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Proposition: A polynomial f(x)F[x] has a root rF if and only if (xr) divides f.

Corollary: A polynomial of degree n over a field F has at most n roots.

Proposition: A finite subgroup of the multiplicative group of a field is cyclic.

Proof: Let U be such a subgroup. By the fundamental, theorem of abelian groups, U is the product of its Sylow p-subgroups. Let U(p) be such a subgroup. If U(p) were not cyclic, then U(p) and hence U would have more than p elements that are solutions to the equation xp=1. But xp1 has at most p roots. Since U(p) is cyclic for each p dividing the order of U, U itself is cyclic.

Corollary: The group of units (Z/pZ)× is cyclic.

Generators of this group are called primitive roots mod p.

Back to the Gaussian integers

The irreducibles in Z[i] are:

  • (1+i)
  • pZ with p3(mod4)
  • a±bi where a2+b2=p for pZ and p1(mod4).

A positive integer is a sum of two squares if and only if it factors n=2kp1e1pkekq1f1qrfr where the pi1(mod4) and the qi3(mod4) and all the fi are even.

The proof follows from the question of when is n=N(x) for some xZ[i].

Algorithm for Fermat’s theorem

Suppose p1(mod4). To write p=a2+b2, find a solution u to the congruence u21(modp). Then use the Gaussian Euclidean algorithm to find a generator π for the ideal (p,u+i) in Z[i]. This generator divides p so its norm is a divisor of p2. If its norm were p2, then π would be an associate of p and this would mean p divides u+i, which it visibly does not. If its norm were 1, then the ideal (p,u+i) would be all of Z[i] and so we would have px+(u+i)y=1 in Z[i]. But in that case, multiplying by (ui) would be px(ui)+(u2+1)y=(ui) and since p divides the left side we’d have p dividing ui, which is not true. So therefore N(π)=p and so if π=a+bi we have a2+b2=p.