Use Exercise 6 to prove the following. If \(f\), \(g\), \(h\), and \(p\) are polynomials over the field \(F\) and \(p \neq 0\), and if \(f \equiv g \bmod p\), then \(h(f) \equiv h(g) \bmod p\).
Solution
Write \[ h(x)=a_0+a_1x+\cdots+a_rx^r. \] From \(f\equiv g\pmod p\), repeated application of Exercise 6 gives \[ f^k\equiv g^k\pmod p \] for every \(k\ge0\). Multiplication by the scalar \(a_k\) preserves congruence, and adding the resulting congruences gives \[ \sum_{k=0}^r a_k f^k \equiv \sum_{k=0}^r a_k g^k \pmod p. \] Therefore \[ h(f)\equiv h(g)\pmod p. \]