- #1

Feryll

- 7

- 0

*is exactly.*

p is a prime.

"We know that p is reducible [ie p=(a+bi)(a-bi), a,b∈Z] iff (p)=pZ

p is a prime.

"We know that p is reducible [ie p=(a+bi)(a-bi), a,b∈Z] iff (p)=pZ

*is not prime.*

Consider the isomorphisms

Consider the isomorphisms

Z

*≅Z[X]/(X*

Z^{2}+1)Z

*/(p)≅Z[X]/(X*

Z^{2}+1,p)Z

*/(p)≅(Z[X]/(p))/(X*

Z^{2}+1)Z

*/(p)≅F*_{p}[X]/(X^{2}+1)

If p≠2, we have

If p≠2, we have

(p) reducible⇔X

⇔-1∈(F

⇔p congruent to 1 mod(4) (Euler's criterion)

"I just barely know where to start with what he's getting at. What is (p), exactly? What is Z[X] and F^{2}+1 factors in F_{p}[X]⇔-1∈(F

_{p}^{*})^{2}, the group of squares in F_{p}^{*}⇔p congruent to 1 mod(4) (Euler's criterion)

_{p}[X]?