Algebras, Characterizing the Tensor Product

Characterizing the Tensor Product

Let R be a commutative ring and let S and T be R algebras, such that R embeds in S and in T. In other words, the homomorphism that carries R into S, and into T, is a monomorphism - perhaps the identity map. Other than R, S and T are assumed disjoint.

I'm going to talk about the generators of S, as a ring, or as an R module. These are quite different, and should not be confused. If S is R[x], the polynomials in x with coefficients in R, then the single indeterminant x generates all of S as a ring. In other words, S is a finitely generated R algebra, or an R algebra of finite type. However, as an R module, S is infinitely generated, with generators 1, x, x², x³, x⁴, and so on.

Thanks to the previous section, we know the tensor product S×T is a ring. We would like to describe this ring in terms of R, and the generators of S and T as rings, but we're going to have to start by looking at the generators of S and T as R modules, since the tensor product acts on modules, not rings.

Let's start with a warmup problem - polynomial rings. Let S be R adjoin x₁ x₂ x₃ etc. These indeterminants may or may not commute with each other. Similarly T is R adjoin y₁ y₂ y₃ etc. If S and T can be described using generators, a procedure builds the tensor product. But we need to generate S and T as R modules, not as rings. The generators of S are all finite products of x₁ x₂ x₃ etc, including the empty product, which is 1. Similarly, the generators of T are the finite products of y₁ y₂ y₃ etc. If rings are noncommutative, then x₃x₅ will be a different generator from x₅x₃. Now, armed with these generators, we can apply the procedure.

Cross each generator of S with a generator of T to get a generator of S×T. This is a product of x indeterminants times a product of y indeterminants. Examples include 1, x₂y₅, x₇³, x₄y₁y₆, etc.

The relations denoted K_S and K_T are folded into the kernel of the tensor product; but S and T are free R modules, and there are no relations. That leaves only the action of R. If c is in R then (xc)y = x(cy). Since R commutes with indeterminants by convention, this is not unexpected. As an R module, S×T is generated by products of indeterminants in S, times products of indeterminants in T.

The previous section describes multiplication within S×T. the product of arbitrary generators x_i and y_j gives x_iy_j, and from there, polynomials are multiplied in the usual way, although the x and y indeterminants commute past each other - even if they do not commute among themselves. If all rings are commutative then the tensor product is simply R[x₁,x₂,x₃…y₁,y₂,y₃…], the ring of polynomials in all indeterminants.

Now suppose S and T are arbitrary ring extensions of R. Write S as the quotient of a free ring. If x₁ x₂ x₃ etc act as generators for S, then S is the image of the polynomial ring R[x₁,x₂,x₃…], having kernel K_S. Similarly, let T be the quotient of a polynomial ring having kernel K_T. As an R module, S is generated by all finite products of these generators - and similarly for T. Take the cross product to find generators in S×T. This time we have kernels K_S and K_T that must be folded into the tensor product. Multiply the relations that generate K_S by the products over y, and the generators of K_T by the products over x, to build a kernel that characterizes the tensor product S×T.

As it turns out, we can use the relations that span K_S as a subring, a much smaller set, and the relations that span K_T as a subring, and put them together, and the resulting subring contains all the relations in the aforementioned set. This gives a remarkably concise description of S×T. Put the indeterminants together and let them generate a polynomial ring (such that x's and y's commute past each other), and put the polynomial relations together and let them generate a subring, then the quotient ring becomes the tensor product. Furthermore, multiplication within the tensor product agrees with multiplication in S and in T, i.e. per component.

Embedding S and T

If S and T are algebras, they need not embed in their tensor product. This is demonstrated by Q and Z_p, which are both Z algebras. However, when R embeds in S and T, which is typical, then S and T embed in S×T.

Suppose u is a nonzero element of S that becomes 0 in S×T. In other words, u cross 1 is equivalent to 0. How can this happen? In the aforementioned example, we moved the action of p in R, from 1 back to u, whence u cross 1 was equivalent to up cross 1/p. And in our example, up became 0. If we try to do the same thing here, p becomes a unit in R, which becomes a unit in S. Since u is nonzero and p is a unit, up cannot equal 0. We can pass the action of R back and forth, but that will not take u to 0. The relations corresponding to addition in S and in T will not carry u to 0 either. Therefore u remains nonzero in the tensor product. Both S and T embed in their tensor product, and the embedding is accomplished by crossing the original ring with 1.

Remember that multiplication in S×T occurs per component, and now we know these components are copies of S and T. Therefore the ring S×T is almost the direct product of S and T. Almost - except S and T share a common instance of R.

When R does not Embed

Let S and T be R algebras, with f(R) in the center of S, and g(R) in the center of T. Let K be the tensor product of f(R) and g(R). This is an R module and an R algebra. As an R module, K is the generators of f(R) cross the generators of g(R), mod the relations of f(R) cross the generators of g(R), and the relations of g(R) cross the generators of f(R), and the relations that pass the action of R between f(R) and g(R). On top of this, S is a free f(R) ring mod some polynomial relations, and T is a free g(R) ring mod some polynomial relations. Let's look at S×T.

A generator of S, as an R module, is a generator of f(R) times a finite product of indeterminants in the free ring that defines S. Cross this with a similar generator of T to find a generator of S×T. Next we need to get a handle on the relations that lead to the tensor product.

The action of R, from S to T, can always be accomplished through f(R) and g(R). In other words, the relations of K are sufficient to pass R between S and T.

Let p be a polynomial relation that defines S as the image of a free ring over f(R). In S×T, p joins with a generator of g(R), and a finite product of indeterminants of T over g(R). The coefficients of p become elements of f(R)×g(R), or K, and p is multiplied by y₁, y₂, y₃, etc in T. The same holds for polynomials in T over g(R). If you put it all together, and I'm glossing over a lot of details here, S×T is K adjoin the generators of S as a f(R) ring, adjoin the generators of T as a g(R) ring, mod the polynomial relations that define S and T. Once again S and T run independently, except for their common base of f(R)×g(R).

This is similar to the previous result. To illustrate, let f and g embed. Thus f(R) = g(R) = R, and their tensor product is simply R. This reproduces the earlier characterization.

Units in the Tensor Product

If S is an extension of R, tensor S with itself. If u and v are units in S, such that uy = 1 and vz = 1, then u,v times y,z = 1,1, and u,v is a unit in S×S. Conversely, assume uy,vz is equivalent to 1,1 in the tensor product. Let h in R accomplish the equivalence. Thus uyh = 1, and vz/h = 1. Multiply these together and uyvz = 1. Thus u and v are units in S.

This generalizes to a finite tensor product of S with itself. A tuple is a unit in the tensor product iff each component is a unit in S.

The same result holds for S×T when S and T are integral over R. With uyh = 1, u is a unit in S. Since h is a unit in S, it is also a unit in R. Now vz = h, which makes v a unit in T.

This generalizes to a finite tensor product of integral extensions of R. To illustrate, consider the tensor product of six rings. After normalizing by the action of R, u_iy_i becomes 1, as i runs from 1 to 6. Assume the first three products are multiplied by h₁ h₂ and h₃ to reach 1. The next three products are divided by h₄ h₅ and h₆. Clearly u₁, u₂, and u₃ are units in their respective rings. Also, h₁ h₂ and h₃ are units, which makes them units in R. Since h₁h₂h₃ = h₄h₅h₆, each of the last three factors is also a unit in R. That makes u₄y₄ a unit in R, whence u₄ is a unit in its ring, and so on for u₅ and u₆.

Of course, every finitely generated extension of R is integral.