Dear David,

> If p is an odd prime, then the group (Z/Zp)*

> of integers from 1 to p-1 is cyclic under

> multiplication modulo p.

That´s right, but does not consider the ring with an adjoined square root.

If p is an odd prime, then the group of natural numbers with an adjoined square root is cyclic with the order of p+1 under multiplication modulo p.

This is the difference to the natural numbers and i will try to

explain it to you.

If jacobi (a, p)=-1 and p a prime and A=sqrt (a), then

(1+A)^p=(1+A^p) mod p (Is this step clear ? The binominal coefficients in the middle are always 0)

A^p=A^(p-1+1)=A^(p-1) * A (a^((p-1)/2)= -1 mod p)

= -1 * A

= -A

Therefore (1+A)^p=1-A mod p

and with the same proof (1-A)^p= 1+A mod p

If you combine these two equalations then you get

(1+A)^(p^2)=1+A mod p , divison by 1+A

(1+A)^(p^2 -1) = 1 mod p this is equal to

(1+A)^((p+1)*(p-1)) = 1 mod p

(1+A)^p = 1 - A , multiplication with (1+A)

(1+A)^(p+1) = 1-A^2 = 1 - a mod p

(1- a) is a natural number, therefore the order of the cyclic group of a natural number with adjoined part is p+1,

in contrast to the cyclic group of natural numbers,

which is sure p-1.

Therefore i reveal a cyclic group with order d | p+1 in the adjoined numbers and show a contradiction.

If you have any question, i will try to answer them.

Nice Greetings from the primes

Bernhard