The Field of Cong. Classes of Polynomials Modulo p(x) over a Field

The Field of Congruence Classes of Polynomials Modulo p(x) over a Field

 Theorem 1: Let $(F, +, \cdot)$ be a field and let $p \in F[x]$ where $p$ is a nonconstant polynomial. The set $F[x] /$ with the operations $+$ and $\cdot$ defined for all $[a(x)]_{p(x)}, [b(x)]_{p(x)}$ by $[a(x)]_{p(x)} + [b(x)]_{p(x)} = [a(x) + b(x)]_{p(x)}$ and $[a(x)]_{p(x)}[b(x)]_{p(x)} = [a(x)b(x)]_{p(x)}$ is a field if and only if $p$ is irreducible over $F$.

We use the notation "$[a(x)]$", "$[b(x)]$", and "$[c(x)]$" in place of "$[a(x)]_{p(x)}$", "$[b(x)]_{p(x)}$", and "$[c(x)]_{p(x)}$" in the following proof.

• Proof: Let $[a(x)], [b(x)], [c(x)] \in F[x]/<p(x)>$. Then:
(1)
\begin{align} \quad [a(x)] + ([b(x)] + [c(x)]) = [a(x)] + [b(x) + c(x)] = [a(x) + (b(x) + c(x))] = [(a(x) + b(x)) + c(x)] = [a(x) + b(x)] + [c(x)] = ([a(x)] + [b(x)]) + [c(x)] \end{align}
(2)
\begin{align} \quad [a(x)]([b(x)][c(x)]) = [a(x)][(b(x)c(x))] = [a(x)(b(x)c(x))] = [(a(x)b(x))c(x)] = ([a(x)b(x)])[c(x)] \end{align}
• So $+$ and $\cdot$ are associative. Also:
(3)
\begin{align} \quad [a(x)] + [b(x)] = [a(x) + b(x)] = [b(x) + a(x)] = [b(x)] + [a(x)] \end{align}
(4)
\begin{align} \quad [a(x)][b(x)] = [a(x)b(x)] = [b(x)a(x)] = [b(x)][a(x)] \end{align}
• So $+$ and $\cdot$ are commutative.
• Furthermore:
(5)
\begin{align} \quad [a(x)]([b(x)] + [c(x)]) = [a(x)][b(x) + c(x)] = [a(x)(b(x) + c(x))] = [a(x)b(x) + a(x)c(x)] = [a(x)][b(x)] + [a(x)][c(x)] \end{align}
• So the distributivity property holds. Also it is easy to show that $[0]$ and $[1]$ are the additive and multiplicative identities respectively.
• For all $[a(x)]$ we have that $[-a(x)]$ is the additive inverse.
• Hence $F[x] / <p(x)>$ has multiplicative inverses for all $[a(x)] \in F[x] / <p(x)>$ with $[a(x)] \neq [0]$ provided that $p$ is irreducible (since then $\gcd (r, p) = 1$ for any polynomial $r$ with $\deg (r) < \deg (p)$. $\blacksquare$