Basic Theorems Reg. Cong. Classes of Poly. Mod. p(x) over a Field

Basic Theorems Regarding Congruence Classes of Polynomials Modulo p(x) over a Field

Recall from the Congruence Classes of Polynomials Modulo p(x) over a Field page that if $(F, +, \cdot)$ is a field and $p \in F[x]$ then for any $a, b \in F[x]$ we said that $a$ is congruent to $b$ modulo $p$ denoted $a(x) \equiv b(x) \pmod {p(x)}$ if $p(x) | (a(x) - b(x))$.

We saw that congruence modulo $p$ is a equivalence relation and we defined the congruence class of $a$ modulo $p$ denoted $[a(x)]_{p(x)}$ by:

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\begin{align} \quad [a(x)]_{p(x)} = \{ b(x) \in F[x] : b(x) \equiv a(x) \pmod {p(x)} \} \end{align}

We noted that any polynomial $b(x) \in [a(x)]_{p(x)}$ is of the form:

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\begin{align} \quad b(x) = a(x) + q(x)p(x) \end{align}

We denoted the set of all congruence classes modulo $p$ by $F[x] / <p(x)>$.

Lastly, we showed that if $p(x) \neq 0$ and $p$ does not divide $a$ then there exists exactly one polynomial $r(x) \in [a(x)]_{p(x)}$ such that $\deg (r) < \deg (p)$.

We now look at some basic theorems regarding congruence classes of polynomials modulo $p$ - all of which are analogous to proof for congruence of integers modulo $m$.

Theorem 1: Let $(F, +, \cdot)$ be a field and let $p \in F[x]$ with $p(x) \neq 0$. Let $a, b, c, d \in F[x]$. If $a(x) \equiv b(x) \pmod {p(x)}$ and $c(x) \equiv d(x) \pmod {p(x)}$ then $a(x) + c(x) \equiv b(x) + d(x) \pmod {p(x)}$.

• Proof: Since $a(x) \equiv b(x) \pmod {p(x)}$ there exists a polynomial $q_1 \in F[x]$ such that:

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\begin{align} \quad p(x)q_1(x) = a(x) - b(x) \end{align}

• Similarly, since $c(x) \equiv d(x) \pmod {p(x)}$ there exists a polynomial $q_2 \in F[x]$ such that:

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\begin{align} \quad p(x)q_2(x) = c(x) - d(x) \end{align}

• Adding these equations together gives us:

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\begin{align} \quad p(x)[q_1(x) + q_2(x)] = (a(x) + c(x)) - (b(x) + d(x)) \end{align}

• Hence:

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\begin{align} \quad a(x) + c(x) \equiv b(x) + d(x) \pmod {p(x)} \quad \blacksquare \end{align}

Theorem 2: Let $(F, +, \cdot)$ be a field and let $p \in F[x]$ with $p(x) \neq 0$. Let $a, b, c, d \in F[x]$. If $a(x) \equiv b(x) \pmod {p(x)}$ and $c(x) \equiv d(x) \pmod {p(x)}$ then $a(x)c(x) \equiv b(x)d(x) \pmod {p(x)}$.

Theorem 3: Let $(F, +, \cdot)$ be a field and let $p \in F[x]$ with $p(x) \neq 0$. Let $a, b, c \in F[x]$. If $a(x)b(x) \equiv a(x)c(x) \pmod {p(x)}$ and $\gcd (a, p) = 1$ then $b(x) \equiv c(x) \pmod {p(x)}$.

• Proof: Since $a(x)b(x) \equiv a(x)c(x) \pmod {p(x)}$ we have that:

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\begin{align} \quad p(x) | (a(x)b(x) - a(x)c(x)) = a(x)[b(x) - c(x)] \end{align}

• Since $\gcd (a, p) = 1$ this means that $p(x) | (b(x) - c(x))$ so $b(x) \equiv c(x) \pmod {p(x)}$. $\blacksquare$