If F is a finite field of order q then the number of subspace of dimension k of an n-dimensional vector space over F is the q-binomial coefficient<http://en.wikipedia.org/wiki/Gaussian_binomial_coefficient>: binomial_q(n,k). The limit as q-->0 of binomial_q(n,k) is binomial(n,k) --the usual binomial coefficient - the number of k element subsets of an n element set. For this and similar reasons people that study q-analogues <http://en.wikipedia.org/wiki/Q-analog>sometimes refer to sets as vector spaces over the field with 1 element and subsets as being subspaces of such a vector space. See http://en.wikipedia.org/wiki/Field_with_one_element or Peter Cameron's blog entry http://cameroncounts.wordpress.com/2011/07/20/the-field-with-one-element/ --Edwin On Tue, Feb 4, 2014 at 11:11 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I think (0,+,*) ought to be considered a field. But apparently algebraists ignore it, I think by requiring that 1 != 0, probably to avoid having a special case.
--Dan
On 2014-02-04, at 8:00 PM, Henry Baker wrote:
Is {0,+,*} a field or not?
All of its non-zero elements have multiplicative inverses (vacuously) !
At 05:39 PM 2/4/2014, James Buddenhagen wrote:
Back in 1914 the number 1 was a prime. At least according to D. N. Lehmer. See D. N. Lehmer, List of primes numbers from 1 to 10,006,721, Carnegie Institution Washington, D.C., 1914
On Tue, Feb 4, 2014 at 1:38 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Tue, Feb 4, 2014 at 2:12 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
According to a friend who volunteers in the Santa Cruz CA Public Schools, the official view is that 1 is a prime, because its only divisors are 1 and itself. However, a teacher did mention that in more advanced mathematics, 1 is not a prime.
Do they also teach that unique factorization is false? Or is it stated as "numbers > 1 have a unique factorization into primes other than 1"?
Andy
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