Hagen
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 May7 comment $CL(O_S) \cong \mathbb{Z}/3\mathbb{Z}$. If $Cl(O_S)$ is meant to be the class group of the ring $O_S$, then there is something wrong: $O_S$ is a principal ideal domain, because it is a localization of $\mathbb{Q}[T]$. Therefore its class group is trivial. May7 comment are connnected components of this scheme irreducible? So then my argument doesn't work. My guess is that your conjecture is true. I'm however struggling with the case of a fibral curve in the intersection of two components. Why should it necessarily be singular? Couldn't we just take two normal surfaces having isomorphic components in their repective closed fibres and glue them together along that componnents? May6 comment are connnected components of this scheme irreducible? Normality ascends along smooth extensions, therefore $X_{R^\prime}$ is normal, which, as you state, seems to imply the property you are asking for (I did not check the latter). May6 answered vector space of continuously differentiable functions is complete regarding a specific norm Apr6 answered Example of ramified and unramified morphism of rings Mar19 comment Uniqueness of the algebraic closure of a field Nevertheless the fields $\mathbb{C}$ and $\mathbb{D}$ are isomorphic. Mar14 answered What is so special about the real and complex numbers? Mar11 comment Let L be a linear operator on a finite-dimensional vector space V over a field F, with p in F[T]. Show that Ker(p(L)) = {0} iff gcd(p, $\mu$) = 1 Yes, you must be careful with factorization arguments, since plugging in $L$ moves things into the ring $F[L]$, which has zero-divisors. So polynomials behave strangely here. Mar11 revised Let L be a linear operator on a finite-dimensional vector space V over a field F, with p in F[T]. Show that Ker(p(L)) = {0} iff gcd(p, $\mu$) = 1 added 501 characters in body Mar10 answered Let L be a linear operator on a finite-dimensional vector space V over a field F, with p in F[T]. Show that Ker(p(L)) = {0} iff gcd(p, $\mu$) = 1 Mar3 comment Restoring 2D properties from 1D data The argument works for any linear projection -- so "no" in this case. But you can use non-linear projections, for example you could define $P_i:=x_i^2+y_i^2$. Then you can see whether the data are lying on a circle just because the $P_i$ are all equal in that case. Looks hillarious, but even if you have a bit of noise on the data, it gives you a criterion showing that the data lie near a circle. Mar3 answered Restoring 2D properties from 1D data Feb28 comment Ideal of $\mathbb{C}[X,Y]$ contained in infinitely many distinct proper ideals There is a more direct way of showing what you want to show: consider the ideals $(X-a,Y-b)$, $a,b\in\mathbb{C}$. What does it mean, that $I$ is contained in such an ideal? Feb27 comment Direct product of subgroups What do you mean by "weak" direct product? If $\phi:G\rightarrow H_1\times H_2$ is an isomorphism of groups, then $G$ is the internal direct product of $\phi^{-1}(H_1\times 1)$ and $\phi^{-1}(1\times H_2)$. Feb25 answered on a proof of the Primitive Element Theorem in zero characteristic Feb25 answered Symmetries of the regular hexagon Feb20 answered Asking for some help in proof of the properties of a special type of metric space Feb13 revised Normalization or integral closure of ring over $\mathbb Z_p$ Mistake fixed. Feb13 comment Normalization or integral closure of ring over $\mathbb Z_p$ There is indeed a mistake / weak point in my argument - I'll correct it in a moment. Feb11 comment (hyper) elliptic curve in characteristic two and the Jacobian criterion The statement that $E$ is singular everywhere seems wrong -- see the example in the extension of my post.