Why the Cartesian product of fields isn’t a field

Kragen Javier Sitaker, 2019-05-02 (2 minutes)

If you have some underlying field like GF(2) you might think you could derive an infinite variety of other fields from it by way of taking arrays of some finite size which operate elementwise.

For example, GF(2) is just bits, using AND as multiplication and XOR as addition; considering the field axioms, these are associative, commutative, have identity, distribute (a & (b ^ c) = (a & b) ^ (a & c)) and have inverses (each element is its own inverse, trivially so in the case of AND: 1 & 1 = 1), so that’s a finite field. You might think that you could extend this elementwise to “bitvectors” (not in the sense of a vector space, just in the sense of arrays of bits) but this fails when we get to the inverse: AND with a “bitvector” containing zeroes is information-destroying, so there can be no inverse. In the single-bit case, we get to escape by pleading division by zero, but not in the multibit case.

So “bitvectors” of some size form a commutative ring with unity, but not a field.

There is a field of size 2⁸, though. It’s GF(256), which is not just ℤ/256ℤ, arithmetic modulo 256, as you might think — that’s, again, just a commutative ring with unity, since there are plenty of pairs of numbers that multiply to zero, like 2·128, so not every member has a multiplicative inverse. No, in some sense it is a vector of 8 bits, but I don’t understand the construction of the operations; it’s some kind of construction with monic irreducible polynomials.

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