| > OK, thanks. No worries. > Your answer has allowed me to follow the paper a little further, though I think your answer may be inconsistent with the other two answers I got. I was trying to be helpful. That said, in lieu of the irreducibility criterion, you could stay in the original context where all the variables a_i, x_i, and alpha_i are indeterminates so that all the extensions are higher-ranked rational function fields over Q. So Q(a1,a2) is a field in 2 indeterminates and Q(x1,x2) is an extension of that. They are isomorphic to subfields of C, but we don't think of them as subfields of C because there don't come with canonical embeddings. You have to choose the algebraically independent transcendentals. > Also I haven't been able to show myself that if the x_i are the roots of an irreducible polynomial, then Q(x_1,...,x_n) is symmetric with respect to x_1,...,x_n, in the sense claimed in the paper without proof. There's no claim. The paper's just defining what it means for a field extension to be "symmetric w.r.t." to the adjoined elements. |