Let F be the set of fractions whose numerators and denominators are fixed by G. Verify that this is a ring. It is a subring of R/S that is fixed by G. In fact a ring homomorphism embeds this ring into the ring that is fixed by G. We only need prove this map is onto, and the two subrings will coincide. In other words, the localization of the fixed ring is the fixed ring of the localization.
Given a fraction n/d, numerator over denominator, multiply top and bottom by all the conjugates of d, ignoring the identity automorphism. This is the same fraction, and the denominator is fixed by G. Thus it is enough to consider a fraction n/d where d is fixed by G.
For each automorphism c in G, c maps n/d to an equivalent fraction. Write an equation u*(c(n)d-nd) = 0, where u depends on c.
Remember that c maps u into S, and S is multiplicatively closed. Multiply u by all its conjugates, so that u is fixed by G. Thus we can assume each uc is fixed by G.
Fold d into uc. Thus, for each c, some uc satisfies uc*(c(n)-n) = 0.
Let v be the product over uc. This is in S, and is fixed by G. Also, v*(c(n)-n) = 0 for each c.
Consider the fraction vn/vd, which is equivalent to n/d. The denominator is fixed by G. Apply c to the numerator and get c(n)v, which is the same as nv. That completes the proof. The fractions of the fixed ring produce the same subring as the fixed ring of the fractions.
Remember that R is a subring of S, but it might not be a subring of T. Let f(R) = R′ in T. Then review the structure of S×T. Basically, S and T are treated as separate rings, except they share an instance of R. For convenience, the action of R is always passed over to T, by vectoring through R′. (You might think of U as an R′ module.) Aside from the shared instance of R → R′, multiplication takes place per component, and terms are added together, like polynomials.
Let c be one of the automorphisms of G, and build a new function d by letting c act on S, and leaving T alone. Note that c fixes R, which is shared between S and T, so there is no contradiction between c moving S about and c fixing T.
Since c respects addition and multiplication, the same holds for d. It is a ring endomorphism on U. Also, d can be reversed by reversing c on the S component of each term, and leaving T alone.
Suppose d(x,y) = 0. This means one of the two components has to become 0. If y is 0 then x,y was already 0. If c(x) becomes 0 then x is 0 (since c is injective). If c(x) is divisible by k in R, such that f(k)*y = 0, and c(x),y is equivalent to 0, then we can reverse c to show that x was already divisible by k, and x,y was already 0. If f(k) pulls out of y, such that k*c(x) = 0, then k*x = 0, and x,y = 0. If d maps a term to 0, then that term was already 0.
Suppose d maps a sum of terms to 0. Collaps the sum of terms as far as possible. If one of the terms is equivalent to 0, throw it out. If something in R can be passed between x1 and y1, so that y1 becomes y2, then this term can be combined with x2,y2. Assume this has already been done, and apply d. We're talking about more than one term, since d is injective on single terms. Somehow c(x1),y1 + c(x2),y2 (and perhaps more terms) collapses to 0. The element k ∈ R that passes between c(x1) and y1, to turn y1 into y2, passes right through c, because c fixes R. The same action passes between x1 and y1, and we should have combined these two terms at the outset. The original sum of terms collapses to one term, which is 0. Therefore d is injective, and c induces a ring automorphism on U.
This holds for every c in G.
If R embeds in T, these induced automorphisms remain distinct. Let c1 and c2 map x to two distinct elements z1 and z2. Now x,1 is a nonzero element of U, and d1 and d2 map x,1 to z1,1 and z2,1, respectively. Suppose these values are equal in U. Subtract them to find (z1-z2),1 = 0. Since R embeds in T, we can't pass k over to the right to make this term equal to 0. Nor can we pass a unit to the left to kill z1-z2. Therefore z1 = z2, which is a contradiction. If R embeds in T, distinct automorphisms in G remain distinct when extended to U.
don't assume that fixed ring and tensor product commute, as was the case with localization. Clearly R×T = T is fixed by each d, but additional members of U may be fixed. Let's look at an example.
Let R be Z[y], the integer polynomials in y. This will be the ring that embeds in S and in T, i.e. the ring common to both components of the tensor product. Let S = R[x], the integer polynomials in x and in y. Let T = R[z], with yz = 0. The tensor product U = S×T consists of all polynomials in x y and z, but yz drops to 0 wherever it appears.
The group G consists of two automorphisms, c and its inverse. Let c replace x with x+y throughout. Clearly c inverse replaces x with x-y. Verify that these are automorphisms on S.
Suppose a polynomial is fixed by G, and let xm be the highest power of x. Replace x with x+y and apply the binomial theorem. This introduces a new term myxm-1 at degree m-1. This cannot be undone by replacing x with x+y in the lower degree terms. The polynomial is not the same. Therefore G fixes precisely R.
Moving to U, consider the element xz. This is not a member of T. However, (x±y)z = xz, and xz is fixed by G.