I'm trying to find number systems with specific properties. If we suppose that $p(x)$, a polynomial in $x$ is irreducible over the finite field with $q$ elements, $q$ prime, I have a few questions.
I believe that we can then use this irreducible polynomial to construct a finite field of size $q^n$ if $n$ is the degree of the polynomial. I want to make sure that I know when I can do this, so that's my first question: "When can I construct this field?"
Now, assuming that we can construct a finite field of size $p^n$ elements in the fashion above, I'm interested in using an analog of Chinese remainder theorem on it. The idea is now to use a smaller prime $r$ such that the polynomial does reduce over $r$. The next idea is that if I then use enough primes $r_1$, $r_2$, $r_3$, \dots that I can represent this "number" exactly. In other words, if the number is $c_1 s_1 + c_2 s_2 + \dots$, where $s_1$ and $s_2$ are roots of the polynomial, and $c_1$ and $c_2$ are non-negative integers, I can possibly recover the exact value of the number using enough primes. Is this possible?
CLARIFICATION
We pick one "exact" polynomial $p(x)$, whose exact value is irreducible modulo $q$, but reducible modulo the smaller primes $r_1$, $r_2$, $\dots$.
COMPLETE EXAMPLE
Here's one complete example; let me know if you would like more.
Let $p(x) = 6 x^3 + 2 x + 3$.
Then, we have, according to Mathematica 11:
$$p(x) \equiv 6 (7 + 9 x + x^3) \bmod 13$$ $$p(x) \equiv 6 (3+x)(9+x)(10+x) \bmod 11$$ $$p(x) \equiv 6 (2+x)(2+5x+x^2) \bmod 7$$ $$p(x) \equiv (1+x^2)(3+x) \bmod 5$$
It's not perfect, but it should do.
Then we let $q = 13$, our largest prime. Note that the polynomial is irreducible modulo $13$.
Now, modulo $11$ is our best example, it has 3 roots: -3, -9, and -10, up to a reversal of the negative sign.
Now the idea is to take equivalence classes, modulo primes, using polynomials and/or field extensions. If we have
$$2 \text{(root 1)} + 3 \text{(root 2)} + 1 \text{(root 3)} \bmod 13$$
Where the roots are taken from $p(x)$, we can rewrite this as:
$$\text{value 1} \equiv 2 \underbrace{(-3)}_{\text{root 1}} + 3 \underbrace{(-9)}_{\text{root 2}} + 1 \underbrace{(-10)}_{\text{root 3}} \bmod 11$$
Furthermore, we can use two more equations to permute the roots and recover the values:
$$\text{value 2} \equiv 2 \underbrace{(-9)}_{\text{root 2}} + 3 \underbrace{(-10)}_{\text{root 3}} + 1\underbrace{(-3)}_{\text{root 1}} \bmod 11$$
$$\text{value 3} \equiv 2 \underbrace{(-10)}_{\text{root 3}} + 3\underbrace{(-3)}_{\text{root 1}} + 1 \underbrace{(-9)}_{\text{root 2}} \bmod 11$$
We can now use linear algebra to recover the coefficients of the roots, modulo $11$.
Assuming that we can do similar for $5$ and $7$, we can do calculations and recover the exact values of each root coefficient up to $5 \cdot 7 \cdot 11 = 385$. We should be able to use this for larger coefficients and polynomials, similarly.
QUESTIONS
My main question is "does this construction work?", and when can I use it?