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Sep 29, 2010 at 3:43 vote accept debapriyay
Sep 28, 2010 at 13:42 comment added Matthew Daws Well, sure, if K is finite then so is $X=K^n$, and so $PK^{n-2}$ is finite. It's just an exercise to write down explicit representatives. If you don't know how to do this, read an introduction to projective space.
Sep 28, 2010 at 12:56 comment added debapriyay $X = K'^{n}$, where $K'$ is a field and $K \subset K'$. This can also be the case. Now, you talked about an equivalence relation, one question is if $K'$ is a finite field, then can it be possible to enumerate all the equivalence classes, or the number of equivalence classes.
Sep 28, 2010 at 12:47 comment added rpotrie Nice. I could't express this fact correctly. This defines exactly all possible maps.
Sep 28, 2010 at 12:28 history answered Matthew Daws CC BY-SA 2.5