# Constructing a field with p^2 elements

How can we construct a field having p^2 elements, p being an odd prime. Please explain in detail.

(Hint: Think about polynomial rings.)

https://brainmass.com/math/basic-algebra/constructing-field-elements-363570

## SOLUTION This solution is **FREE** courtesy of BrainMass!

There is a standard way of constructing a field with p^n elements, for a prime number p.

You start with Z_p = Z/pZ, which is a field with p elements (integers modulo p) and consider the splitting field of the polynomial f(x) = x^{p^n} - x over Z_p.

Such a polynomial has precisely p^n distinct roots over Z_p, since its derivative is p^n x-1 = -1 (since p^n=0 in Z_p) and is never zero. The set F of roots is a field:

0 and 1 are in F, because 0^{p^n}-0=0, and 1^{p^n}-1 = 0.

For all x,y in F we have

(x+y)^{p^n} = x^{p^n} + y^{p^n} (all other terms contain the coefficient divisible by p, which is zero in Z_p) = x+y

Since x and y are solutions, so (x+y) is in F for any x and y in F.

(xy)^{p^n} = x^{p^n} y^{p^n} = xy,

so F is closed under multiplication and addition.

Finally, for a non-zero x in F we have (x^{-1})^{p^n} = (x^{p^n})^{-1} = x^{-1}, so the inverse of x is in F.

So, F is a field.

In a slightly different approach, we can consider the polynomial ring Z_p[x] over Z_p and find a monic irreducible polynomial q(x) of degree 2 over it, which always exists. Then, we consider the quotient Z_p[x]/(q(x)) of Z_p[x] by the ideal generated by q(x). This quotient is a field, because of irreducibility of q(x), and it has precisely p^2 elements.

For a reference, use http://en.wikipedia.org/wiki/Finite_field .

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