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

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

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

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


For correct definiteness...if $p \mid (f-g)$, then $p \mid (f(\alpha)-g(\alpha))$ I suppose.

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

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

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

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


Then I remember $\phi$ was defined on $\frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]}$, so I get that $ker(\phi)=\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$, 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
$A=\frac{\mathbb{Z}[X]}{p\mathbb{Z}[X]}$, $B=\frac{(X^2+tX+q, p)\mathbb{Z}[X]}{p\mathbb{Z}[X]}$
And I am looking at
$\frac{A}{B}\simeq \frac{\mathbb{Z}[\alpha]}{p\mathbb{Z}[\alpha]}$ .

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


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