How to prove that two principal ideals are equal

Question

Background info

Provided that the general form of a polynomial is $(a_0X^{0}+...+a_nX^{n})$ where $X$ is an element of the field provided.

an ideal generated by an element is the set $(a*r s.t. a\in V, r\in ring)$

Question

For $F$ a field, and $q(x)$ a polynomial in the polynomial ring $F[X]$, with $a\in F$ where $a \neq 0$, show that $\langle q(x) \rangle= \langle aq(x) \rangle$,

How do you prove this by saying that they are both subsets of each other?

Answer 1

The principal ideal $\langle q(x) \rangle$ may be rewritten as $$\langle q(x) \rangle = q(x) F[x],$$ where: $$q(x) F[x] = \{ q(x) p(x) : p(x) \in F[x] \}.$$ Similarly, we have that: $$\langle a \cdot q(x) \rangle = (a \cdot q(x)) F[x] = \{ a \cdot q(x) r(x) : r(x) \in F[x] \}.$$ Letting $q(x) p(x)$ be an arbitrary element in $q(x) F[x]$, we have that $$q(x) p(x) = a \cdot q(x) \left(\frac{1}{a} p(x) \right) \in \langle a \cdot q(x) \rangle$$ and we thus have that $\langle q(x) \rangle \subseteq \langle a \cdot q(x) \rangle$. Conversely, letting $a \cdot q(x) r(x)$ be an arbitrary element in $\langle a \cdot q(x) \rangle$, we have that $$a \cdot q(x) r(x) = q(x) \left( a r(x)\right) \in \langle q(x) \rangle,$$ and we thus have that $\langle q(x) \rangle = \langle a \cdot q(x) \rangle$ by mutual inclusion.

Answer 2

Hint: Prove that the generators of the ideals divide each other and use the facts that in any ring, $x\in\langle y\rangle \iff y\mid x $, and $x\in \langle y\rangle \implies \langle x\rangle \subseteq \langle y\rangle $.