Algebras, Tensor Product of Algebras

Algebra iff Associative Homomorphism

We're going to prove that the tensor product of two R algebras is another R algebra, but first we need a lemma about associative homomorphisms.

Throughout this page, R is commutative, else the tensor product of R modules wouldn't even be an R module, much less an R algebra.

Let S be an R algebra. Let T be S×S, the tensor product of S and S. If S was commutative, and if T was viewed as an S module, it would simply be S, because S×S = S. But S need not be commutative, and in any case, we're viewing S and T as R modules. For instance, if S is R2, the direct product of R and R, then T becomes R4.

Remember that S is a ring, possibly without 1, and multiplication in S implements a bilinear map from S cross S into S. Use (xa)y = x(ay) in S to pass the action of R between the two operands, as required by bilinearity.

Since T is universal, there is a module homomorphism h(T) into S that agrees with multiplication in S. Furthermore, h(h(xy)z) = h(xh(yz)) = xyz in S, whence the application of h is associative. As long as S is an R algebra, such an h exists.

Now assume the converse. Let S be an R module, and let T = S×S. Let h be an R module homomorphism from T into S, such that for every x y and z in S, h(h(xy)z) = h(xh(yz)).

Define multiplication in the module S as x*y = h(xy). Remember that xy is a pair generator in the tensor product T. Multiplication actually takes x and y in S cross S, builds xy in T, and applies h to find the product xy in S.

We need to verify the properties of a ring. Addition is already an abelian group, since S is an R module, so look at multiplication.

Consider xy+xz in S. Every tensor product is bilinear, whence xy+xz = x(y+z) in T. Since h is a module homomorphism it respects addition, thus h(xy)+h(xz) = h(xy+xz) = h(x(y+z)). Translate to multiplication in S, and xy+xz = x*(y+z). Multiplication distributes over addition.

Since 0 crossed with anything yields 0 in T, 0 times anything in S equals 0.

Use h to derive the product xy, then multiply by z, running through h a second time. This is the same as x times yz, whence multiplication is associative. That's enough to prove S is a ring, although it may not be a ring with 1.

Now let's show S is an R algebra. If x and y are in S and a is in R, (xa)y is equal to x(ay) in T. This because we can pass the action of R between x and y in the tensor product. Applying h produces the same value in S.

Next look at the action of a on xy. Since h is a module homomorphism this is the same as h(axy). Now a acts on T by modifying either of the two components, so this is the same as h((ax)y), or (ax)*y. This satisfies the criteria of an R algebra, as described in the introduction.

The Tensor Product of two Algebras

We can use the above to prove an important theorem - the tensor product of finitely many algebras yields another algebra.

Let S1 S2 S3 … Sn be R algebras, whose tensor product is T. Let U = T×T. Since tensoring is commutative and associative, we can regroup.

U = (S1×S1) × (S2×S2) × (S3×S3) … (Sn×Sn)

We're really talking about an isomorphism here. In other words, T×T is isomorphic to U, which is given by the above formula.

Since S1 is an R algebra there is a module homomorphism h1 from S1×S1 into S1 with the associative property. find such a homomorphism hi for each algebra Si.

Tensor the domains of these homomorphisms to get U, and tensor the ranges to get T. This induces a new homomorphism h(U) into T that is compatible with all the component homomorphisms.

Let g be the isomorphism from T×T onto U, followed by h.

A point x in T can be represented by x1x2x3…xn, elements drawn from S1, S2, S3 and so on up to Sn. Similarly, y = y1y2y3…yn. Consider g(xy). The isomorphism rearranges generators, giving x1y1x2y2x3y3…xnyn in U. Then h is applied, giving h1(x1y1)h2(x2y2)h3(x3y3)…hn(xnyn). This gives an element in T.

Cross this with z = z1z2z3…zn, another element of T, and apply g again. The ith component is hi(hi(xiyi)zi). Yet this is the same as hi(xihi(yizi)). Put this all together and g(g(xy)z) = g(xg(yz)), whence T is an R algebra.

In summary, the tensor product of finitely many R algebras gives another R algebra. Furthermore, multiplication in the tensor product, as defined by g, is compatible with multiplication in each ring, as defined by hi.

Notice that the rings do not interact with each other. For all practical purposes, multiplication is performed per component. This does not mean the individual rings Si are commutative, but they do commute with each other in the tensor product.

Multiplicative Identity

If each ring Si contains 1, let e be the element in T represented by 1 cross 1 cross 1 … over all n rings. Now e times x in T is defined by g(ex), which must agree with hi(1xi) within each Si, which is xi. Therefore e times x = x, and e is the identity element, i.e. 1 in the ring T. If the component rings contain 1 then so does their tensor product. Restrict h to hi to prove the converse.

An Si Algebra

We showed T is an R algebra, but it is also an Si algebra for each Si. (This assumes Si is commutative, whence T is an Si module.)

The aforementioned homomorphism g is still associative. If it is an Si module homomorphism then that proves T is an Si algebra.

Let v be an element of Si and consider g(vxy). The action of v on T×T, or on T, is implemented by multiplying xi by v. Thus xi becomes vxi. The isomorphism onto U rearranges generators, but doesn't change vxi. Now h is compatible with hi, which implements multiplication in Si, hence the ith component becomes vxiyi. this is once again the action of v on T. Therefore g respects the action of v, and is an Si module homomorphism, whence T is an Si algebra.