Tristan

Graduate student mostly dabbling in algebraic geometry.

 15h comment Pullback and Pushforward Isomorphism of SheavesSee also exercise 5.1.1.(a) on page 171 of Q. Liu's "Algebraic Geometry and Arithmetic Curves". Mar 18 asked Kernel of $p$-adic logarithm. Mar 4 accepted A proof that every projective morphism is proper? Mar 4 asked A proof that every projective morphism is proper? Mar 2 revised If $\gcd(m,n)$=1, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is cyclic.Changed some round brackets to pointy brackets. Aug 20 awarded Analytical Aug 20 asked The ring of integers of a number field is finitely generated. Aug 20 accepted Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. Aug 20 answered Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. Aug 20 asked Notation $[A]_q$ for finite abelian group $A$ and prime power $q$. Aug 12 asked Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. Jul 1 comment Embedding torsion units of an order into torsion units of the reduced order.Keen observation! I overlooked this minor detail myself. Your phrasing makes for a much nicer proof. And I never noticed that the sum of a zero divisor and a nilpotent is again a zero divisor. Jun 28 accepted Embedding torsion units of an order into torsion units of the reduced order. Jun 28 comment Embedding torsion units of an order into torsion units of the reduced order.I see now, thank you for your help (and patience). I failed to notice that in the equation $$\sum_{i=1}^n{n\choose i}x^i=nx\cdot\sum_{i=0}^n{n\choose i+1}\frac{x^i}{n}=0,$$ the right most summation is of the form $1+y$ with $y\in\sqrt{0_A}$, and hence a unit. Jun 27 comment Embedding torsion units of an order into torsion units of the reduced order.If $a$ is in the kernel of this map, then $a=1+x$ for some $x\in\sqrt{0}_A$, and it is a torsion unit. Hence there exist $m,n\in\mathbb{Z}_{>0}$ such that $x^m=0$ and $a^n=1$. The latter yields $$(1+x)^n=x^n+nx^{n-1}+\ldots+nx+1=1,$$ so $x^n+nx^{n-1}+\ldots+nx=0$, and we may reduce this to $$\sum_{i=1}^k{n\choose i}x^i=0,$$ where $k=\min\{m-1,n\}$. On the other hand, we may also write $$nx^{n-1}+n\frac{n-1}{2}x^{n-2}+\ldots+nx=-x^n,$$ which shows that $x^n\in(n)$, the ideal generated by $n$. But I don't see how this helps me. Jun 25 comment Embedding torsion units of an order into torsion units of the reduced order.I'm sorry to say that I don't see where it leads. Jun 25 revised Embedding torsion units of an order into torsion units of the reduced order.added 293 characters in body Jun 24 asked Embedding torsion units of an order into torsion units of the reduced order. Jun 10 awarded Scholar Jun 10 awarded Supporter