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Jun 25, 2014 at 18:23 vote accept user54347
Jun 25, 2014 at 18:23 vote accept user54347
Jun 25, 2014 at 18:23
Jun 25, 2014 at 18:04 comment added user27920 Even with finitely many variables we get counterexamples, using irreducible $f \in R = k[x_1,\dots,x_n]$ with $f(0) = 0$ (so $f$ is irreducible in the local ring at 0) such that $f$ is reducible in $k[\![x_1,\dots,x_n]\!]$. Consider the invertible ideal $J$ generated by an irreducible formal factor of $f$. If it comes from an ideal of $R$, hence of the local ring $R_0$ at $0$, by faithful flatness of completion that ideal in $R_0$ would have to be invertible and hence provide a proper factor of $f$ in $R_0$, contradiction.
Jun 25, 2014 at 8:46 history answered Misha Verbitsky CC BY-SA 3.0