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Ladder

Finite fields — the ladder

2 distinct arguments against one idea, from the one that introduces it to the one that assumes the rest.
  1. +01αα+101αα+101αα+110α+1ααα+101α+1α10×01αα+101αα+1000001αα+10αα+110α+11α× mod 40000012302020321the 4 elements of GF(4), built as polynomials in α modulo α+1 + α² — every product of two non-zero elements is non-zerobeside it the integers mod 4, where 2 × 2 = 0 and 1 element has no reciprocal

    The field with four elements

    The integers modulo four are not a field: two times two is zero and two has no reciprocal. There is nevertheless a field with four elements, and building it means giving up on counting as the way to make arithmetic finite.

    rung 1 · computation
  2. 1αα²α³α+1α²+αα³+α²α³+α+1α²+1α³+αα²+α+1α³+α²+αα³+α²+α+1α³+α²+1α³+1GF(16)*15 elementspowerelementα⁰1α¹αα²α²α³α³α⁴α+1α⁵α²+αα⁶α³+α²α⁷α³+α+1α⁸α²+1α⁹α³+αα¹⁰α²+α+1α¹¹α³+α²+αα¹²α³+α²+α+1α¹³α³+α²+1α¹⁴α³+1every non-zero element of GF(16) is a power of α — the 15 of them, once each, around the wheel8 of the 15 elements generate the whole group, which is φ(15) = 8, and the product rule waschecked on all 225 pairs

    Every element is a power of one of them

    Pick the right element of a finite field and its powers run through every other non-zero element exactly once before returning to one. Multiplication becomes addition of exponents, and a table of q − 1 entries replaces the whole multiplication table.

    rung 2 · computation

All ladders