(Sorry for posting two proof-verification questions twice in a row within such short span of time - I'm not posting any other proof-verification questions soon. )
I'm learning abstract algebra and there's an exercise which states
Let $F$ be a field, and $q(x) \in F[x]$. If $\displaystyle q(x) = \prod_{0 \leq i \leq k} [p_i(x)]^{e_i}$ where $p_i(x)$ are irreducible over $F[x]$, then prove that $$ \displaystyle \frac{F[x]}{q(x)} \simeq \frac{F[x]}{(p_1(x))^{e_1}} \oplus \frac{F[x]}{(p_2(x))^{e_2}} \oplus \cdots \oplus \frac{F[x]}{(p_k(x))^{e_k}}$$
I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?
Let $A = \frac{F[x]}{q(x)}$ and $B = \frac{F[x]}{(p_1(x))^{e_1}} \oplus \frac{F[x]}{(p_2(x))^{e_2}} \oplus \cdots \oplus \frac{F[x]}{(p_k(x))^{e_k}}$. To show that $A \simeq B$, we will construct two rings homomorphisms $\phi_1, \phi_2$ such that $\phi_1$ maps $A$ to $B$ and $\phi_2$ maps $B$ to $A$, which would imply the conclusion.
Construction of the ring homomorphism $\phi_1$ is relatively easier, for $h(x) \in F[x]$, let $f_i(h(x))$ denote the remainder of $h(x)$ when divided by $[p_i(x)]^{e_i}$. Then it's easy to see the mapping $$ \phi_1 \overset{\text{def}}{:=}h(x) \rightarrow (f_1(h(x)), f_2(h(x)), \cdots, f_k(h(x)))$$ is a ring homomorphism from $A$ to $B$.
For the other way, since $\gcd([p_i(x)]^{e_i}, \frac{q(x)}{[p_i(x)]^{e_i}}) = 1$, we can always find polynomials $m_i(x), n_i(x) \in F[x]$ such that $m_i(x)\frac{q(x)}{[p_i(x)]^{e_i}}+n_i(x)[p_i(x)]^{e_i} = 1$. Then it's not hard to see the mapping $$\displaystyle \phi_2 \overset{\text{def}}{:=}(h_1(x), h_2(x), \cdots, h_k(x)) \rightarrow \sum_{0 \leq i\leq k} \frac{q(x)}{[p_i(x)]^{e_i}} m_i(x)h_i(x) \mod q(x) $$ is a ring homomorphism from $B$ to $A$, so we're done.