Caitlin Beecham

Adam Freilich

Keaton Naff

Matt Weaver

February 6, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework02
Let R be a commutative ring and G be a finite group.

The tensor product R ⊗

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 ⊗_{ℤ}R → R and
m_{ℤ[G]} : ℤ[G] ⊗_{ℤ}ℤ[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 ⊗_{ℤ}R) ⊗_{ℤ}(ℤ[G] ⊗_{ℤ}ℤ[G]) → R ⊗_{ℤ}ℤ[G]. Because of the commutativity and associativity of the
tensor product, there exists an isomorphism α : (R ⊗_{ℤ}ℤ[G]) ⊗ (R ⊗_{ℤ}ℤ[G]) → (R ⊗_{ℤ}R) ⊗_{ℤ}(ℤ[G] ⊗_{ℤ}ℤ[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 ⊗_{ℤ}ℤ[G]) ⊗ (R ⊗_{ℤ}ℤ[G]) → R ⊗_{ℤ}ℤ[G] is a linear map, which, via the universal property of tensor
products, yields a bilinear map m : (R ⊗_{ℤ}ℤ[G]) × (R ⊗_{ℤ}ℤ[G]) → R ⊗_{ℤ}ℤ[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 ∑
_{i}r_{i} ⊗ x_{i},∑
_{j}s_{j} ⊗ y_{j} ∈ R ⊗_{ℤ}ℤ[G]

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]}

Thus R ⊗_{ℤ}ℤ[G] has a ring structure.

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 ⊗

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,r_{1},r_{2} ∈ R,
x,x_{1},x_{2} ∈ R[G], and n ∈ ℤ

Let M and N be two left R-modules over a non-commutative ring R. Define M ⊙

Define α : M × N → M ⊙

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 ⊗_{ℤ}N,Q)
such that γ = γ′∘ i_{1}.

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

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

Let M and N each be left R-modules of rank 1 with generators m and n, respectively. Then for the R-module M ⊙

to light, from which we deduce that

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.

Begin with a non-commutative ring, say M

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

Because F is a field, then the group rings F[G×G] and F[G] are vector spaces over F with dimension (#G)

Again, because F is a field, F[G] and F[G] ⊗

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 ⊗

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

where n = #G and (a_{ij}) = A. Therefore we finally arrive at

The rings ℤ[x,y] and ℤ[x] ⊗

Define α : ℤ[x] × ℤ[x] → ℤ[x,y] by f(x),g(x)f(x)g(y). Then for all f,f_{1},f_{2},g,g_{1},g_{2} ∈ ℤ[x] and n ∈ ℤ we have the
following through heavy use of ring properties of ℤ[x,y].

Let c : ℤ[x] → ℤ[x] ⊗

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

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