Irreducible polynomials over the field $GF(2)$ corresponds to prime numbers. Is this a known theorem? Conjecture: Consider the field $GF(2) = {(0, 1)}$. An irreducible polynomial over this field corresponds to a prime number.
For example: $x^4 + x^0$ is irreducible and corresponds to $2^4 + 2^0 = 16 + 1 = 17$, which is prime.
I'll take a prime number $11 = 2^3 + 2^1 + 2^0$. Converting to binary: 1011. The corresponding polynomial over the field $GF(2)$ is $x^3 + x^1 + x^0$. This polynomial is irreducible.
I suspect there is a theorem for this that could be more general. If there is a theorem for this, what is it called?
Please note: I am not a Mathematics Major.
 A: The conjecture is unfortunately false; we can see that
$5 = 2^{2}+1$ is prime, but the corresponding polynomial $(x^2 + 1) = (x+1)(x+1)$ over $\mathbb{F}_{2}$ so is not irreducible; conversely, $x^{4}+x^{3}+1$ is irreducible over $\mathbb{F}_{2}$, but $2^{4}+2^{3}+1 = 16+8+1=25$ is not prime.
I think you have likely found a coincidence for some small numbers, since the numbers corresponding to an irreducible polynomial will always be odd, and most of the odd numbers less than 32 (so corresponding to polynomials of degree 4 or less) are prime. I would not expect there to be anything interesting to discover here.
In fact, if you consider the irreducible polynomials of degree 8 or less, roughly half (36 of 71) correspond to primes and half correspond to composites. Polynomials of degree 15 or less, about 1/4 correspond to primes (1156 of 4720). Once we consider polynomials of degree 20 or less, we are down to about 16% corresponding to primes (17772 out of 111013). I would guess that as the degree of the polynomial gets large, the percentage of irreducible polynomials corresponding to prime numbers approaches 0.
(On the other side of this, there are 82025 primes less than $2^{20}$ only about 22% of which will give an irreducible polynomial.)
