Clarification on this submodule over $\mathbb{Z}[t]$

Question

I am having trouble making the following calculation consistent. Consider the map from a $\mathbb{Z}[t]$-module to itself:

$$\phi:\mathbb{Z}[t]/I\rightarrow \mathbb{Z}[t]/I,\\p(t)+I\mapsto (p(t)+I)(t^2-1),$$

where $I=(t+1)(t-n+1)\mathbb{Z}[t]$.

We know $\phi$ is a homomorphism. Thus im $\phi$ is a submodule.

Let $x=at+b+I\in\mathbb{Z}[t]/I$. Then \begin{align} \phi(x) & = (at+b)(t^2-1)+I \\ & = (at+b)(t+1)(t-n+1+n-2)+I \\ & = (n-2)(at+b)(t+1)+(at+b)(t+1)(t-n+1)+I \\ & = (n-2)(a(t-n+1)+a(n-1)+b)(t+1)+I\\ & = (n-2)(a(n-1)+b)(t+1)+I \end{align}

Then I concluded that im $\phi=(n-2)(t+1)\mathbb{Z}[t]/I$.

However, every submodule of $\mathbb{Z}[t]/I$ should be a quotient like $J/I$ such that $I$ is a submodule of $J$ and $J$ is a submodule of $\mathbb{Z}[t]$. But in my calculation $I$ is not a submodule of $(n-2)(t+1)\mathbb{Z}[t]$ for $n>3$, this quotient even doesn't make sense.

Where does the flaw in my reasoning occur? Any help will be appreciated.

Answer 1

The map can be written $\phi(p(t)+I)=(t^2-1)p(t)+I$. Thus the image is $$ \bigl((t^2-1)\mathbb{Z}[t]+I\bigr)\big/I $$ An element of the top submodule $J$ is of the form $$ (t^2-1)f(t)+(t+1)(t-n+1)g(t)= (t+1)\bigl((t-1)f(t)+(t-n+1)g(t)\bigr) $$ Now \begin{align} (t-1)f(t)+(t-n+1)g(t) &=(t-n+1+n-2)f(t)+(t-n+1)g(t)\\[4px] &=(t-n+1)(f(t)+g(t))+(n-2)f(t) \end{align} so $J=(t-n+1)\mathbb{Z}[t]+(n-2)\mathbb{Z}[t]$ is generated by $t-n+1$ and $n-2$.

Answer 2

Your first step fails; in fact $$\phi(x)=(at+b)(t^2-1)+I(t^2-1).$$ The map $\phi$ is given by multiplication by $t^2-1$, hence its image consists of all multiples of $t^2-1$. That is to say $\operatorname{im}\phi=(t^2-1)\Bbb{Z}[t]/I$.