Ring isomorphism involving polynomials

[MathNugget]

I'm running out of creative names. I decided to skip some stuff from the paper I mentioned earlier .

I'd like to prove: [imath]\frac{\mathbb{Z}[X]/p\mathbb{Z}[X]}{(X^2+tX+q, p)\mathbb{Z}[X]/p\mathbb{Z}[X]}\simeq \frac{\mathbb{Z}[\alpha]}{p\mathbb{Z}[\alpha]}[/imath]

now, the exercise needs a bit more input: the polynomial ([imath]X^2+tX+q[/imath] ) is irreducible in [imath]\mathbb{Z}[X][/imath] (of course, it is in [imath]\mathbb{Q}[X][/imath] too). [imath]\alpha[/imath] is one of the roots (it is in [imath]\mathbb{C}\backslash \mathbb{Q}[/imath] ).

I am trying to find a surjective morphism [imath]\phi: \frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]} \rightarrow \frac{\mathbb{Z}[\alpha]}{p\mathbb{Z}[\alpha]}[/imath].

[imath]\phi(\widehat{f})=\widehat{f(\alpha)}[/imath]. I'll assume believed that [imath]\mathbb{Z}[\alpha][/imath] are [imath]a+b\alpha[/imath], with [imath]a, b \in \mathbb{Z}[/imath] . There is the surjectivity, and then there is the morphism: [imath]\phi(\widehat{f+g})=\widehat{(f+g)(\alpha)}=\widehat{f(\alpha)+g(\alpha)}=\widehat{\phi(f)+\phi(g)}[/imath]. Multiplication seems to go similarly. I realize I didn't really put the hats properly, maybe...


For correct definiteness...if [imath]p \mid (f-g)[/imath], then [imath]p \mid (f(\alpha)-g(\alpha))[/imath] I suppose.

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

If [imath]\phi(\widehat{f})=\widehat{0}[/imath], then [imath]p \mid f(\alpha)[/imath] in [imath]\mathbb{Z}[\alpha][/imath]. Notice: that means either [imath]f(\alpha)=0[/imath], or [imath]f(X)[/imath] has all coefficients divisible by p.

(Suppose [imath]g(\alpha)=kp[/imath]; then [imath]\alpha[/imath] would be root of g-kp would be a polynomial of same degree as f, with root [imath]\alpha[/imath], which contradicts the uniqueness of f, which I didn't prove here. I suppose we can find some integers c, d such that [imath]deg(cf-d(g-kp))\leq 1[/imath], has integer coefficients, and root [imath]\alpha[/imath], which contradicts it being irrational etc).

As such, [imath]f(X) \in (X^2+tX+q, p)\mathbb{Z}[X][/imath] (I guess this can be proven with double inclusion: on one hand, I proved that f satisfies at least on of these: [imath]f(\alpha)=0[/imath] or [imath]p \mid f(x) \: \forall x[/imath]; on other hand, an element of [imath](X^2+tX+q, p)\mathbb{Z}[X][/imath] is of the form [imath]g_1(X^2+tX+q)+g_2p[/imath], and obviously [imath](g_1(X^2+tX+q)+g_2p)(\alpha)=0+pg_2(\alpha)[/imath], which is divizible by p... [imath]g_1, g_2[/imath] are arbitrary elements of [imath]\mathbb{Z}[X][/imath], by the way...


Then I remember [imath]\phi[/imath] was defined on [imath]\frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]}[/imath], so I get that [imath]ker(\phi)=\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}[/imath], and it should be complete...

 

[blamocur]

I wish you success in your exercise, but if you need help from us you will have to formulate your questions in a more self-contained way which does not require reading an esoteric article. Particularly, I have no clue what you second line means

 

[MathNugget]

I wish you success in your exercise, but if you need help from us you will have to formulate your questions in a more self-contained way which does not require reading an esoteric article. Particularly, I have no clue what you second line means

Well...
It means
[imath]A=\frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]}[/imath], [imath]B=\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}[/imath]
And I am looking at
[imath]\frac{A}{B}\simeq \frac{\mathbb{Z}[\alpha]}{p\mathbb{Z}[\alpha]}[/imath] .

p is a prime number (in [imath]\mathbb{Z}[/imath] ), [imath]X^2+tX+q \in \mathbb{Z}[X][/imath], with roots [imath]\alpha, \overline{\alpha} \notin \mathbb{Z}[/imath].


And I decided to approach the problem by finding a morphism [imath]\phi: A \rightarrow \frac{\mathbb{Z}[\alpha]}{p\mathbb{Z}[\alpha]}[/imath] that is surjective and has [imath]ker(\phi)=B[/imath]