Timeline for What conditions are needed for $-\otimes_A B$ to be faithful?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Aug 19 at 5:47 | history | suggested | Nicolas Hemelsoet | CC BY-SA 4.0 |
parenthesis removed in latex (+ invisible garbage for the 6 characters limit)
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| Aug 18 at 21:32 | review | Suggested edits | |||
| S Aug 19 at 5:47 | |||||
| Nov 15, 2011 at 8:28 | comment | added | Martin Brandenburg | A more easy counterexmaple: The zero section $\mathrm{Spec}(k) \to \mathrm{Spec}(k[e]/e^2)$. | |
| Nov 12, 2011 at 15:44 | comment | added | Martin Brandenburg | Here is a counterexample to the statement "f surjective implies $f^*$ faithful": It is enough to find a nil ideal $I$ of some ring $A$ and an $A$-module $M$ such that $M = IM$, but $M \neq 0$; then $\mathrm{Spec}(A/I) \to \mathrm{Spec}(A)$ is a homeomorphism, but the corresponding pullback functor is not faithful. One specific example is $A = k[x_1,x_2,\dotsc]/(x_i^i = 0, x_i = x_{i+1} x_{i+2})_{i \geq 1}, M = I = (x_1,x_2,\dotsc)$. | |
| Nov 12, 2011 at 12:40 | comment | added | Martin Brandenburg | Also the last step of the proof seems to be problematic if the point is not closed: How do you prove that the skyscraper sheaf pulls back to $0$? | |
| Nov 12, 2011 at 12:13 | comment | added | Martin Brandenburg | I don't believe that 2. is true. Since $\mathcal{O}_Z$ is a quotient of $\mathcal{O}_Y$, already $\Gamma(X,M) = 0$ implies $\mathrm{Hom}_X(\mathcal{O}_Z,M)=0$. Or do you mean the Hom-sheaf? | |
| Dec 2, 2010 at 18:19 | comment | added | Ketil Tveiten | I'd give this an extra +1 if I could, as it also answers the problem I had which motivated the thread. Unfortunately, the answer was negative. | |
| Dec 2, 2010 at 17:52 | history | answered | Greg Muller | CC BY-SA 2.5 |