GCDs and PIDs -- Abstract Examples 23

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[llchan]

29:43 Don’t understand why such a complicated argument. Someone also already pointed out the mistakes in his argument.

We want to show that if p does not divide f, then it must divide g. If p does not divide f, then gcd(p, f)=1 or otherwise it contradicts with p being irreducible. So ap+bf=1 for some a, b in F[x]. Multiplying both sides by g yields apg+bfg=g, and because p divides the left hand side, p therefore divides g.

[alegal]

At 29:59 Justin wrote F [ x1, ..., xk, xk+1 ] = F [ x1, ..., xk ] F [ xk+1 ]
But I think it should be F [ x1, ..., xk, xk+1 ] = F [ x1, ..., xk ] [ xk+1 ]