ⋅ and show that (R ⊗_ℤℤ[G],⋅,1_R ⊗ 1_ℤ[G]) is a monoid as well as the distributive law holds.

Constructing Multiplication We first recognize that multiplication in a ring R′ is some associative bilinear operation from R′× R′→ R′. It’s bilinear because of the distributive law, and associative so that the demands of the aforementioned monoid are met.

Thus because R and ℤ[G] are both rings, there exist such associative, bilinear maps m_R : R × R → R and m_ℤ[G] : ℤ[G] × ℤ[G] → ℤ[G]. But then the universal property of tensor products yields m_R : R ⊗_ℤR → R and m_ℤ[G] : ℤ[G] ⊗_ℤℤ[G] → ℤ[G] through which m_R and m_ℤ[G] factor, respectively. Thus we can combine these two to form the linear map m_R ⊗m_ℤ[G] : (R ⊗_ℤR) ⊗_ℤ(ℤ[G] ⊗_ℤℤ[G]) → R ⊗_ℤℤ[G]. Because of the commutativity and associativity of the tensor product, there exists an isomorphism α : (R ⊗_ℤℤ[G]) ⊗ (R ⊗_ℤℤ[G]) → (R ⊗_ℤR) ⊗_ℤ(ℤ[G] ⊗_ℤℤ[G]) such that α((r ⊗ x) ⊗ (s ⊗ y)) = (r ⊗ s) ⊗ (x ⊗ y) for all r,s ∈ R and x,y ∈ ℤ[G]. Now the composition m_R ⊗m_ℤ[G] ∘ α : (R ⊗_ℤℤ[G]) ⊗ (R ⊗_ℤℤ[G]) → R ⊗_ℤℤ[G] is a linear map, which, via the universal property of tensor products, yields a bilinear map m : (R ⊗_ℤℤ[G]) × (R ⊗_ℤℤ[G]) → R ⊗_ℤℤ[G] through which m_R ⊗m_ℤ[G] ∘ α factors. The map m is our desired multiplication.

Associativity of Multiplication Because m is bilinear, it satisfies the distributive laws and thus we need only show that m is associative. Also because m is bilinear, for arbitrary ∑ _ir_i ⊗ x_i,∑ _js_j ⊗ y_j ∈ R ⊗_ℤℤ[G]

( ) ∑ ∑ ∑ ∑ m ( ri ⊗ xi , sj ⊗yj) = m (ri ⊗ xi,sj ⊗ yj) i j i j

which implies that we need only determine that m(r ⊗x,m(s⊗y,t⊗z)) = m(m(r ⊗x,s⊗y),t⊗z) for all r,s,t ∈ R and x,y,z ∈ ℤ[G]. So finally, through the use of the associativity of m_R and m_ℤ[G] the following sequence of equations shows m to be associative. For clarity we set φ = m_R ⊗m_ℤ[G]

m(r⊗ x,m (s⊗ y,t⊗ z)) = φ∘ α((r ⊗ x)⊗ φ∘ α((s ⊗ y)⊗ (t ⊗z))) = φ∘ α((r ⊗ x)⊗ φ((s ⊗ t)⊗ (y-⊗z))) = φ∘ α((r ⊗ x)⊗ (mR(s ⊗t)⊗ m ℤ[G](y⊗ z))) = φ∘ α((r ⊗ x)⊗ (mR(s,t)⊗ mℤ[G](y,z))) = φ((r ⊗ m (s,t))⊗ (x ⊗ m (y,z))) --- R ----ℤ[G] = mR(r ⊗mR (s,t))⊗ m ℤ[G ](x⊗ m ℤ[G](y,z)) = mR(r,mR (s,t))⊗ mℤ[G](x,m ℤ[G](y,z)) = mR(mR (r,s),t)⊗ mℤ[G](m ℤ[G](x,y),z) = m-(m (r,s) ⊗t)⊗ m----(m (x,y)⊗ z) R R ℤ[G] ℤ[G] = φ((mR (r,s) ⊗ t)⊗(m ℤ[G](x,y)⊗ z)) = φ∘ α((mR (r,s) ⊗ mℤ[G](x,y))⊗ (t⊗ z)) = φ∘ α((mR-(r⊗ s)⊗ m-ℤ[G](x⊗ y))⊗ (t⊗ z)) = φ∘ α(φ((r ⊗ s) ⊗(x ⊗ y))⊗ (t⊗ z)) = φ∘ α(φ∘ α((r ⊗ x)⊗ (s ⊗ y))⊗ (t⊗ z)) = φ∘ α(m(r⊗ x,s⊗ y)⊗ (t⊗ z)) = m(m (r ⊗ x,s⊗ y),t⊗ z)

Thus R ⊗_ℤℤ[G] has a ring structure.

(b) Are R ⊗_ℤℤ[G] and R[G] isomorphic?

Because R is a free R-module of rank one, ℤ[G] is a free ℤ-module of rank #G, and R[G] is a free R-module of rank #G, then we know immediately that R ⊗_ℤℤ[G] and R[G] are isomorphic as R-modules, since their ranks are the same. Furthermore, any homomorphism which takes basis elements to distinct basis elements will be an R-linear isomorphism. Thus if we can find such an R-module isomorphism and go on to show that it preserves multiplication between elements of the domain and codomain, then it will also be a ring isomorphism. We endeavor to find such an isomorphism.

So define α : R × ℤ[G] → R[G] to be the map (r,x)rx. Through use of the properties of R[G], for all r,r₁,r₂ ∈ R, x,x₁,x₂ ∈ R[G], and n ∈ ℤ

α(r + r ,x) = (r + r)x = r x+ r x = α (r ,x)+ α(r ,x) 1 2 1 2 1 2 1 2 α(r,x1 + x2) = r(x1 + x2) = rx1 + rx2 = α(r,x1)+ α (r,x2) α(nr,x ) = (nr)x = signum (n)(r+-⋅⋅⋅+-r)x = signum (n)r◟x+-⋅◝⋅◜⋅+-rx◞= signum (n)r(x+-⋅⋅⋅+-x)= r(nx) = α(r,nx) ◟ |n|◝ t◜imes ◞ |n| times ◟ |n◝|◜ times ◞

by which α is ℤ-bilinear. Therefore the universal properties of tensor products yields α : R ⊗ ℤ[G] → R[G] such that α = α ∘ i where i is the inclusion map. Moreover, α is an isomorphism of modules due to it’s mapping basis elements to distinct basis elements: α(1 ⊗ [g]) = α(1,[g]) = [g] for each g ∈ G. It remains to be shown that α preserves the operation of multiplication. The following yields that fact for arbitrary elements ∑ _ir_i ⊗ x_i and ∑ _js_j ⊗ y_j in R ⊗ ℤ[G]

( (∑ ) (∑ ) ) ( ∑ ∑ ) α-( ri ⊗ xi ( sj ⊗ yj) ) = α( (ri ⊗ xi)(sj ⊗yj)) i j i j ( ) -( ∑ ∑ ) = α i j (risj ⊗ xiyj) ∑ ∑ = α(risj ⊗ xiyj) i j = ∑ ∑ α(rs ,x y) i j ij ij ∑ ∑ = risjxiyj i j ( ) (∑ ) ∑ = rixi ( sjyj) i j ( ) ( ) = ∑ α(ri,xi) ( ∑ α(sj,yj)) i j ( ) ( ) ∑ -- ∑ -- = α(ri ⊗ xi) ( α(sj ⊗ yj)) i ( j ) ( ∑ ) ∑ = α- ri ⊗ xi α-( sj ⊗yj) i j

Hence, R ⊗_ℤℤ[G] and R[G] are isomorphic rings.

2

Let M and N be two left R-modules over a non-commutative ring R. Define M ⊙_RN to be the quotient of M ⊗_ℤN by it’s submodule which is generated by all elements of the form (r ⋅ m) ⊗ n - m ⊗ (r ⋅ n) where m ∈ M, n ∈ N, and r ∈ R. We will refer to this submodule as S.

(a)

Define α : M × N → M ⊙_RN to be the compositions of the canonical map i₁ : M × N → M ⊗_ℤN and the quotient map i₂ : M ⊗_ℤN → M ⊙_RN.

Let Q be an R-module and γ : M × N → Q be a R-bilinear map. Therefore Q is an abelian group and γ is also a ℤ-bilinear map. Thus the universal property of tensor products gives us the existence of a unique γ′∈ Hom_grp(M ⊗_ℤN,Q) such that γ = γ′∘ i₁.

Now since S is generated by elements of the form (r ⋅ m) ⊗ n - m ⊗ (r ⋅ n) where m ∈ M, n ∈ N, and r ∈ R and

γ′((r⋅m )⊗ n - m ⊗ (r ⋅n)) = γ′((r⋅m )⊗ n)- γ′(m ⊗ (r⋅n)) = γ(r ⋅m,n)- γ(m, r⋅n) = rγ (m, n)- rγ(m,n) = 0

then S ≤ kerγ′, by which the universal property of quotient modules yields a unique map β ∈ Hom_R(M ⊙_RN,Q) such that γ′ = β ∘ i₂. Hence γ = γ′∘ i₂ = β ∘ i₂ ∘ i₁ = β ∘ α.

Finally, the existence and uniqueness of both γ′ and β, demand that the map ββ ∘ α is bijective.

(b) What is M ⊙_RN when M, N are R-modules of rank one

Let M and N each be left R-modules of rank 1 with generators m and n, respectively. Then for the R-module M ⊙_RN and for an arbitrary element ∑ _im_i ⊙ n_i ∈ M ⊙_RN

∑ ∑ mi ⊙ ni = rim ⊙ sin i i = ∑ sr m ⊙ n i ii ( ) = ∑ siri (m ⊙ n) i

while similarly

∑ ∑ mi ⊙ ni = rim ⊙ sin i ∑i = m ⊙ risin i = ∑ ((r s)m ⊙ n) i i i ( ) = ∑ risi (m ⊙ n) i

Either one of these implies that M ⊙_RN is isomorphic as a R-module to a subring of R, not necessarily proper. However, combining the two results brings

( ) ( ) ∑ ∑ siri (m ⊙ n) = risi (m ⊙ n) i i

to light, from which we deduce that

M ⊙ N ~ R∕S R =

where S is the subring of R generated by elements of the form rs-sr. Note that since R is non-commutative, then S will not simply be zero.

(c) Give an R such that M ⊙_RN is zero for M,N in (b)

Begin with a non-commutative ring, say M₂(ℝ), and let M = N = ℝ. Set R to be the ring generated by elements of the form AB - BA for A,B ∈ M₂(ℝ). Then certainly, given the previous part of this problem, the R-module M ⊙_RN will be zero because it will be isomorphic to R ∕ R in this case.

3

Let G be a finite group with subgroup H. Let F be a field.

(a)

Because F is a field, then the group rings F[G×G] and F[G] are vector spaces over F with dimension (#G)² and #G, respectively. Hence F[G] ⊗_FF[G] is also a vector space of dimension (#G)². So F[G × G] and F[G] ⊗_FF[G] are isomorphic. Furthermore, because {[(x,y)]|x,y ∈ G} is a basis for F[G × G] and {[x] ⊗ [y]|x,y ∈ G} is a basis for F[G] ⊗_FF[G], then the map from F[G×G] to F[G] ⊗_FF[G] defined by [(x,y)]

[x] ⊗ [y] is an isomorphism, and uniquely so.

(b)

Again, because F is a field, F[G] and F[G] ⊗_FF[G] are vector spaces over F. Furthermore, because {[x]|x ∈ G} and {[x] ⊗ [y]|x,y ∈ G} are bases for F[G] and F[G] ⊗_FF[G], respectively, then defining α : F[G] → F[G] ⊗_FF[G] by [x]

[x] ⊗ [x] makes α the unique injective linear homomorphism from F[G] to F[G] ⊗_FF[G]. In order for α to be an F-algebra homomorphism, it remains only to prove that α([x][y]) = α([x])α([y]) for all [x],[y] ∈ F[G]. The following yields that property:

α ([x][y]) = α([xy]) = [xy]⊗ [xy] = [x][y]⊗ [x][y] = ([x]⊗ [x])([y]⊗ [y]) = α([x])α ([y])

(c)

Since V and W are left F[G]-modules, then they a free modules each of rank #G. Since the tensor product of free modules is also a free module, then V ⊗_FW is also a left R[G]-module, however, it has rank (#G)².

Fix an x ∈ G. Given that ρ_V(x) ∈ GL(V ) and ρ_W(x) ∈ GL(W) then we can define ρ_{V ⊗W} : G → GL(V ) by ρ_{V ⊗W}(x) = ρ_V(x) ⊗ ρ_W(x)

So set A to be the matrix representation of ρ_V(x) in the basis {[g]|g ∈ G}. Similarly, set B to be the matrix representation of ρ_W(x) in the same basis. Also set C to be the matrix representation of ρ_{V ⊗W}(x) in the basis {[g] ⊗ 1|g ∈ G}∪{1 ⊗ [g]|g ∈ G}. Then

( a11B a12B ⋅⋅⋅ a1nB ) | a21B a22B | C = || . . || ( .. .. ) an1B an2B annB

where n = #G and (a_ij) = A. Therefore we finally arrive at

TrF(ρV⊗W (x )) = TrF (C( ) ) ∑n = TrF aiiB n i=1 = ∑ a Tr (B) i=1 ii F ∑n = TrF (B ) aii i=1 = TrF (B )TrF(A) = Tr (A )Tr (B) F F = TrF (ρV (x))TrF(ρW (x))

(d) Extra Credit

4

(a) Show that there is a natural ring isomorphism between ℤ[x,y] and ℤ[x] ⊗_ℤℤ[x]

The rings ℤ[x,y] and ℤ[x] ⊗_ℤℤ[x] are free ℤ-modules with bases {xⁿ|n ∈ ℤ_≥0}∪{yⁿ|n ∈ ℤ_≥0} and {xⁿ ⊗ 1|n ∈ ℤ_≥0}∪{1 ⊗ xⁿ|n ∈ ℤ_≥0}, respectively. These two bases have the same cardinality, so any homomorphism that maps the elements of one basis to distinct elements of the other will be a ℤ-module isomorphism. So once we find such a module isomorphism, we need only show that it preserves the multiplication operation in order to obtain a ring isomorphism.

Define α : ℤ[x] × ℤ[x] → ℤ[x,y] by f(x),g(x)f(x)g(y). Then for all f,f₁,f₂,g,g₁,g₂ ∈ ℤ[x] and n ∈ ℤ we have the following through heavy use of ring properties of ℤ[x,y].

α(f1(x)+ f2(x),g(x)) = (f1(x) +f2(x))g(y) = f1(x)g(y)+ f2(x)g(y) = α(f (x),g(x))+ α(f (x),g(x)) 1 2 α (f (x),g1(x)+ g2(x)) = f(x)(g1(y)+ g2(y)) = f(x)g1(y)+ f(x)g2(y) = α(f(x)g1(y))+ α(f(x)g2(y)) α(nf(x),g(x)) = signum (n)(f(x )+ ⋅⋅⋅+ f(x))g(y) ◟------◝◜------◞ |n| times = signum (n)f(x)g(y)+-⋅⋅⋅+-f(x)g(y) ◟ |n|◝◜ times ◞ = signum (n)f(x )(g◟(y)+-⋅◝⋅◜⋅+-g(y))◞ |n| times = α(f(x),ng(x ))

which implies that α is ℤ-bilinear. Thus the universal property of tensor products gives us α : ℤ[x] ⊗ ℤ[x] → ℤ[x,y] such that α = α ∘ i where i is the inclusion map. Now α(xⁿ ⊗ 1) = α(xⁿ,1) = xⁿ and α(1 ⊗ xⁿ) = α(1,xⁿ) = yⁿ, making α a ℤ-module isomorphism. We now only to multiplication to be preserved by α. This is shown by the following for arbitrary elements ∑ _if_i ⊗ g_i, ∑ _jf_j ⊗ g_j ∈ ℤ[x] ⊗_ℤℤ[x]

( (∑ ) (∑ ) ) ( ∑ ∑ ) α-( fi ⊗ gi ( fj ⊗ gj) ) = α( (fi ⊗ gi)(fj ⊗ gj)) i j i j ( ) -( ∑ ∑ ) = α i j (fifj ⊗ gigj) ∑ ∑ = α(fifj ⊗ gigj) i j = ∑ ∑ α(ff ,gg ) i j ij ij ∑ ∑ = fi(x)fj(x)gi(y)gj(y) i j ( ) (∑ ) ∑ = fi(x)gi(y) ( fj(x )gj(y)) i j ( ) ( ) = ∑ α(fi,gi) (∑ α (fj,gj)) i j ( ) ( ) ∑ -- ∑ -- = α(fi ⊗ gi) ( α(fj ⊗ gj)) i ( j ) ( ∑ ) ∑ = α- fi ⊗ gi α-( fj ⊗ gj) i j

Hence ℤ[x] ⊗_ℤℤ[x] is naturally isomorphic to ℤ[x,y].

(b)

Let c : ℤ[x] → ℤ[x] ⊗_ℤℤ[x] be a ring homomorphism such that c(x) = x⊗ 1 + 1 ⊗x for the polynomial x ∈ ℤ[x]. Therefore, for n ∈ ℤ, c(xⁿ) = (c(x))ⁿ. Since {1,x,x²,…} is a basis for ℤ[x], then c(x) = x ⊗ 1 + 1 ⊗ x completely defines c. Hence, c is unique.

(c)

First note that for n ∈ ℤ we have a formula for c(xⁿ)

∑n ( ) n∑ ( ) ∑n ( ) c(xn) = (x ⊗ 1+ 1 ⊗x )n = n (x⊗ 1)n-i(1 ⊗ x)i = n (xn- i ⊗1)(1⊗ xi) = n xn-i ⊗ xi i=0 i i=0 i i=0 i

Note that the last line could also be written as ∑ _i=0ⁿ ( )
ni xⁱ ⊗x^n-i. We will make use of both. Then, for any f(x) =∈ ℤ[x], c(f(x)) = f(x⊗ 1 + 1 ⊗x) = ∑ _na_n ∑ _i=0ⁿ (n)
i x^n-i ⊗xⁱ when f(x) = ∑ _na_nxⁿ. Given these results, we obtain the following

α∘ (1⊗ c)∘c(f) = α(1(⊗ c(f((x⊗ 1+ 1 ⊗ x))) ) ) ∑ ∑n (n ) = α 1⊗ c an i xn- i ⊗ xi ( ( n i=0 )) ∑ ∑n ( (n) n-i) i = α 1⊗ c an i x ⊗ x ( (n i=(0) ) ) ∑ ∑n n n-i ( i) = α an i x ⊗ c x ( n i=n0( ( ) ) ( i ( ) )) ∑ ∑ n n-i ∑ i i-k k = α n i=0 an i x ⊗ k x ⊗ x ( n i ( )( ) k=0 ) = α ∑ ∑ ∑ a n i (xn- i ⊗(xi-k ⊗xk)) n i=0k=0 n i k n i ( )( ) = ∑ ∑ ∑ an n i ((xn-i ⊗ xi-k)⊗ xk) n i=0k=0 i k

???? Seems like there should be a way to manipulate the coefficients above so that (c⊗ 1) ∘c results, but I can’t figure out how.