Let y be a basis for this vector space. Now each product yiyj is a linear combination of the basis elements of y, with coefficients taken from S.
Assume S is in the center of R. In other words, S commutes with the basis elements. Now the product of any two elements in R can be computed by representing each operand as a linear combination of basis elements and expanding the product. Each yiyj is replaced with its image, and the coefficients are brought in. Group all the terms together to create a new linear combination of basis elements, a new element in R. This algorithm is mandated by the properties of addition and multiplication in a ring, and if addition and multiplication obey these rules, the result is a ring. Thus we have characterized the (possibly noncommutative) rings with a division ring at the center.
As a corollary, a field is isomorphic to a vector space over any of its subfields, and the resulting field is completely characterized by the pairwise products of the basis elements in that vector space. This is sometimes used recursively to build a field from the ground up, where the ground is either the integers or the integers mod p. But that's another story.
If T is a subring of S is a subring of R, and R is a division ring, let x be a basis for S as a T vector space, and let y be a basis for R as an S vector space. Let z be the cross product of x and y, elements of x times elements of y. Consider linear combinations of z, with coefficients ttaken from T, and multiplied on the left. For any element w in R, w is spanned by y, using coefficients in S. Each of these coefficients is spanned by x, using coefficients in T. Thus z spans w for every w, and z spans all of R. If z spans 0, group terms together to write 0 as a linear combination of basis elements of y. Since y is a basis, all coefficients are 0. Each coefficient is a linear combination of elements of x, and all those coefficients are 0, hence all the coefficients on z are 0. Therefore z is a basis for R, written as a T vector space.
the product of the dimensions R over S and S over T gives the dimension of R over T. The dimension of R over T is infinite iff at least one of the subdimensions is infinite.