# Tom Bachmann

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 Name Tom Bachmann Member for 3 years Seen 2 days ago Website Location Age
 May14 comment The pth power of a distance function is twice continuously differentiable, for $p>2$?I don't think that makes a difference, just consider the same example in the plane and "close up" the two bits via a "connection far enough out" (i.e. a something like a thickened semi-circular arc in the upper half plane). Apr11 comment Prime ideals in polynomial rings over integersLet $k$ be a field and $P$ a prime of $k[X,Y].$ Then $p$ contains an irreducible element $f$, hence $(0) \subset (f) \subset P \subset M,$ where $M$ is a maximal ideal containing $P.$ The first inclusion is strict. Since $k[X,Y]$ has dimension two, $P=M$ or $P=(f).$ This reduces the problem to classifying irreducible polynomials and maximal ideals. Maximal ideals of $k[X_1, \dots, X_n]$ are well-known, by the Nullstellensatz. Apr11 comment Prime ideals in polynomial rings over integersLet $P$ be a prime ideal of $R := \mathbb{Z}[X,Y]$. Then $P' := P \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}.$ If $P' = (p)$ for a rational prime $p,$ then $P/p$ is a prime of $\mathbb{F}_p[X,Y]$. Otherwise, let $S = \mathbb{Z} \setminus {0} \subset \R.$ It follows that $S^{-1}P$ is a prime of $S^{-1}R = \mathbb{Q}[X,Y].$ This reduces the problem to polynomial rings over fields. Apr8 comment Hypercohomology of a complex via Cech cohomologyIndeed regarding "sufficiently nice": it suffices that for each $n$, the complex $C^{n, \bullet}$ computes the cohomology of $\scr{F}^n$ (consider the spectral sequence taking "vertical cohomology" first).