Thread: Algebra isomorphism View Single Post
 Sci Advisor HW Helper P: 9,488 That seems to agree with what I got, namely (u,v) goes to (1/2)tens(u+vbar) - (i/2)tens(u-vbar)i. The idea I used (of yours) was that since the map from C[x] to CxC which induces the isomorphism, takes a polynomial to its pair of values at i and -i, to invert it we are trying, as you say, to find a polynomial in C[x]/(x^2+1) whose value at i and -i gives us u and v. But there is a standard trick for this, using euclid's algorithm for gcd's. I.e. to invert the chinese remainder map Z/(mn)-->Z/m x Z/n we solve an+bm = 1, and then an goes to (1,0) and bm goes to (0,1). Since (x-i) - (x+i) = 2/i, we get (i/2) (x-i) - (i/2) (x+i) = 1, hence the polynomials (i/2)(x-i) and (-i/2)(x+i) having the values (0,1) and (1,0) at i and -i. But this was apparently off by a conjugate, so I fudged it at the end, adding bar to the v's.