Let p be a prime congruent to -1 mod 4. Show that X^2 + 1 is irreducible in Z_p[X], and hence K = Z_p[X] / (X^2 + 1) is the field of order p^2. Note that K has a multiplication similar to that of the complex numbers.Proof:
Since p = -1 (mod 4), then -1 is not a quadratic residue of p. It means that the equation x^2 = -1 (mod p) has no solution.
It also means that x^2 + 1 = 0 (mod p) has no solution. Thus x^2 + 1 is irreducible in Z_p[x].
Hence the corresponding field Z_p[x] / (x^2 + 1) has p^2 elements, or has the order p^2.
Each element has the form ax + b, where a, b can be 0, 1, 2, …, p-1.
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