I need k (field) card(k)=n, n has not the form p^q how do i build k? is there a k to satisfy this?

Question

I need k (field) card(k)=n, n has not the form p^q where p is prime and q is natural how do i build k? is there a k to satisfy this?

for p^q whith that character we have Zp[x]/(poly)
where poly is in Zp[x] and poly is irreductible and grad(poly)=q

for p^q whith that character we have Zp[x]/(poly)
where poly is in Zp[x] and poly is irreductible and grad(poly)=q, poly=ax^n+bx(^n-1)+......z, a....z in Zp[x] and grad(poly)=n, where "/" is the factorization of the field to a poly( with f=gq+r where r are the elements)

for p^q whith that character we have Zp[x]/(poly)
where poly is in Zp[x] and poly is irreductible and grad(poly)=q, poly=ax^n+bx(^n-1)+......z, a....z in Zp[x] and grad(poly)=n, where "/" is the factorization of the field to a poly( with f=gq+r where r are the elements of serched field)
if there is no k field to satisfy this pls mail me a proof of [e-mail]

cs@[e-mail]

poly is irreductible over Zp[x]
means: Zp[x]=0 has no solution for x in Zp
any other details?

thank you for your grat answers!

Answer 1

I'm writing up a proof right now that this is not possible.

Check out the pdf at:
http://www.geocities.com/euleratuo/

Answer 2

Every finite field has cardinality a power of a prime. If you look at the multiplicative identity, 1, and consider the sequence 1, 1+1, 1+1+1, ..it has to repeat. So there is some natural number p with p*1=0. But the field axioms show that this p must be a prime, so Zp is a subfield of k. But then, k is a vector space over Zp. If there is a basis with q elements, then the cardinality of k is p^q.

In other words, you can't get a field of the type you are asking for.

Answer 3

The number of elements in all finite fields is the power of a prime, so it is not possible.

Answer 4

How did you become so smart?????
Very nice...

Answer 5

Please identify the main ideas of your question in plain English, if you'd please.