Timeline for Characterization of locally free modules via exterior powers
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2011 at 15:44 | comment | added | Martin Brandenburg | Thus $v_1,...,v_n$ span a free submodule $U$ of $F$ and for $V:=\ker(\phi)$ we have $F = U \oplus V$. Thus $\wedge^n F$ is the direct sum of $\wedge^n U, \wedge^{n-1} U \otimes V, ... , \wedge^n V$. However, by construction $\wedge^n U \to \wedge^n F$ is an isomorphism. Thus all the other summands are zero, in particular $\wedge^{n-1} U \otimes V = 0$. Since $\wedge^{n-1} U$ is a free module of rank $\binom{n}{n-1}=n>0$, it follows $V=0$. Thus $F=U$ is free. | |
| Jul 19, 2011 at 15:42 | vote | accept | Martin Brandenburg | ||
| Jul 19, 2011 at 15:31 | comment | added | Martin Brandenburg | Neil, your proof generalizes to arbitrary $n$: Let $\wedge^n F$ be free of rank $1$. Writing a generator as sums of pure wedge products and localizing, we may arrange that the generator is pure, say $v_1 \wedge ... \wedge v_n$. Define $\phi : F \to (\wedge^n F)^n \cong A^n$ by $x \mapsto (v_1 \wedge ... \wedge v_{i-1} \wedge x \wedge v_{i+1} \wedge ... \wedge v_n)_{1 \leq i \leq n}$. Then $\phi$ is linear and satisfies $\phi(v_i)=e_i$. | |
| Jul 19, 2011 at 10:30 | comment | added | Martin Brandenburg | @Baptiste: We may assume that $X$ is affine and I think that Neil means $k$ just to be a ring (not a field, in which case nothing has to be shown). @Neil: Thanks! I'll try to generalize this method of proof. | |
| Jul 19, 2011 at 9:52 | comment | added | Baptiste Calmès | Can you expand/explain how you reduce to the case $X=\mathrm{Spec}(k)$? | |
| Jul 19, 2011 at 9:39 | history | answered | Neil Strickland | CC BY-SA 3.0 |