Finding the Kernel of a ring morphism

[MathNugget]

The morphism is: $\phi: \frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]} \rightarrow \frac{\mathbb{Z}_p[X]}{(X^2+tX+q)\mathbb{Z}_p[X]}$, $\phi(\overline{f(x)})=\widehat{f(x)}$ (mod p).

Some introductions:
$p\mathbb{Z}[X]$ is the ideal generated by p in $\mathbb{Z}[X]$.
Similarly for $(X^2+tX+q)\mathbb{Z}_p[X]$ .

$\overline{f(x)}$ is an element (class) of $\frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]}$. (1st ring in the definition of the morphism)
$\widehat{f(x)}$ is an element of $\frac{\mathbb{Z}_p[X]}{(X^2+tX+q)\mathbb{Z}_p[X]}$. (2nd ring in the definition of the morphism)



I am trying to prove $Ker(\phi)=\{\overline{f(x)} \mid \phi(\overline{f(x)})=\widehat{0} \}$ to be:

$ker(\phi)=\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$


I attempted to use the double inclusion method. It's easy to show that if $\overline{f(x)} \in \frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$, then $f(x)=(X^2+tX+q)g(X)+p h(X)$, with $g(X), h(X) \in \mathbb{Z}[X]$, and $\widehat{f(x)}=\widehat{(X^2+tX+q)g(X)}=\widehat{0}$.


I have trouble proving that if $\widehat{f(X)}=\widehat{0} \rightarrow \overline{f(X)} \in \frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$.

I suppose
$$\widehat{f(X)}=\widehat{0}$$$$f(x) \in (X^2+tX+q)\mathbb{Z}_p[X]$$$$f(x) \equiv (X^2+tX+q) g(x) \: (mod p)$$$$f(x) -(X^2+tX+q) g(x)=pk$$$$f(x) =(X^2+tX+q) g(x)+pk$$$$f(x) \in (X^2+tX+q) g(x)+pk$$with $g(x)\in \mathbb{Z}[X], k \in \mathbb{Z}$, so
$$f(x) \in (X^2+tX+q, p)\mathbb{Z}[X]$$. And when we go into $\overline{f(x)}$, I suppose we get that $\overline{f(x)}\in \frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$ (that which I was trying to prove). Could I get a 2nd opinion about what I did here?

One thing I am unsure about, is when I switched from mod p to next line, whether $f(x) \in (X^2+tX+q) g(x)+pk$ or $$f(x) \in (X^2+tX+q) g(x)+pk(x)$$. I am starting to think it's the latter, but I don't think I understand it well enough.

 

[fresh_42]

You are on the right track but I think you could make your life easier. $\mathbb{Z}[X]/p\mathbb{Z}[x]\cong \mathbb{Z}_p[X]$ which makes the homomorphism to be

$$\phi\, : \,\mathbb{Z}_p[X]\twoheadrightarrow\mathbb{Z}_p[X]/\bigl\langle X^2+tX+q \bigr\rangle$$
and its kernel is the ideal $\bigl\langle X^2+tX+q \bigr\rangle \subseteq \mathbb{Z}_p[X]$ which is what you wrote.

 

[MathNugget]

You are on the right track but I think you could make your life easier. $\mathbb{Z}[X]/p\mathbb{Z}[x]\cong \mathbb{Z}_p[X]$ which makes the homomorphism to be

$$\phi\, : \,\mathbb{Z}_p[X]\twoheadrightarrow\mathbb{Z}_p[X]/\bigl\langle X^2+tX+q \bigr\rangle$$
and its kernel is the ideal $\bigl\langle X^2+tX+q \bigr\rangle \subseteq \mathbb{Z}_p[X]$ which is what you wrote.

Thank you. I didn't know I can do that. Much appreciated.

I tried to get a sense of how these things go some weeks/months ago, and posted some threads, but I wasn't quite able to understand the results.