Can someone point me to someone that can prove the following, or help me find something similar that is provably correct? See the link near the bottom of this question for a perhaps easier introduction.
Suppose that we have a field with $p^k$ elements. As an example of performing computations in this field, if we have an irreducible polynomial, we can take $x$ as one of the roots of the field, and we can record elements of the field as
$$c_0 x^0 + c_1 x^1 + c_2 x^2 + \dots + c_{k-1} x^{k-1}$$
where $0 \le c < p$ for each $c$ above.
Now, if we multiply this value by another value
$$d_0 x^0 + d_1 x^1 + d_2 x^2 + \dots + d_{k-1} x^{k-1}$$
we can record the result as
$$\underbrace{(c_0 d_0)}_{r_0}x^0 + \underbrace{(c_1 d_0 + c_0 d_1)}_{r_1}x^1 + \dots$$
for all powers of the root $x$ up to $k-1$, and then we have to multiply further constants by a function of an irreducible polynomial. For example, if our irreducible polynomial is $x^2+x+1$, we multiply this by $x$ and find
$$\begin{align} x^2+x+1 &= 0 \\ x^3+x^2+x &= 0 \\ x^3 &= -x^2 -x \end{align} $$
We can proceed similarly for higher powers.
This will give us a result
$$r_0 x^0 + r_1 x^1 + r_2 x^2 + \dots + r_{k-1} x^{k-1}$$
with each $r$ possibly greater than or equal to $p$. If we take each $r$ coefficient and find what the result or remainder is modulo $p$, this will give us the correct result in our field once again.
But we can expand this field by recording each coefficient modulo a set of primes. for example, if $r_0=10$, then $r_0 \equiv 0 \bmod 2$, $r_0 \equiv 1 \bmod 3$, and so on. If we want the actual $r$ value, then we simply use the Chinese remainder theorem to get the actual value of $r_0$, and then take this result modulo $p$ to find the value of $r_0$ in the field.
Now, if we're careful, we can find a prime $q$ where the polynomial that is irreducible modulo $p$ factors completely modulo $q$. In fact, we may be able to find many primes where the polynomial completely factors. If we're careful, and we can use enough of these primes, I believe we can use these primes, and not $p$, to simulate calculations in the field of $p^k$ elements, which I will attempt to explain. I'm wondering where I can find someone that can prove this. That's my question... How can I prove this, or if nobody here can prove this, where can I find someone that can?
Now to explain this a bit further, if we have a prime where our polynomial completely factors, we can construct the matrix
$$\begin{bmatrix} \begin{array}{ccccc|c} {x_1}^0 & {x_1}^1 & {x_1}^2 & \dots & {x_1}^{k-1} & v_1\\ {x_2}^0 & {x_2}^1 & {x_2}^2 & \dots & {x_2}^{k-1} & v_2\\ {x_3}^0 & {x_3}^1 & {x_3}^2 & \dots & {x_3}^{k-1} & v_3\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ {x_k}^0 & {x_k}^1 & {x_k}^2 & \dots & {x_k}^{k-1} & v_k\\ \end{array} \end{bmatrix}$$
where each $x_j$ for some $j$ is one of the roots modulo $q$. Then we can solve for the coefficients of each $x_j$ using this matrix and our recorded values $v_j$, which we calculate by substituting $x_j$ in for the root in our calculations. This will give the value of each $r$, modulo $q$. We then take
$$y_1 \bmod q_1 \\ y_1 \bmod q_2\\ y_1 \bmod q_3\\ \dots$$
and use the Chinese remainder theorem to find the "technical" value of $r_1$, and then take this value modulo $p$ to find the final value or actual (coefficient) value modulo $p$.
ANOTHER (MORE CONCRETE) EXAMPLE
You can also refer to An analog of Chinese remainder theorem using field extensions. This may be easier to understand. Go through the comments for more description.