Problem 1577

1. Define the Frobenius Endomorphism:
Let $K$ be a finite field with $p$ elements, where $p$ is a prime. For any polynomial $f(x) = \sum_{i=0}^n a_i x^i \in K[x]$, define $\overline{f(x)} = \sum_{i=0}^n a_i x^{p^i}$. This transformation is related to the Frobenius endomorphism $\sigma: K[x] \to K[x]$ defined by $\sigma(f) = f^p$.

2. Homomorphism Construction:
Using the universal property of $K[x]$, we can construct a $K$-algebra homomorphism $\varphi: K[x] \to \text{End}_K(K[x])$ such that $\varphi(x) = \sigma$. For any polynomial $f(x) = \sum_{i \ge 0} a_i x^i \in K[x]$, we have:
$$\varphi(f(x))(x) = \left( \sum_{i \ge 0} a_i \sigma^i \right)(x) = \sum_{i \ge 0} a_i x^{p^i} = \overline{f(x)}$$

3. **Division Algorithm in $K[x]$:**
For any polynomials $f(x), g(x) \in K[x]$ with $f(x) \neq 0$, there exist polynomials $q(x), r(x) \in K[x]$ such that:
$$g(x) = q(x) f(x) + r(x) \quad \text{with} \quad \deg(r(x)) < \deg(f(x))$$

4. Applying the Homomorphism:
Applying the homomorphism $\varphi$ to the equation $g(x) = q(x) f(x) + r(x)$, we get:
$$\overline{g(x)} = \varphi(g(x))(x) = \varphi(q(x) f(x) + r(x))(x) = \varphi(q(x))(\varphi(f(x))(x)) + \varphi(r(x))(x)$$
Since $\varphi$ is a homomorphism, we have:
$$\overline{g(x)} = \varphi(q(x))(\overline{f(x)}) + \overline{r(x)}$$

5. **Expressing $\varphi(q(x))$:**
Let $q(x) = \sum_{i \ge 0} c_i x^i$. Then:
$$\varphi(q(x))(x) = \sum_{i \ge 0} c_i \sigma^i(x) = \sum_{i \ge 0} c_i x^{p^i}$$
Therefore:
$$\varphi(q(x))(\overline{f(x)}) = \sum_{i \ge 0} c_i \overline{f(x)}^{p^i}$$

6. Combining Results:
Thus, we have:
$$\overline{g(x)} = \sum_{i \ge 0} c_i \overline{f(x)}^{p^i} + \overline{r(x)} = \overline{f(x)} \left( \sum_{i \ge 0} c_i \overline{f(x)}^{p^i - 1} \right) + \overline{r(x)}$$
Since $\deg(\overline{r(x)}) = p^{\deg(r(x))} < p^{\deg(f(x))} = \deg(\overline{f(x)})$, $\overline{r(x)}$ is the remainder of the division of $\overline{g(x)}$ by $\overline{f(x)}$.

7. Conclusion:
Therefore, $f(x) \mid g(x)$ if and only if $r(x) = 0$, which implies $\overline{r(x)} = 0$. Hence, $\overline{f(x)} \mid \overline{g(x)}$.

$\blacksquare$