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Sep 3, 2015 at 13:18 vote accept User
Sep 3, 2015 at 0:11 comment added darij grinberg Another source (though with far less useful answers): math.stackexchange.com/questions/1004341/…
Sep 3, 2015 at 0:07 comment added darij grinberg @PeterSamuelson: Sure it would, but two people beat me to it before I could even remind myself of how the proof went.
Sep 3, 2015 at 0:01 comment added Peter Samuelson It would be more constructive to explain why the characteristic 0 assumption isn't needed. One proof I've seen of this uses the averaging operator for $S_n$, which isn't defined in characteristic less than $n$.
Sep 2, 2015 at 19:45 answer added David E Speyer timeline score: 4
Sep 2, 2015 at 17:37 answer added David Hill timeline score: 8
Sep 2, 2015 at 17:05 comment added user43326 By writing down the Poincare Series, we see that the dimension is correct. So you need to prove either it is free or it generates, whichever is easier.
Sep 2, 2015 at 15:32 history edited darij grinberg
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Sep 2, 2015 at 15:31 comment added darij grinberg The characteristic-$0$ assumption is useless.
Sep 2, 2015 at 15:19 history asked User CC BY-SA 3.0