The Twin Primes Conjecture Gets Solved for Finite Fields
wired.com
wired.com
I'm not a mathematician, genuinely curious.
Your intuition is correct, but doesn't go far enough! In fact, the notion of a prime in general doesn't make much sense (in Galois fields) :)
The article explains it fairly well. Basically, what was proved was the fact that there exist an infinite number of twin prime polynomials (i.e. polynomials that cannot be factored and differ by a fixed gap) in finite number systems.
It's often referred to as the "s_box" and the "reverse s_box" (for de-encryption).
I'm out of practice, so maybe you're staying that because you'd essentially have prime equivalence classes that the meaning is different?
If you are curious about some of the details, we can add that “prime” is not quite the same as “irreducible”, although the prime elements and irreducible elements are the same in the more familiar rings.
The reason why “prime” doesn’t make much sense in Galois fields is precisely because it’s a field—technically, there are no primes in Galois fields, because every nonzero element divides 1, therefore every nonzero element is a unit, therefore not a prime. So there are no primes in Galois fields. This is mentioned in the article.
(The other answer to your question is more complete but also a bit more advanced, figured this was worth surfacing.)
You are talking about the polynomial ring, which is not a field.
The finite field F2 has four polynomials: 0, 1, x, and x+1. Other polynomials do not exist, because x^2=x.
As far as I can tell you're referring to the integer finite field containing the values {0, 1}.
2 % 2 = 0
3 % 2 = 1
x^n % 2 = x
2x % 2 = 0
(2n + 1)x % 2 = x
So there are no (distinct) constants other than 0 or 1 and no multiples or exponents of x other than 1.These facts might be obvious to someone who understands the jargon and theory of the mathematics in question, but probably need a bit more clarity when the target is the general populace.
They are different things really, but close enough to have much in common.
upd. formatting
Consider for example Grothendieck-Ax theorem: if f: C^n -> C^n is an injective (that is, one-to-one) multiple variable polynomial function over complex numbers, then it's also surjective. One of the ways to prove it is to show via model theory magic that if there was a counterexample to this theorem, there would also be a counterexample g: F^n -> F^n, where F is some finite field. But an injective function from finite set to finite set of the same size is also surjective.