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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 7 months |
| seen | Feb 12 at 13:43 | |
| stats | profile views | 18 |
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Sep 14 |
comment |
About the proof of the existence of a decomposition of subset of $\mathbb{A}^n$ @Zhen, thank you for your help. I cannot understand the last part of your proof showing that any maximal irreducible subset is closed. What does the expression $U \cap Y \subset Z \cap Y$ mean ? |
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Sep 13 |
comment |
About the proof of the existence of a decomposition of subset of $\mathbb{A}^n$ Thank you for answering my question. Could you explain the last line of your answer ? "Note that the only property of Y we need is it has finitely many irreducible components." |
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Sep 12 |
asked | About the proof of the existence of a decomposition of subset of $\mathbb{A}^n$ |
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Jul 24 |
accepted | Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? |
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Jul 24 |
comment |
Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? I think I get it thanks to your additional comment "we don't care what the original form of $\alpha$ is/was". |
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Jul 24 |
comment |
Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? Thank you, @GeorgesElencwajg. Now, I see that my assumption was wrong. |
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Jul 24 |
comment |
Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? It is the revised third edition. |
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Jul 24 |
asked | Is it true that if $\alpha \in \operatorname{Frac}(A)$ and $s\alpha \in A$, then $\alpha \in S^{-1}A$? |
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Jan 27 |
accepted | Containment of extension fields in a Bourbaki proof: $K(S)\subset K(B)$? |
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Jan 27 |
asked | Containment of extension fields in a Bourbaki proof: $K(S)\subset K(B)$? |
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Jan 15 |
comment |
Why is it biholomorphic? Thank you for your detailed explanation. Then, I made a mistake in using the chart $(U_1,z)$ to evaluate the function $\phi_2$. And, each chart $\psi:U \rightarrow V$ on any Riemann surface is biholomorphic since $\psi \circ \psi^{-1} = \mbox{id}_V$ is biholomorphic. Am I right ? |
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Jan 13 |
comment |
Why is it biholomorphic? I thought that $\phi_2'(\infty)\neq 0$ is a necessary condition for the existense of the holomorphic inverse mapping of$\phi_2$ at $\infty$. |
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Jan 13 |
asked | Why is it biholomorphic? |
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Dec 23 |
comment |
Why is it trivial for compact Riemann surfaces to have a countable topology? Thank you ! Now it is clear for me too. |
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Dec 23 |
comment |
Why is it trivial for compact Riemann surfaces to have a countable topology? Thank you @ChrisEagle and ZevChonoles for quick answers. The above is the direct quotation from the book. I understand it as second countable or "has a countable base for its topology" as you suspect. |
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Dec 23 |
asked | Why is it trivial for compact Riemann surfaces to have a countable topology? |
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Oct 19 |
accepted | Why does the continuity of $(x,y) \rightarrow x-y$ mean the commutativity of a topological group? |
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Oct 19 |
awarded | Scholar |
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Oct 19 |
accepted | Is it always true that |1+1|>1 in an Archimedean valuated field? |
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Oct 19 |
awarded | Supporter |
