Ed and I came up with this simple construction for fields of order p^2 yesterday. We thought we had one for fields of order p^n, but then we found out that we were using a theorem that only holds in characteristic 0. Anyway:

Suppose p=2. Let alpha be a root of x^2+x+1=0 in Z_2[x]. This polynomial is irreducible, as 0 and 1 do not satisfy it. Thus Z_2[alpha] is a field with four elements.

Now suppose p>2. Let k be a quadratic nonresidue modulo p. Then x^2-k is irreducible in Z_p[x], and let alpha be a root of x^2-k=0. Thus Z_p[alpha] is a field with p^2 elements.

(I had to break off the p=2 case because there are no quadratic nonresidues modulo 2.) The construction for p>2 seems a bit unfortunate, since quadratic reciprocity is a fairly powerful theorem, and I had wanted to use something less powerful to apply this construction. This construction doesn’t really seem to lend itself to constructions of fields of order p^n in general, but we can sometimes get higher ones.

If k_1, k_2, …, k_m are the quadratic nonresidues modulo p, then x^2-k_i is irreducible in Z_p[x] for each k_i. If alpha_i is a root of x^2k_i, then Z_p[alpha_1, alpha_2, …, alpha_m] will be a field with p^(2^m) elements.

I like this approach better than the normal way of doing things like Z_2[x]/(x^2+1) for a field of order 4, which seems very unnatural to me.

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## About Simon

Hi. I'm Simon Rubinstein-Salzedo. I'm a mathematics postdoc at Dartmouth College. I'm also a musician; I play piano and cello, and I also sometimes compose music and study musicology. I also like to play chess and write calligraphy. This blog is a catalogue of some of my thoughts. I write them down so that I understand them better. But sometimes other people find them interesting as well, so I happily share them with my small corner of the world.

Awesome.

You aren’t actually using anything nearly as strong as quadratic reciprocity; you are just using the existence of a quadratic nonresidue mod p for p>2, which follows immediately from x^2 = (-x)^2. But your construction of a field of order p^(2^m) does not work, because Z_p(alpha_1, alpha_2, … , alpha_m) = Z_p(alpha_1); this is the case because for each j, (alpha_j/alpha_1)^2 = k_j/k_1 is a quadratic residue mod p, so alpha_j/alpha_1 is in Z_p.

I agree that Z_2[x]/(x^2+1) is not a natural construction for a field, because it is not a field, as x^2+1 = (x+1)^2… As you noted, the existence of a field of order p^n follows immediately from the existence of an irreducible polynomial of degree n in Z_p[x], so I believe that the standard way of showing the existence of a field of order p^n is to prove that there are exactly (1/n)*sum_{d divides n}(mu(d)p^(n/d)) monic irreducible polynomials of order n in Z_p[x], and this number is positive.