Existence and uniqueness of a morphism that completes an alternate path Let be $p,q\in\mathbb N^*$,
$$\left\{\begin{align}
A&\triangleq \mathbb Z_p[x]\\
B&\triangleq A/(x^q-\bar1)
\end{align}\right.$$
and the morphisms
$$\begin{align}
\varphi:&A\longrightarrow A\\
&f(x)\longmapsto f(x^m)
\end{align}\quad,$$
with $m\in\mathbb N$ fixed, and
$$\begin{align}
\pi:&A\longrightarrow B\\
&f\longmapsto\hat f
\end{align}\quad,$$
which maps each element of $A$ onto the class of $B$ it represents. Prove that
a. there exists a morphism $\psi:B\to B$ such that $\psi\circ\pi\equiv\pi\circ\varphi$ if and only if $\pi(f)=\hat0$ implies $\pi\circ\varphi(f)=\hat0$.
b. if it exists, $\psi$ is unique.
 A: a. $$\begin{matrix}
A&\xrightarrow\varphi&A\\
\downarrow\scriptstyle\pi&&\downarrow\scriptstyle\pi\\
B&\overset\psi\Rightarrow&B
\end{matrix}$$
We will call the proposition about the existence of $\psi$ (i) and the one about the kernel of $\pi$ (ii). Firstly, let's assume (i) holds to deduce (ii): $\pi\circ\varphi(f)=\psi\circ\pi(f)$, and if $\pi(f)=\hat0$, since $\psi$ is a morphism, $\psi\circ\pi(f)=\hat0$, and so $\pi\circ\varphi(f)=\hat0$.
For the converse implication, let $\hat g\in B$ such that, for some $f_0\in A$, $\pi(f_0)=\hat g$. Therefore, $\psi(\hat g)=\pi\circ\varphi(f_0)$. We must check if $\psi$ is well defined. Let us choose $f_1\in A\setminus\{f_0\}$ with $\pi(f_1)=\hat g$ as well.
$$\begin{align}
&\pi(f_1)=\pi(f_0)\\
&\pi(f_1)-\pi(f_0)=\pi(f_1-f_0)=\hat0\quad,
\end{align}$$
which, by (ii), means that 
$$\begin{align}
&\pi\circ\varphi(f_1-f_0)=\hat0\\
&\pi(\varphi(f_1)-\varphi(f_0))=\pi(\varphi(f_1))-\pi(\varphi(f_0))=\hat0\\
&\pi\circ\varphi(f_0)=\pi\circ\varphi(f_1)\quad,
\end{align}$$
so that for every $f\in A$ such that $\pi(f)=\hat g$, $\psi(\hat g)=\pi\circ\varphi(f)$.
It lacks us to prove $\psi$ is a ring morphism. Let $\star\in\{+,\cdot\}$ and $i\in\{1,2\}$, and let us define $\pi(f_i)\triangleq\hat g_i$. From $\psi(\hat g_i)=\pi\circ\varphi(f_i)$, we conclude that
$$\begin{align}
\psi(\hat g_1)\star\psi(\hat g_2)&=\pi(\varphi(f_1)\star\varphi(f_2))=\\ &=\pi(\varphi(f_1\star f_2))=\\
&=\psi\circ\pi(f_1\star f_2)=\\
&=\psi(\pi(f_1)\star\pi(f_2))=\\
&=\psi(\hat g_1\star\hat g_2)\quad,
\end{align}$$
quod erat demonstrandum. The propositions (i) and (ii) are equivalent.
b. If there is a $\psi$ with the property required, its uniqueness comes from the 2nd paragraph of the previous item.
A: I just want to point out that there's something very general going on here: you have a ring homomorphism $h\colon A \to B$ (for you this is $\pi \circ \varphi$) and you'd like to check whether it factors through a quotient $q\colon A \to A/I$ of $A$ by an ideal, in the sense that there exists an $h_*\colon A/I \to B$ such that $h = h_* \circ q$. Such a thing exists if and only if $I$ is contained in the kernel of $f$ (equivalently, $q(a) = 0$ implies $h(a) = 0$), and in that case it is unique—modulo notation, you've shown this.
In other words, to give a homomorphism out of $A/I$ is equivalent to giving a homomorphism out of $A$ which vanishes on $I$. It would be interesting to find the $p, q, m$ for which your condition (ii) holds; the identity
\[
x^{mq} - 1 = (x^q - 1)(x^{q(m - 1)} + x^{q(m - 2)} + \cdots + x^q + 1)
\]
might help.
