## Math 503: Abstract Algebra

Homework 2 <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 [G] is already an abelian group, so we need to find a multiplication operation and show that (R [G],,1R 1[G]) is a monoid as well as the distributive law holds.

Constructing Multiplication We first recognize that multiplication in a ring Ris 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 [G] are both rings, there exist such associative, bilinear maps mR : R × R R and m[G] : [G] × [G] [G]. But then the universal property of tensor products yields mR : R R R and m[G] : [G] [G] [G] through which mR and m[G] factor, respectively. Thus we can combine these two to form the linear map mR m[G] : (R R) ([G] [G]) R [G]. Because of the commutativity and associativity of the tensor product, there exists an isomorphism α : (R [G]) (R [G]) (R R) ([G] [G]) such that α((r x) (s y)) = (r s) (x y) for all r,s R and x,y [G]. Now the composition mR m[G] α : (R [G]) (R [G]) R [G] is a linear map, which, via the universal property of tensor products, yields a bilinear map m : (R [G]) × (R [G]) R [G] through which mR m[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 iri xi, jsj yj R [G]

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

Thus R [G] has a ring structure.

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

Because R is a free R-module of rank one, [G] is a free -module of rank #G, and R[G] is a free R-module of rank #G, then we know immediately that R [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 × [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

by which α is -bilinear. Therefore the universal properties of tensor products yields α : R [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 iri xi and jsj yj in R [G]
Hence, R [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 N 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 N and the quotient map i2 : M N 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 -bilinear map. Thus the universal property of tensor products gives us the existence of a unique γ′∈ Homgrp(M N,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

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

Finally, the existence and uniqueness of both γand 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 imi ni M RN
while similarly
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

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.

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

Begin with a non-commutative ring, say M2(), and let M = N = . Set R to be the ring generated by elements of the form AB - BA for A,B M2(). 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:

#### (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 ρV (x) GL(V ) and ρW(x) GL(W) then we can define ρV W : G GL(V ) by ρV W(x) = ρV (x) ρW(x)

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

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

### 4

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

The rings [x,y] and [x] [x] are free -modules with bases {xn|n 0}∪{yn|n 0} and {xn 1|n 0}∪{1 xn|n 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 -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 : [x] × [x] [x,y] by f(x),g(x)f(x)g(y). Then for all f,f1,f2,g,g1,g2 [x] and n we have the following through heavy use of ring properties of [x,y].

which implies that α is -bilinear. Thus the universal property of tensor products gives us α : [x] [x] [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 -module isomorphism. We now only to multiplication to be preserved by α. This is shown by the following for arbitrary elements ifi gi,  jfj gj [x] [x]
Hence [x] [x] is naturally isomorphic to [x,y].

#### (b)

Let c : [x] [x] [x] be a ring homomorphism such that c(x) = x1 + 1 x for the polynomial x [x]. Therefore, for n , c(xn) = (c(x))n. Since {1,x,x2,} is a basis for [x], then c(x) = x 1 + 1 x completely defines c. Hence, c is unique.

#### (c)

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

Note that the last line could also be written as i=0nxi xn-i. We will make use of both. Then, for any f(x) =[x], c(f(x)) = f(x1 + 1 x) = nan i=0nxn-i xi when f(x) = nanxn. Given these results, we obtain the following

???? Seems like there should be a way to manipulate the coefficients above so that (c1) c results, but I can’t figure out how.