Tensor Algebra and Vector Spaces

Austin Alderete

Jul. 23rd, 2017

Let R be a ring with 1. 10.4.1) Let f : R → S be a ring homomorphism from the ring R to the ring S with f(1R) = 1S. Verify the details that sr = sf(r) deﬁnes a right R-action on S under which S is an (S,R)−bimodule.

Proof. Let f : R → S be a ring homomorphism with f(1R) = 1S. Deﬁne an right action on S by R by sr = sf(r). Observe that

s1R = sf(1R) = s1S = s

and for all a, b ∈ R

(sa)b = (sf(a))b = (sf(a))f(b) = sf(a)f(b) = sf(ab) = s(ab)

by the properties of a ring homomorphism. These satisfy the axioms for a right action on S. We will now check that this makes S an (S,R)−bimodule. We have that S is an abelian group, is naturally a left module over itself, and have veriﬁed that the above operation sr is a right R-action. All that remains to be checked are the distributive laws and the compatibility of the two structures. We have, for a, b ∈ R and s, t ∈ S, s(a + b) = sf(a + b) = s(f(a) + f(b)) = sf(a) + sf(b) = sa + sb

(s + t)a = (s + t)f(a) = sf(a) + tf(a) = sa + ta

and s(ta) = s(tf(a)) = stf(a) = (st)f(a) = (st)a

which completes the proof.

1 10.4.2) Show that the element 2 ⊗ 1 is 0 in Z ⊗Z Z/2Z but is nonzero in 2Z ⊗Z Z/2Z.

Proof. Consider 2 ⊗ 1 in Z ⊗Z Z/2Z. Then

2 ⊗ 1 = 1 ⊗ 1 + 1 ⊗ 1 = 1 ⊗ 2 = 1 ⊗ 0 or equivalently 2 ⊗ 1 = (1 · 2) ⊗ 1 = 1 ⊗ (2 · 1) = 1 ⊗ 0.

Now instead consider 2 ⊗ 1 in 2Z ⊗Z Z/2Z. We begin by considering the map

φ : 2Z × Z/2Z → Z/2Z ;(a, b) 7→ (a/2)b.

This is clearly linear in both factors, and moreover is well-deﬁned assuming that a is divided by 2 prior to descending to the quotient. Note that φ(2, 1) = 1 6= 0. As this map must factor through 2Z ⊗Z Z/2Z we have that 2 ⊗ 1 is not zero in 2Z ⊗Z Z/2Z.

10.4.3) Show that C ⊗R C and C ⊗C C are both left R−modules but are not isomorphic as R-modules. ∼ Proof. First, note that C ⊗C C = C. As for any simple tensor x ⊗ y ∈ C ⊗C C we can write

x ⊗ y = 1 · x ⊗ y = 1 ⊗ xy, that is the map

φ : C ⊗C C → C ; φ(x ⊗ y) = xy is an R-module isomorphism where C⊗C C is an R module by multiplication of the left factor (as the map φ shows, it’s really just multiplication by any factor.

Now consider the following elements in C ⊗R C: 1 ⊗ i and i ⊗ 1. These two elements are linearly independent and in fact, any element in C ⊗R C can be written uniquely in the form

a ⊗ i + bi ⊗ 1.

To show this, consider (x + iy) ⊗ (z + iw) ∈ C ⊗R C where x, y, z, w ∈ C. Then

(x + iy) ⊗ (z + iw) = x ⊗ (z + iw) + iy ⊗ (z + iw) = x ⊗ z + x ⊗ iw + iy ⊗ z + iy ⊗ iw

= xz ⊗ 1 + xw ⊗ i + iyz ⊗ 1 + iyw ⊗ i = (xz + iyz) ⊗ 1 + (xw + iyw) ⊗ i

2 ∼ so in fact, C⊗R C = C×C, but this isn’t a full proof of that. Rather, it’s just a demonstration that the two are diﬀerent- namely that there are linearly independent elements in the latter tensor product.

10.4.4) Show that Q ⊗Z Q and Q ⊗Q Q are isomorphic left Q−modules.

∼ a c Proof. Observe that, similar to before, Q⊗QQ = Q. Now for the simple tensor b ⊗ d ∈ Q⊗ZQ observe that we can write

a c a c b a c 1 a c 1 c 1 c a ⊗ = ⊗ = ⊗ b = b ⊗ a ⊗ = 1 ⊗ b d b d b b d b b d b d b d b

and so this is just Q in disguise (when we note that any arbitrary element is just the sum of simple tensors). We should note for future use that if we have the tensor product of fields of fractions over an integral domain, we can clear denominators easily and so it is no different than the tensor product over the field of fractions.

10.4.5) Let A be a ﬁnite abelian group of order n and let pk be the largest power of the k prime p dividing n. Prove that Z/p Z ⊗Z A is isomorphic to the Sylow p-subgroup of A.

Proof. Let A be a ﬁnite abelian group of order n and let pk be the largest power of the prime p dividing n. Note that A has a unique p-Sylow subgroup, call it P . For a ∈ A, if the order of a is a power of p, then a ∈ P . Moreover, as A is abelian, it factors as A = P × N for some subgroup N for which p does not divide |N|. k Now consider the map φ : A → Z/p Z ⊗Z A deﬁned by φ(a) = 1 ⊗ a. We will show that this map is onto with kernel N. k (Onto): Suppose we have z ⊗ a ∈ Z/p Z ⊗Z A. Then φ(a + ..(z) + ..a) = φ(za) = 1 ⊗ za = z ⊗ a. As it is onto its simple tensors, by the property of homomorphisms, it is onto the entire set. (Kernel): Suppose that n ∈ ker(φ). Then 1 ⊗ n = 0. We will show that this occurs exactly when |n| is coprime to p. Recall that |n| is a unit of Z/pkZ if and only if |n| is coprime to p. In other words,

1 ⊗ n = (|n|−1|n|) ⊗ n = |n|−1 ⊗ |n|n = 0

3 is a valid equation only when |n|−1 exists in Z/pkZ which is exactly when |n| is coprime to p. Thus, N = ker(φ). One can then write this as a map out of P × N with kernel 1 × N which implies that ∼ k P = Z/p Z ⊗Z A.

10.4.6) If R is any integral domain with quotient ﬁeld Q, prove that (Q/R)⊗R(Q/R) = 0.

Proof. Let R be an integral domain and let Q be its quotient ﬁeld. We consider an arbitrary a c simple tensor of (Q/R) ⊗R (Q/R), ( b + R) ⊗ ( d + R) where a, b, c, d ∈ R with b, d 6= 0. Then a c a + bR c + dR + R ⊗ + R = ⊗ b d b d

and recall that, as shown in 10.4.4, we can ‘pass’ fractions so that this is just

c + dR 1 ⊗ (a + bR) bd

and as a + bR ∈ R for all a ∈ R this is zero.

10.4.7) If R is any integral domain with quotient ﬁeld Q and N is a left R-module, prove

that every element of the tensor product Q ⊗R N can be written as a simple tensor of the form (1/d) ⊗ n for some nonzero d ∈ R and some n ∈ N.

a Proof. Let b ⊗ n be a simple tensor in Q ⊗R N. From previous work, it should be clear that 1 this is b ⊗ an. Therefore, any element in Q ⊗R N is of the form J J J X 1 X b1b2...bi−1bi+1...bJ X 1 ⊗ a n = ⊗ a n = ⊗ (b b ...b b ...b )a n b i i b b b ...b b ...b i i b ...b 1 2 i−1 i+1 J i i i=1 i i=1 i 1 2 i−1 i+1 J i=1 1 n

J ! 1 X = ⊗ b b ...b b ...b a n b ...b 1 2 i−1 i+1 J i i 1 n i=1 and the result is clear.

2 10.4.11) Let {e1, e2} be a basis of V = R . Show that the element e1 ⊗ e2 + e2 ⊗ e1 in 2 V ⊗R V cannot be written as a simple tensor v ⊗ w for any v, w ∈ R .

4 2 Proof. Let {e1, e2} be a basis of V = R . We will eventually see that dim(V ⊗ V ) =

dim(V ) dim(V ) = 4 and a basis for V ⊗ V is given by {e1 ⊗ e1, e1 ⊗ e2, e2 ⊗ e1, e2 ⊗ e2}. Suppose that there exist v, w ∈ R2 such that

e1 ⊗ e2 + e2 ⊗ e1 = v ⊗ w

in V ⊗ V . As e1, e2 form a basis for V we can write

v ⊗ w = (ae1 + be2) ⊗ (xe1 ⊗ ye2)

for a, b, x, y ∈ R. We can expand this using the rules of tensor products to obtain

v ⊗ w = (ae1 + be2) ⊗ (xe1 ⊗ ye2) = ae1 ⊗ xe1 + ae1 ⊗ ye2 + be2 ⊗ xe1 + be2 ⊗ ye2

which, under our hypothesis, forces the equality

e1 ⊗ e2 + e2 ⊗ e1 = ae1 ⊗ xe1 + ae1 ⊗ ye2 + be2 ⊗ xe1 + be2 ⊗ ye2

and equivalently

e1 ⊗ e2 + e2 ⊗ e1 = ax (e1 ⊗ e1) + ay (e1 ⊗ e2) + bx (e2 ⊗ e1) + by (e2 ⊗ e2) .

Therefore ax = 0, ay = 1, bx = 1, and by = 0. There are no such a, b, x, y ∈ R satisfying this system of equations and so v ⊗ w cannot exist (if a = 0, then ay = 0 6= 1; if x = 0, then bx = 0 6= 1).

For what follows, R is a commutative ring with 1, M is an R-module, F is a ﬁeld, and V is a ﬁnite dimensional vector space over F . 11.5.1) Prove that if M is a cyclic R-module then T(M) = S(M), i.e. the tensor algebra T(M) is commutative.

Proof. Let M be a cyclic R-module. Then there exists a ∈ M such that Ra = M, so every element is of the form ra for some r ∈ R. Observe that a k-tensor takes the form

v1 ⊗ ... ⊗ vk = r1a ⊗ ... ⊗ rka = (r1...rk)a ⊗ ... ⊗ a.

And so every simple tensor takes the form of a scalar product of the k-fold tensor product of a with itself. Observe that there is no distinction between the ith component and the jth component. In other words, every tensor is invariant under permutation. That is, S(M) = T(M). Alternatively, one can observe that the ideal to be quotiented is trivial.

5 11.5.3) Show that the image of the map Sym2 for the Z-module Z consists of the 2- tensors a(1 ⊗ 1) where a is an even integer. Conclude in particular that the symmetric

tensor 1 ⊗ 1 in Z ⊗Z Z is not contained in the image of the map Sym.

Proof. Let x ⊗ y ∈ Z ⊗Z Z. Then

Sym2(x ⊗ y) = (x ⊗ y + y ⊗ x) = (xy ⊗ 1 + yx ⊗ 1) = 2xy (1 ⊗ 1) .

The image of the map Sym2 is therefore a(1 ⊗ 1) for even a. The number 1 is not even and

so (1 ⊗ 1) 6∈ Im(Sym2). However, 1 ⊗ 1 is in S(Z).

11.5.4) Prove that if M is a free R-module of rank n then Vi(M) is a free R-module of n rank i for i = 0, 1, 2, ....