q(x)F[x] is in p(x)F[x] iff p(x) divides q(x)

Dependencies:

1. Polynomial divisibility

Let $R$ be a ring and $p(x), q(x) \in R[x]$.

Then $p(x) \mid q(x) \iff q(x)F[x] \subseteq p(x)F[x]$.

Proof

\begin{align} & p(x) \mid q(x) \\ &\Rightarrow \exists r(x) \in F[x], q(x) = p(x)r(x) \\ &\Rightarrow \forall f(x) \in F[x], q(x)f(x) = p(x)r(x)f(x) \in p(x)F[x] \\ &\Rightarrow q(x)F[x] \subseteq p(x)F[x] \end{align}

\begin{align} & q(x)F[x] \subseteq p(x)F[x] \\ &\Rightarrow q(x) \in p(x)F[x] \\ &\Rightarrow \exists r(x) \in F[x], q(x) = p(x)r(x) \\ &\Rightarrow p(x) \mid q(x) \end{align}

Dependency for:

1. The ideal generated by an irreducible polynomial is maximal

Info:

• Depth: 4
• Number of transitive dependencies: 4

Transitive dependencies:

1. Group
2. Ring
3. Polynomial
4. Polynomial divisibility