Algebras, Compositum

Compositum

Assume S and T are ring extensions of R, that are contained in an even larger ring E. The compositum of S and T, sometimes written S+T, is the smallest subring of E that contains S and T. Note that the compositum always exists; it is the intersection of all subrings containing S and T.

If there is no larger ring containing S and T, yet they still share a common subring R, let E be their tensor product. (Refer to the previous page for a characterization of S×T.) The generators of S and the generators of T necessarily lie in the compositum, and that brings in all of E. Therefore S+T within S×T is S×T. However, if S and T are embedded in some other ring E, the compositum could be different.

Tensor Product as Compositum

We're going to ask whether S×T equals the compositum S+T, relative to a larger ring E. What do we mean by equal? The tensor product is constructed from S and T, so we say it equals the compositum if it is isomorphic to the compositum, with S mapping onto S and T mapping onto T. In other words, the tensor product embeds into E and becomes the compositum.

If S×T embeds in E, with S and T fixed, then S+T is contained in the image of S×T. We saw above that the compositum within the tensor product is the tensor product. Therefore the compositum is isomorphic to the tensor product iff the tensor product embeds in E.

Within the tensor product, S and T commute past each other. If this is not the case in E, the tensor product will never embed, so we may as well assume E is commutative.

Let's have a closer look at the map from S×T into E. Of course S and T map onto themselves. Multiplication in the tensor product occurs per component, so a typical pair generator like xy is the product of x1 and 1y. If the map is a ring homomorphism, then the image of xy in E has to be x times y in E. In other words, the pair generator xy in S×T maps to xy in E. This holds for all pair generators, and hence the entire embedding has been defined.

Is this map well defined? Suppose u and v represent the same element in S×T, yet they map to different elements in E. Subtract v, and u becomes a representative of 0 in S×T, yet it maps to something nonzero in E. Since the tensor product is essentially the cross product of S and T, with R in common, every pair generator used to build u has to have an S component of 0 or a T component of 0. A pair that is nonzero in both components is nonzero in the tensor product, and persists in u, and makes u nonzero. Since every term of u is 0, u equals 0 after all. The map is well defined.

You may prove, if you wish, that the map is a ring homomorphism. It pretty much falls out from the definition. So - the tensor product equals the compositum iff S×T embeds in E, iff the aforementioned map is 1-1, iff 0 is the preimage of 0.

A Tensor Product that Won't Embed

Let's look at an example where S×T ≠ S+T. Let S equal Z adjoin the cube root of 2, a real extension, and let T = Z adjoin a complex cube root of 2. Suppose these rings have some element c in common. This means the corresponding field extensions, based on Q, have c in common. Both field extensions have dimension 3, and they are clearly not equal. (One is real and the other complex.) If c spans a subfield of both these field extensions its dimension over Q is 1 or 3. It can't be 3, since that would force both extensions to contain each other, whence they would be equal. The dimension is 1, and c is in the base ring after all. The extensions are disjoint, other than Z, and we can apply the formulas on the previous page.

If x and y are the respective cube roots of 2, i.e. the elements that generate S and T as rings, the tensor product is Z[x,y] mod x³-2 mod y³-2. This is a free Z module of dimension 9. Let's compare this with S+T.

The compositum is Z adjoin the cube root of 2, adjoin the cube root of 1, which is a free Z module of dimension 6. since dimension is an invariant, the tensor product does not equal the compositum, even up to isomorphism.

Let's show this another way. Suppose the tensor product embeds in the complex plane, with x mapping to the real cube root of 2 and y mapping to the complex cube root of 2. Within the complex plane, y+y² = -x, which is not the case in S×T. In other words, y+y²+x is a nonzero element of the tensor product that maps to 0 in E. The tensor product will not embed, and is not equal to the compositum. Let's try to generalize this result. But first we need a lemma about simple ring extensions inside an integral domain.

When R[x] is an Integral Domain

Let the ring S be characterized as R adjoin x, where x satisfies the polynomial p(x), with coefficients in R. Further assume S is an integral domain.

In what follows, x sometimes refers to an indeterminant, as in p(x), and at other times x is a specific element of S. This comes from the very description of S, as R[x]/p(x). Here S is R adjoin x (x the element), which is isomorphic to R[x]/p(x) (x the indeterminant). Hope this isn't too confusing.

If p, as a polynomial over R, can be factored, then S contains some zero divisors. Yet S is an integral domain, hence p is irreducible over R.

As a bonus, p is irreducible over F, the fraction field of R. If p = gh, multiply through by the common denominator of g, and of h, and a scale multiple of p is the product of a multiple of g and a multiple of h. The two factors on the right are nonzero in R[x]/p(x), yet their product is zero. Once again we have a contradiction, so p remains irreducible over F.

I have to admit, this lemma isn't terribly helpful in practice. Most of the time we know S is isomorphic to R[x]/p(x) because p is irreducible over F. That guarantees linear independence amount the powers of x, so that S does indeed agree with its polynomial characterization. So it's probably a given anyways, but let's move forward with p irreducible over F.

A Simple Ring Extension and a Field

Let S and T be disjoint R algebras, with R common to S and T, such that S and T are part of a larger integral domain E. Let T be a field and let S be a simple algebraic extension of R. In particular, S is R[x] mod p(x), where p has coefficients in R. As mentioned earlier, p is irreducible. When does S×T = S+T?

If p is not monic we run into trouble right away. Set p = 2x⁵-3, where 2 is not a unit in R. Within S, x⁵ is not part of R, but in E, which includes T, which includes the fraction field of R, x⁵ is the same as 3/2. In other words, x⁵-3/2 is nonzero in S×T, where S and T run independently, but it becomes 0 in E. To avoid this problem, p must be monic. Thus S is a simple integral extension of R.

Suppose p factors in T[x]. since p(x) becomes 0 in E, one of its factors, say g(x), is 0 in E. Yet g(x), a linear combination of powers of x and elements of T, is nonzero in S×T. This prevents S×T from embedding in E.

Conversely, assume p(x) remains irreducible over T, and suppose some element g(x) in S×T maps to 0 in E. The field extension T(x)/T has dimension equal to the degree of g, or less, yet x is a root of p, which is irreducible, hence the dimension of T(x)/T equals the degree of p. This is a contradiction, hence S×T embeds. The tensor product equals the compositum iff p is monic and remains irreducible over T.

Now consider S tensor U, where U is a ufd, and S = R[x]/p(x), and p is irreducible over U. Let T be the fraction field of U. By gauss' lemma, p is irreducible over T. Extend E by bringing in the nonzero elements of U as denominators. Now E contains T. Since U embeds in T, S×U embeds in S×T. By the above theorem, S×T embeds in E, hence S×U embeds in E. Furthermore, S×U embeds without using any of the denominators drawn from U. Thus it embeds in the original ring. The previous result is replicated here; The tensor product equals the compositum iff p is monic and irreducible over U.

Disjoint from Normal Closure

We could continue the above when S is a finite tower of integral extensions, but this is rarely practical. Here is a more useful criterion for S×T = S+T.

Let S and T be integral extensions of R inside an integral domain E, such that S is disjoint from U, the normal closure of T. Take a generator x in T, with irreducible polynomial p, and adjoin all the conjugates. Do this for all the generators of T, building U. Finally, assume either S or T is a finite R module.

Let E′ be the fraction field of E, with T′ the fraction field of T, and so on. The generators of T, as a ring extension of R, generate the field extension T′/R′. (Remember that each polynomial remains irreducible over R′.) When their conjugates are brought in, the result is U. When these conjugates are folded into T′, the result is U′. The extension U′/R′ is a splitting field, splitting the polynomial p(x) for each generator x in T. Therefore U′ is a normal extension of R′, and is the normal closure of T′.

For example, T might be R[y,z], where y and z satisfy monic irreducible polynomials over R, and over R′. Adjoining y and z to R′ produces a field which is the fraction field of R[y,z], which is T′. Adjoin all the conjugates of y and z to R′ and the result is a field, which is the fraction field of U, and a normal extension of R′, and the normal closure of T′.

Suppose U′ and S′ are not disjoint. Some w is common to both, and lies outside of R′. Write S′ and U′ as R vector spaces. A linear combination of basis elements from S, with coefficients in R′, equals a linear combination of basis elements of U, with coefficients in R′. Multiply through by a common denominator, and S and U intersect. This is a contradiction, hence S′/R′ and U′/R′ are disjoint.

Review the theorems describing the compositum of fields. The tensor product of S′ and T′ equals their compositum in E′. This means S′×T′ embeds in E′. Since S×T embeds in S′×T′, it also embeds in E′. In fact it embeds without using any of the denominators, hence S×T embeds in E, and is equal to S+T.