(another) exercise on isomorphisms of rings

[MathNugget]

I am trying to prove [imath]\frac{\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]}\simeq \frac{(\mathbb{Z}[X]/p(X^2+tX+q)\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]/(X^2+tX+q)\mathbb{Z}[X]}[/imath]. It's not very clear how I'd get from one to another, but I can do this:

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

I do not know what to do from here (I am trying to avoid the process of finding a surjective morphism from classes to classes, then finding its ker)

 

[MathNugget]

I have an idea for a different thread (that is unrelated but will save the forum from me asking a 2-3 more questions). I still don't know how to solve this though (I have tried with the surjective morphism, somehow it always ends up having to prove

[imath]\frac{\mathbb{Z}[X]}{p(X^2+tX+q)\mathbb{Z}[X]} \simeq \frac{\mathbb{Z}[X]}{(X^2+tX+q)\mathbb{Z}[X]}[/imath]

Maybe the morphism isn't the identity, would make sense why I cannot force the ker to be as in the isomorphism theorem...

 

[blamocur]

As I mentioned in my reply in your other thread, you need to make your post more understandable to an average Joe. You cannot expect your readers to know the context.

 

[MathNugget]

As I mentioned in my reply in your other thread, you need to make your post more understandable to an average Joe. You cannot expect your readers to know the context.

Well...the context is

[imath]p \in \mathbb{Z}[/imath] prime number, [imath]X^2+tX+q \in \mathbb{Z}[X][/imath] with roots [imath]\alpha, \overline{\alpha} \notin \mathbb{Z}[/imath].
Trying to prove:

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

the "/" means factorisation, just like the fraction. I could also write it like this:
[imath]\frac{\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]}\simeq \frac{\frac{\mathbb{Z}[X]}{p(X^2+tX+q)\mathbb{Z}[X]}}{\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{(X^2+tX+q)\mathbb{Z}[X]}}[/imath]
if it makes it clearer


There's not much context? Unless the problem is currently impossible to solve, because the premises are not enough (which I don't know right now)...

 

[blamocur]

Does [imath](X^2+tX+q,p) \mathbb Z[X][/imath] stand for an ideal generated by [imath]X^2+tX+q[/imath] and [imath]p[/imath] ?

 

[MathNugget]

Does [imath](X^2+tX+q,p) \mathbb Z[X][/imath] stand for an ideal generated by [imath]X^2+tX+q[/imath] and [imath]p[/imath] ?

Yes... and the polynomial is irreducible (just to be clear). I am using this notation, so we know in which ring we consider the ideal. Here, it's the ideal generated by [imath]X^2+tX+q[/imath] and [imath]p[/imath] in [imath]\mathbb{Z}[X][/imath].

 

[blamocur]

Well...the context is

[imath]p \in \mathbb{Z}[/imath] prime number, [imath]X^2+tX+q \in \mathbb{Z}[X][/imath] with roots [imath]\alpha, \overline{\alpha} \notin \mathbb{Z}[/imath].
Trying to prove:

the "/" means factorisation, just like the fraction. I could also write it like this:
[imath]\frac{\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]}\simeq \frac{\frac{\mathbb{Z}[X]}{p(X^2+tX+q)\mathbb{Z}[X]}}{\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{(X^2+tX+q)\mathbb{Z}[X]}}[/imath]
if it makes it clearer


There's not much context? Unless the problem is currently impossible to solve, because the premises are not enough (which I don't know right now)...

[imath]\gdef\zx{\mathbb Z[X]}[/imath]
[imath]\gdef\px{X^2+tX+q}[/imath]
Wouldn't you first have to show that [imath]\frac{(\px,p)\zx}{(\px)\zx}[/imath] is isomoprphic to an ideal of [imath]\frac{\zx}{p(\px) \zx}[/imath] so that the right hand side can be defined?

 

[MathNugget]

[imath]\gdef\zx{\mathbb Z[X]}[/imath]
[imath]\gdef\px{X^2+tX+q}[/imath]
Wouldn't you first have to show that [imath]\frac{(\px,p)\zx}{(\px)\zx}[/imath] is isomoprphic to an ideal of [imath]\frac{\zx}{p(\px) \zx}[/imath] so that the right hand side can be defined?

I suppose you're right, but how would that work?
if [imath]B \simeq C[/imath], [imath]B, C \subseteq A[/imath] are ideals, is this true -> [imath]\frac{A}{B} \simeq \frac{A}{C}[/imath] ?

I was trying to ask this some time ago, with my first question on isomorphisms (but I guess I didn't express it that well). Could I replace
[imath]\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{(X^2+tX+q)\mathbb{Z}[X]}[/imath] with a single line ideal after doing the factorisation? I'd say that is isomorphic to [imath]p\mathbb{Z}[X][/imath], but I dont think I can just switch them around if they're isomorphic.

 

[blamocur]

I don't know how this would work. Nor do I understand how it is related to your "A,B,C" hypothesis.
I remember next to nothing about ring theory (half-century is a long time ), so I can only work from basic definitions. I'll let you know if I manage to come up with any useful ideas later, but I wouldn't hold my breath if I were you -- sorry

 

[MathNugget]

I don't know how this would work. Nor do I understand how it is related to your "A,B,C" hypothesis.
I remember next to nothing about ring theory (half-century is a long time ), so I can only work from basic definitions. I'll let you know if I manage to come up with any useful ideas later, but I wouldn't hold my breath if I were you -- sorry

At worst, it's back to the first ring isomorphism theorem: finding a surjective morphism [imath]\phi: \frac{\mathbb{Z}[X]}{p(X^2+tX+q)\mathbb{Z}[X]} \rightarrow \frac{\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]}[/imath]

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

I guess we're better off trying to solve the exercise on my other post (there I have some ideas). Here it looks so weird (almost mistaken). Wouldn't be surprised if the isomorphism is even true (there's a chain of isomorphisms on the paper I am trying to understand, and I suppose as long as the elements of interest are correct, no one minds a little "x=2=3=2=y" as long as all that is needed is x=y...

Also no need to apologize, you helped me a lot . I don't have much (or any) experience with isomorphisms of rings looking this bad.