Timeline for Finite separable extension of fields imply the number of intermediate subfield is finite
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| Feb 18, 2014 at 3:56 | comment | added | Benjamin Steinberg | Can't you use that separable means that tensoring with the algebraic closure is a finite direct product of copies of the algebraic closure? This has finitely many subalgebras. Tensoring with the algebraic closure doesn't identify any intermediate fields and so you had finitely many to start with. (I'm a bit sleepy so this might be wrong.) | |
| Feb 18, 2014 at 2:53 | comment | added | user76758 | The characterization of when a finite extension (separable or not) admits finitely many intermediate fields is exactly that it is a primitive extension. This is all far more elementary than any categorical formalism. | |
| Feb 18, 2014 at 2:21 | comment | added | Will Sawin | 1. To prove this theorem in the etale site one needs some osrt of finiteness criterion on the base, I think. Imagine the base an infintie union of points, and the cover a 2-to-1 cover at each point. Then there are infinitely many subcovers, corresponding to which points are 2-to-1 and which are 1-to-1. But even a connectedness condition deals with this problem. 2. Have you tried to figure out a proof in a Grothendieck's Galois Theory direction? I'm not sure what geometric facts about finite etale covers one needs to show they form a Galois category, but once you do you get finiteness. | |
| Feb 18, 2014 at 2:03 | history | asked | user41650 | CC BY-SA 3.0 |