You're right - I left a key aspect of the tensor product on the cuttingroom floor. The tensor product consists of all linear combinations of cross products. So a typical element is the sum of xiyi over a finite set of pairs. Let's start the analysis again.
If y is yayb in the cross product R*R, then xy is really xya + xyb. Similarly, separate x into xa and xb. Thus the tensor product consists of linear combinations of xy where each x comes from one of the two copies of R in the first instance of M, and y comes from one of the two copies of R in the second instance of M. Then replace x cross y with xy cross 1. Finally, as an R module, this element is spanned by 1 cross 1. Therefore M×M is spanned by 4 generators, selecting a 1 in either component of the first instance of M, crossed with 1 in either component of the second instance of M. This defines an isomorphism from R4 onto M×M.
Generalize this to M = Rn. Each of n generators of M combines with each of n generators of M to become a generator of M×M. Use an n×2 matrix to represent this map. The matrix is all zeros, except for a 1 in the first column and a 1 in the second. This can be done in n2 ways. Sure enough, R to the n2 is isomorphic to M×M.
Recall that E is spanned by tuples of generators that follow a certain, arbitrary order. In our matrix, every generator of E is a strictly decreasing arrangement of ones against a background of zeros. Each column contains one 1, and it is lower than the 1 in the previous column. These vectors in Rnj span E.
Now I'm in danger of being misled by my own notation. Let n = 7 and j = 3. The first generator of E places a 1 in the upper left and follows the main diagonal from there. The three ones look like the face of a die, although there are four more rows of zeros below. To get the next generator, take the 1 in the third column and push it down a notch. Now if we want the sum of these two generators, you might be tempted to add the two matrices, giving two 2's and two 1's. But don't do that, because it's not right. The generators are all independent of each other. If a generator is indicated by the location of each 1, relative to the top row, the first two generators are {1,2,3} and {1,2,4}, and their sum is simply {1,2,3}+{1,2,4}. The expression cannot be simplified. Thus the generators of E spann a free submodule of T, which maps onto E.
Let a tuple equal z in its third and fifth components. Let the second component equal x, which is an element of M, which can be separated into n pieces according to M = Rn. To keep things simple, assume only two of these pieces are nonzero. Thus x = xa+xb. Replace x with xa, and with xb, and the two simpler tuples span our original tuple. In fact we can cut xa down to 1, because that tuple, within T, can be scaled by xa to bring xa back again. Do this for all the components and we only require tuples that are equal in two components, with ones elsewhere. Imagine z down the third and fifth columns of our matrix, with a one in every other column. This is a basic generator of K. Everything in K is a linear combination of basic generators.
Let everything outside of z, all columns other than 3 and 5, act as the context for the basic generator of K. Then take z apart piece by piece. For illustration, assume z has 3 nonzero pieces, za zb and zc. Place a 1 in position a b or c in column 3, and place a 1 in position a b or c in column 5. Nine generators of T combine to build this basic generator of K. Their coefficients look like za2, zbzc, zcza, and so on through all 9 combinations. Every basic generator of K can be disassembled in this way.
Since R is commutative, zazb = zbza. Take apart a basic generator, and there is a certain symmetry in its representation, relative to T. Flip the ones in columns 3 and 5, so they slant up instead of down, and the coefficient is the same.
Assume there are three basic generators that equate columns 3 and 5, and have the same context, i.e. they place 1's in the same rows for the columns outside of 3 and 5. Let them set Rn, in columns 3 and 5, to x y and z respectively. Represent these using the basis of T. Consider the coefficients when there is a 1 in row a in column 3, and row b in column 5, and a 1 in row b in column 3 and row a in column 5. These two coefficients are equal when z is considered, as shown above, and the same is true for x and y. Add them up and the two coefficients are equal.
In any finite linear combination of basic generators, i.e. in any element of K, select any two columns such as 3 and 5, and select any context outside of 3 and 5 - then combine the basic generators that share this context, represent the results using the basis of T, and find a symmetry among the resulting coefficients. For any two rows a and b, swap the 1's in columns 3 and 5, between a and b, and the coefficient does not change. This coefficient may well be 0, that's ok. Also, if a = b, the swap changes nothing, whence the coefficient is certainly preserved.
Suppose a linear combination of descending generators, i.e. an element spanned by the basis of E, also lies in K. This happens iff E is not free in the quotient T/K.
For notational convenience, assume g is one of the generators of E that is used to build the offending element, also present in K. Furthermore, let g place 1's down the main diagonal, starting at the upper left and following a 45 degree line. There may be rows of zeros below, if j < n, that's ok. If g is really something else, the following argument does not change in any substantial way.
Let the trivial permutation {1,2,3,4..j} represent g. These are the rows, for each column, that hold 1. A permutation of these rows, e.g. {2,1,3,4…j}, is a cousin of g. There are j! cousins, all generators of T, but only one belongs to E, namely g.
Remember that g is part of a linear combination that is common to E and K. Somehow, g is created using the basic generators of K. for some pair of columns, say 3 and 5, apply the context of g {1,2,?,4,?,6,7…j}, and the basic generators of K that fit this criteria combine to produce a nonzero coefficient on {1,2,3,4,5,6,7…j}. By symmetry, the coefficient on {1,2,5,4,3,6,7…j} is nonzero. A cousin has been brought in, and it doesn't belong. It has to go away.
To get rid of the unwanted cousin, another pair of columns, with the proper context, must be invoked. Perhaps it is columns 1 and 4. The basic generators of K that equate columns 1 and 4, in context, combine to produce a nonzero coefficient on {1,2,5,4,3,6,7…j}. However, by symmetry, we now have a new cousin {4,2,5,1,3,6,7…j}. This is not part of E, and it has to go away. Let's see where this leads.
Restrict attention to the submodule of T generated by g and all its cousins. This projection is nonzero, since g is assumed to be present. All entries come from basic generators of K that equate two columns against a particular context, and this forces cousins to enter in pairs. One such pair might be v*{7,6,1,2,3,4,5…j} + v*{6,7,1,2,3,4,5…j}. Each pair consists of a permutation and one of its transpositions.
Place the even permutations on the left and the odd permutations on the right. Each pair joins a cousin from the left with a cousin from the right. They walk into the system hand in hand, with equal coefficients. Add up all the coefficients on the left and the result is nonzero, namely, the coefficient on g. Remember that g is the only cousin to appear in E. Add up the coefficients on the right and get 0. When everything is taken into account, none of those cousins should remain. Therefore something nonzero equals 0, and that is a contradiction.
The module spanned by E in T is disjoint from K. Mod out by K and E is a free R module. The jth exterior power of a free module Rn is free, having rank (n:j).