From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: (ir)reducibility of polynomials Date: 6 Oct 1998 16:58:34 GMT Roberto Maria Avanzi wrote: > Suppose f(x,y) is a polynomial in two variables x,y over the field > K (can be the rationals, or a number field). > Suppose further that for many, but finitely many, distinct values > y_1,y_2,...,y_n of y the polynomial f(x,y_1) (which is now a polynomial in > the solve variable x) is irreducible > Then f(x,y) is irreducible (or the only factors you can factor > out are of degree 0 in x, i.e. polynomials in the sole y). Obviously if f(x,y) were reducible, say f(x,y) = g(x,y)*h(x,y), then f(x,y_1) would be reducible for _every_ y_1, the factors being g(x,y_1) and h(x,y_1). If your concern is that one or both of these factors are of degree zero in x, you need only take any y_1 not in the union of the zero loci of the coefficients of the highest powers of x in each of g and h. >Moreover I would like also to know if there is a (related ?) result >like > > f(x,y) as above and with no factors which are polynomials in > the sole y. if for many values y_i of y the polynomial f(x,y_i) is > reducible, then the polynomial f(x,y) itself is reducible. This is a little harder, but the Hilbert Irreducibility Theorem implies that if f(x,y) were irreducible, then f(x,y_1) would be irreducible in K[x] for almost all y_1 , if K is a subfield of the complex numbers (using the usual topology on C). I don't really know what algorithms are used by computer algebra systems to factor multivariate polynomials, but in principle one could use this idea: to factor f in K[x,y], factor f(x,y_i) completely for several y_i, then consider all possible ways of choosing a factor g_i(x) of f(x,y_i) and use Lagrange interpolation to construct a candidate factor g(x,y) with g(x,y_i)=g_i(x) for all i; at some point you will either succeed or know that this set of choices g_i is incompatible (since the degree of g in y has exceeded the degree of f, for example). dave