Finding the Kernel of a ring morphism

[MathNugget]

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

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

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



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

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


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


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

I suppose
[math]\widehat{f(X)}=\widehat{0}[/math][math]f(x) \in (X^2+tX+q)\mathbb{Z}_p[X][/math][math]f(x) \equiv (X^2+tX+q) g(x) \: (mod p)[/math][math]f(x) -(X^2+tX+q) g(x)=pk[/math][math]f(x) =(X^2+tX+q) g(x)+pk[/math][math]f(x) \in (X^2+tX+q) g(x)+pk[/math]with [imath]g(x)\in \mathbb{Z}[X], k \in \mathbb{Z}[/imath], so
[math]f(x) \in (X^2+tX+q, p)\mathbb{Z}[X][/math]. And when we go into [imath]\overline{f(x)}[/imath], I suppose we get that [imath]\overline{f(x)}\in \frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}[/imath] (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 [imath]f(x) \in (X^2+tX+q) g(x)+pk[/imath] or [math]f(x) \in (X^2+tX+q) g(x)+pk(x)[/math]. 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. [imath] \mathbb{Z}[X]/p\mathbb{Z}[x]\cong \mathbb{Z}_p[X] [/imath] which makes the homomorphism to be

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

 

[MathNugget]

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

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