quasi wrote: > Sandy wrote: >> >> This: >> >> the geometry of an algebraic curve can be reduced to the >> algebra of a certain corresponding field; specifically, >> if k denotes the field of all complex numbers, and if a >> curve is defined in the plane by an *irreducible* >> polynomial equation f(x,y)=0, then the corresponding >> field K is the totality k(x,y) of all rational functions >> z = g(x,y)/h(x,y) with complex coefficients...  >> >> puzzles me. Specifically, what does "irreducible" mean? I >> know, or I think I do, that a polynomial with coefficients >> in F (some structure) is said to be irreducible if it cannot >> be factored into non-constant polynomials with coefficients >> in F. > > Right. > >> So whether a polynomial is irreducible or not depends on what >> F is (real field, ring of integers, etc). > > Right. > >> Mac Lane doesn't say what F is, > > Yes, but from the context it's clear that irreducibility is > intended as irreducibility over the field of complex numbers. > >> but there are two fields "in sight", one is k the complex field, > > Yes -- that's the one. > >> the other is R the real field > > But there's no mention of real numbers. > >> since curves in the plane (R^2?) are being discussed. > > Ah, I see how you got misled. > > The intended space for the graph is C^2, not R^2. > > True, he called it "the plane", but in this context, he meant > "the plane C^2" (which is a 2-dimensional vector space over > the field C).
Thank you. I had considered that it might be any field. The "plane" then being what I think is called an affine plane. I did not consider the possibility that it "had to be" k. The fellow might have said so!