show that for fgh in fx f a field if fg1 then fhghgSolutionN

show that for f,g,h in f[x], f a field, if (f,g)=1, then (fh,g)=(h,g)

Solution

Note that (h, g) | (fh, g). This is a result of Bezout’s identity, that is, by Bezout, there exists p, q F[x] such that fhp + gq = (fh, g). But (h, g) clearly divides the left side of this equality, hence it also divides the right. We also know by Bezout’s identity that there exists r, s F[x] such that fr + gs = 1. Thus multiplying both side by h, fhr + ghs = h whence it is clear that (fh, g) | h. Then (fh, g) | h and (fh, g) | g, so (fh, g) | (h, g). Of course, (fh, g) | (h, g) means that deg(fh, g) deg(h, g) and (h, g) | (fh, g) means that deg(h, g) deg(fh, g). Thus degree fh, g) equals degree (h, g), so that (fh, g) = k(h, g) for some k F. Because (fh, g), and (h, g) are both monic this implies that k = 1 and (fh, g) = (h, g) as required