,
,
,
, ,
, , ,
,
,
Over a ring whose characteristic is a prime number it is true that
because all the non-trivial binomial coefficients are multiples of .
If is prime, then divides neither nor if . So, since divides
it must divide .
If is composite, write where is any prime dividing , so that . If the factor appears with multiplicity in , then it appears with multiplicity in and with multiplicity in , and so it appears with multiplicity in . It follows that cannot divide , i.e., .
Last revised on March 16, 2017 at 14:18:06. See the history of this page for a list of all contributions to it.