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15h
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comment |
Pullback and Pushforward Isomorphism of Sheaves See also exercise 5.1.1.(a) on page 171 of Q. Liu's "Algebraic Geometry and Arithmetic Curves". |
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Mar
18 |
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asked | Kernel of $p$-adic logarithm. |
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Mar
4 |
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accepted | A proof that every projective morphism is proper? |
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Mar
4 |
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asked | A proof that every projective morphism is proper? |
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Mar
2 |
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revised |
If $\gcd(m,n)$=1, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is cyclic. Changed some round brackets to pointy brackets. |
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Aug
20 |
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awarded | Analytical |
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Aug
20 |
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asked | The ring of integers of a number field is finitely generated. |
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Aug
20 |
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accepted | Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. |
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Aug
20 |
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answered | Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. |
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Aug
20 |
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asked | Notation $[A]_q$ for finite abelian group $A$ and prime power $q$. |
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Aug
12 |
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asked | Prove that for an order $R$, the ranks of $\sqrt{0_R}$ and $1+\sqrt{0_R}$ are equal. |
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Jul
1 |
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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. |
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Jun
28 |
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accepted | Embedding torsion units of an order into torsion units of the reduced order. |
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Jun
28 |
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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. |
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Jun
27 |
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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. |
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Jun
25 |
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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. |
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Jun
25 |
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revised |
Embedding torsion units of an order into torsion units of the reduced order. added 293 characters in body |
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Jun
24 |
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asked | Embedding torsion units of an order into torsion units of the reduced order. |
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Jun
10 |
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awarded | Scholar |
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Jun
10 |
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awarded | Supporter |
