Timeline for When is tensoring with a module representable by a scheme?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2010 at 16:21 | comment | added | user2035 | Monomorphisms are injective (you just gave a proof). | |
| Dec 28, 2009 at 17:08 | comment | added | Martin Brandenburg | you seem to use: if $X \to Y$ is a $S$-morphism, which is a monomorphism, where $Y/S$ is separated, then also $X/S$ is separated. is this true? it's well-known that it's true when $X \to Y$ is injective. since the underlying set of a k-scheme can be recovered as the filtered(!) colimit of the K-points, where $K/k$ is a field extension, it's enough to assume that all $X(K) \to Y(K)$ are injective. here, $X(K)=K^{(I)}$ injects to $\mathbb{A}^I(K)=K^I$ and everything is fine. | |
| Nov 25, 2009 at 18:42 | history | edited | user2035 | CC BY-SA 2.5 |
sketched alternative argument
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| Nov 25, 2009 at 12:48 | history | answered | user2035 | CC BY-SA 2.5 |