Math 503: Abstract Algebra

Homework 1
Lawrence Tyler Rush
<me@tylerlogic.com>
February 1, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework01

Some Language. Similar to Dummit and Foote [DF04, p. 365], for left R-module N, right R-module M and abelian group L, we will call a map α : M × N → L R-balanced if it satisfies all three of

α(m1 + m2,n) = α (m1, n)+ α(m2,n) α(m,n1 + n2) = α(m, n1)+ α(m,n2) α(mr,n) = α(m,rn)
for m,m1,m2 ∈ M, n,n1,n2 ∈ N, and r ∈ R.

1

Let V and W be modules over a commutative ring R. Then according to the definition of tensor products, we have the existence of balanced maps α : V ×W → V ⊗RW and α′ : W ×V → W ⊗RV . Since V ×W and W × V are isomorphic, we have an isomorphism i : V × W → W × V , and so setting β = α′∘ i we have that β is a balanced map from V ×W to W ⊗RV . Hence, as W ⊗RV is an abelian group (it’s an R-module), the universal property of the tensor product gives us that there exists a unique homomorphism s : V ⊗RW → W ⊗RV such that the diagram
V × W V ⊗R W βαs W ⊗R V

commutes. Furthermore, since α(v,w) = v ⊗ w and β(v,w) = w ⊗ v, then s(v ⊗ w) = w ⊗ v for each v ∈ V and w ∈ W.

Finally, since R is commutative, the above argument symmetrically holds when V and W are swapped as each is a left and right R-module. This implies the existence of a homomorphism s′ : W ⊗RV → V ⊗RW such that s′(w ⊗ v) = v ⊗ w. Hence, s above is invertible and therefore an isomorphism. 1

2

(a)

Since under standard scalar multiplication, Rn is a left F-module. Furthermore for a ∈ F, r ∈ Rn and A ∈ Mn(F) we have that a(rA) = (ar)A by standard linear algebra rules. Hence Rn is an F-Mn(F)-bimodule. This then gives Rn ⊗FCn structure as a right F-module, but since F is a field, then Rn ⊗FCn is a F-vector space.

(b)

First let us define φ : Rn ×Cn → F by φ(r,c) = rc for each r ∈ Rn and c ∈ Cn. This is essentially the dot product, and so we know it to be bilinear, but just for the sake of completeness:
φ (rA,c) = (rA )c = r(Ac ) = φ(r,Ac) φ(r1 + r2,c) = (r1 + r2)c = r1c+ r2c = φ(r1,c)+ φ(r2,c) φ(r,c1 + c2) = r(c1 + c2) = rc1 + rc2 = φ(r,c1)+ φ(r,c2)
This then implies that φ is Mn(F)-balanced. Hence the universal property of tensor products allots us the existence of a unique ϕ : Rn ⊗Mn(F)Cn → F such that φ = ϕ ∘ i where i is the normal “inclusion” map. With this we have two helpful lemmas.

Lemma 2.1. Let ϕ be defined as it is above. For all r ∈ Rn and c ∈ Cn, if ϕ(r ⊗ c) = 0 then r ⊗ c = 0 ∈ Rn ⊗Mn(F)Cn.

Proof. Let the initial conditions of the Lemma’s statement stand. Denote the components of r and c by r1,…,rn and c1,…,cn, respectively. Since ϕ(r ⊗c) = rc = 0 then for each i, ri = 0 or ci = 0. Let i1,…,ik be the indices of the components of r which are zero. Therefore cj = 0 for j ⁄∈{i1,…,ik} Let A ∈ Mn(R) be the matrix with ones along the diagonal, except in rows i1,…,ik which shall be completely zero. Then we have that

r ⊗ c = rA ⊗ c = r⊗ Ac = r⊗ 0 = 0

__

Lemma 2.2. For every element x ∈ Rn ⊗Mn(F)Cn there exist elements r ∈ Rn and c ∈ Cn such that x = r ⊗ c.

Proof. As Rn ⊗Mn(F)Cn is made up of finite linear combinations of elements of the form r ⊗ c, it suffices to only show that any two such elements can be combined into one. So let r1 ⊗ c1 + ⋅⋅⋅ + r2 ⊗ c2 be arbitrary in Rn ⊗Mn(F)Cn. Then by denoting the individual components of r1 by r11,…,r1n and likewise for r2, we can define A ∈ Mn(F) to be

( -1 ) | r21 r11 | A = ( ... ) r-2n1r1n

Note that for clarity, we neglect to address the case of a component of r2 being zero, in which case we would set the corresponding diagonal entry in A to zero, and the remainder of the proof holds. Therefore we have

r1 ⊗ c1 + r2 ⊗ c2 = r2A ⊗ c1 + r2 ⊗ c2 = r2 ⊗ Ac1 + r2 ⊗ c2 = r2 ⊗ (Ac1 + c2)

__

Now let r1 ⊗c1 + ⋅⋅⋅ + rn ⊗cn in Rn ⊗Mn(F)Cn be such that ϕ(r1 ⊗c1 + ⋅⋅⋅ + rn ⊗cn) = 0. By Lemma 2.2 there are r ∈ Rn and c ∈ Cn such that r1 ⊗ c1 + ⋅⋅⋅ + rn ⊗ cn = r ⊗ c. By Lemma 2.1 r1 ⊗ c1 + ⋅⋅⋅ + rn ⊗ cn = 0 since ϕ(r1 ⊗ c1 + ⋅⋅⋅ + rn ⊗ cn) = ϕ(r ⊗ c) = 0.

Hence ϕ is an injective linear transformation with a 1-dimensional vector space as a codomain. Then, since ϕ is not the zero map, e.g.

ϕ((1,0,...,0)⊗ (1,0,...,0)t) = 1 ⁄= 0

Rn ⊗Mn(F)Cn must also be a one dimensional vector space. Furthermore, in light of Lemma 2.2, ϕ is our explicit isomorphism we need, mapping r ⊗ c to rc.

3

Let F be a field and U, V , and W be vectors spaces over F. Furthermore let u,u1,u2 ∈ U, v,v1,v2 ∈ V , and w,w1,w2 ∈ W.

(a)

Let β : U × V × W → U ⊗F(V ⊗FW) be the map defined by β(u,v,w) = u ⊗ (v ⊗ w). By properties of the tensor product, we have
β (au + u′,v,w ) = (au + u′)⊗ (v ⊗w ) ′ = ((au)⊗ (v⊗ w )) +(u ⊗ (v⊗ w )) = a(u⊗ (v⊗ w ))+ β(u′,v,w) = aβ(u,v,w )+ β(u′,v,w)
β(u,av+ v′,w) = u ⊗ ((av+ v′)⊗ w) = u ⊗ ((av⊗ w )+ v′ ⊗ w) ′ = u ⊗ (a(v⊗ w )+ v ⊗ w) = u ⊗ (a(v⊗ w ))+ u⊗ (v′ ⊗ w) = a(u⊗ (v⊗ w ))+ β(u,v′,w) = aβ(u,v,w)+ β(u,v′,w)
and
β(u,av+ v′,w ) = u⊗ (v⊗ (aw + w′)) ′ = u⊗ ((v⊗ aw) +v ⊗ w ) = u⊗ (a(v⊗ w) +v ⊗ w′) = u⊗ (a(v⊗ w))+ u ⊗ (v ⊗ w′)) = a(u⊗ (v⊗ w))+ β(u,v,w′) ′ = aβ(u,v,w )+ β(u,v,w )

(b)

Let T : U ×V ×W → X be a F-trilinear map where X is some F-vector space. Then for a fixed u ∈ U, we can define Tu : V × W → X by Tu(v,w) = T(u,v,w). Since T is F-trilinear, then Tu is F-balanced. Thus, since X is an abelian group, the universal property of tensor products admits a unique group homomorphism φu : V ⊗ W → X such that
Tu = φu ∘i
(3.1)

where i : V × W → V ⊗ W is the “inclusion” map (v,w)↦→v ⊗ w; i.e. the diagram

V × W V ⊗ W Tiφuu X

commutes.

Next, we will use these φu maps to obtain the f for which we’re looking. So define ϕ : U × (V ⊗ W) → X by ϕ(u,v,w) = φu(v ⊗ w). Then equation 3.1 gives us

ϕ(u +u ,v ⊗ w) = φ (v⊗ w) 1 2 u1+u2 = Tu1+u2(v,w ) = T(u1 + u2,v,w) = T(u1,v,w)+ T (u2,v,w) = T (v,w )+ T (v,w ) u1 u2 = φu1(v⊗ w) + φu2(v ⊗ w) = ϕ(u1,v⊗ w) +ϕ (u2,v ⊗w )
through the use of the F-trilinearity of T. We also have
ϕ(au,v⊗ w ) = φau(v⊗ w) = Tau(v,w ) = T(au,v,w) = T(u,av,w) = Tu(av,w) = φu(av⊗ w ) = ϕu(a(v ⊗ w))
by the F-trilinearity of T, properties of ⊗, and equation 3.1. Finally, the homomorphic properties of each φu give us
ϕ(u,v1 ⊗ w1 +v2 ⊗ w2) = φu(v1 ⊗ w1 +v2 ⊗ w2) = φu(v1 ⊗ w1)+ φu(v2 ⊗ w2) = ϕ(u,v1 ⊗ w1)+ ϕ(u,v2 ⊗ w2)
Given these three equations above, we conclude that ϕ is F-balanced. Hence the universal property of tensor products yields a unique group homomorphism f : U ⊗ (V ⊗ W) → X (i.e. a linear map) such that ϕ = f ∘ i′ is the inclusion map (u,v ⊗ w)↦→(u ⊗ (v ⊗ w)). So by defining ψ : U × (V × W) → U × (V ⊗ W) as ψ(u,v,w) = (u,v ⊗ w), we have ϕ ∘ ψ = f ∘ i′∘ ψ. However, since
ϕ∘ e(u,v,w ) = ϕ(u,v⊗ w) = φu(v⊗ w) = Tu(v,w ) = T(u,v,w)

and

i′ ∘ψ(u,v,w) = i′(u,v ⊗ w) = u ⊗ (v ⊗ w)

the equation ϕ ∘ ψ = f ∘ i′∘ ψ implies T = f ∘ β.

(c)

Due to the symmetry of the situation, it is a laborious plug-and-chug operation to prove that

Given a F-trilinear map T : U × V × W → X where X is some F-vector space, there exists a unique F-linear map f : (U ⊗ V ) ⊗ W → X such that T = f ∘ β

given that we already have the above proof in our hands. So we omit the proof and simply admit the above statement as fact. As there is already grounds for multi-linear universal property of tensor products [Lan02, p. 603], we will overload “the universal property of tensors products” by referring to the above statement, the statement proven in the previous part of this problem, and the original two-dimensional property as “the universal property of tensors products”.

So by part (a) of this problem, we have that β : U ×V ×W → U ⊗ (V ⊗W) is F-trilinear. The same argument holds, by shuffling around parentheses, for β′ : U ×V ×W → (U ⊗V ) ⊗W, implying that it is F-trilinear. Therefore, the universal property of tensor products implies that there exist unique group homomorphisms α′ : (U ⊗V ) ⊗W → U ⊗ (V ⊗W) and α : U ⊗ (V ⊗W) → (U ⊗V ) ⊗W such that α′((u⊗v) ⊗w) = β(u,v,w) and α(u⊗ (v ⊗w)) = β′(u,v,w), i.e. the following diagrams commute

U × V ×W U ⊗ (V ⊗ W ) U × V × W (U ⊗ V )⊗ W ββα ′ (U ⊗ V) ⊗ W ββα′′ U ⊗ (V ⊗ W )

But the uniqueness implies that α′ and α are inverses of each. Thus α is the desired isomophism between U ⊗ (V ⊗W) and (U ⊗ V ) ⊗ W.

4

(a)

Let’s start off by defining Φφϕ : U × V → F for φ ∈ U∨ and ϕ ∈ V ∨ by Φφϕ(u,v) = φ(u)ϕ(v). This map is F-bilinear by the following three equations.
Φφϕ(u1 + u2,v) = φ(u1 + u2)ϕ(v) = φ(u1)ϕ (v)+ φ (u2)ϕ(v) = Φφϕ(u1,v)+ Φφϕ(u2,v) Φ (u,v + v ) = φ(u)ϕ(v + v) φϕ 1 2 1 2 = φ(u)ϕ(v1)+ φ (u)ϕ(v2) = Φφϕ(u,v1)+ Φφϕ(u,v2) Φφϕ(au,v) = φ(au)ϕ(v) = rφ(u)ϕ (v) = φ(u)ϕ(av) = Φφϕ(u,av)
Thus the universal property of tensor products gives us a unique linear map Φφϕ : U ⊗V → F through which Φ factors. We use this in the following argument.

Now define f : U∨× V ∨→ (U ⊗ V )∨ by f(φ,ϕ) = Φφϕ. We yet again have an F-bilinear map here by the following equations, using what we have shown above.

f(φ + φ ,ϕ)(u,v) = Φ (u,v) 1 2 (φ1+φ2)ϕ = φ1(u )ϕ(v)+ φ2(u)ϕ(v) = Φφ1ϕ(u,v) +Φ φ2ϕ(u,v) = f(φ1,ϕ )(u,v)+ f(φ2,ϕ)(u,v) f(φ,ϕ + ϕ )(u,v) = Φ (u,v) 1 2 φ(ϕ1+ϕ2) = φ(u)ϕ1(v)+ φ(u)ϕ2(v) = Φφϕ1(u,v) +Φ φϕ2(u,v) = f(φ, ϕ1)(u,v)+ f(φ,ϕ2)(u,v) f(aφ,ϕ)(u,v) = Φ (u,v) (aφ)ϕ = (aφ )(u)ϕ(v) = a(φ (u)ϕ(v)) = φ(u)(aϕ)(v) = Φ (u,v) φ(aϕ) = f(φ, aϕ)(u,v)
Thus the universal property of tensor products gives us a unique linear map f : U∨⊗ V ∨→ (U ⊗ V )∨ through which f factors.

Now we define g : (U ⊗ V )∨→ U∨⊗ V ∨ (this will be our inverse of f) to be the map

-- α ↦→ (u ↦→ α(u,1))⊗ (v ↦→ α (1,v))

where α : U ×V → F is the map associated with α according to the universal property of tensor products. By the following we see that g is the inverse of f

g(f(φ ⊗ ϕ)) = g(Φ ) φϕ = (u ↦→ Φ φϕ(u,1))⊗ (v ↦→ Φφϕ(1,v)) = (u ↦→ φ (u))⊗ (v ↦→ ϕ(v)) = φ⊗ ϕ
and so we have that f is the isomorphism which we desire.

(b) Extra Credit

References

[DF04]    D.S. Dummit and R.M. Foote. Abstract Algebra. John Wiley & Sons Canada, Limited, 2004.

[Lan02]   S. Lang. Algebra. Graduate Texts in Mathematics. Springer New York, 2002.