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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.

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Minor point: have you checked that the map $\psi$ is well-defined? –  Dylan Moreland Dec 13 '11 at 21:41
    
@DylanMoreland It turns out that's not minor; it's a pretty big part of the solution. I'll post it in a few days. –  Luke Dec 14 '11 at 21:05
    
Interesting. I'll try to find some time to think about this today. –  Dylan Moreland Dec 14 '11 at 21:08
    
@DylanMoreland In order to make sense, $p$ and $q$ have to be integers; but $\mathbb Z_{-k}=\mathbb Z_k$, and $\mathbb Z_0$ is useless (empty, I think), so I restricted $p$ and $q$ to the positive integers. –  Luke Dec 16 '11 at 19:24
    
I just think people will find it strange. $\mathbf N^+$ or something like that might be better. Good on you for answering your own question, by the way. –  Dylan Moreland Dec 16 '11 at 19:28
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2 Answers 2

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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.

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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.

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