I would like, given a prime number $p$ in $\Bbb{N}$, to prove that $$\Bbb{Z}[i]/\langle p\rangle\simeq \Bbb{Z}[x]/\langle X^2+1,p\rangle $$

Given this, I can conclud with the third isomorphism theorem because $\Bbb{Z}[i]\simeq \Bbb{Z}[x]/\langle x^2+1\rangle$ which is a consequence of the euclidean algorithm.

But I would like to use the first one (just for "training"), Here we go :

Let $\phi:\Bbb{Z}[x]\to \Bbb{Z}[i]/\langle p\rangle$ such that $\phi(f(x)=f(i)+(p).$

  1. $\phi$ is surjective

Let $\overline{x}\in\Bbb{Z}[i]/\langle p\rangle$, I have to find an element of $g\in\Bbb{Z}[x]$ such that $\phi{(p(x))}=\overline{x}.$ Such a polynom is $(a+bx)+p(c+dx)$ because $\phi\bigl((a+bx)+p(c+dx)\bigr)=(a+bi)+p(c+di).$

  1. $\ker\phi=\{f(x)\in\Bbb{Z}[x]: \phi(f(x))=\overline{0} \}.$

Now I am not sure how can continue, I now that if $f(i)=0$ then $f(-i)=0$; Plus, $x^2+1$ can be view as polynomial in $\Bbb{Z}[x]$ and monic, I can perform the euclidean algorithm. But I am stuck proving that ther kernel is include $\langle x^2+1\rangle+\langle p\rangle:=\langle x^2+1,p\rangle.$

  • $\begingroup$ Your analysis of the situation in #2 is not correct. After all, $X^2+1$ is certainly in the kernel, isn’t it? So the kernel is definitely larger than $\Bbb Z[X]\cap (p)$. $\endgroup$ – Lubin Nov 25 '15 at 15:26
  • $\begingroup$ @Lubin clearly Yep, I don't know why I wrote that. $\endgroup$ – JeSuis Nov 25 '15 at 15:29

For a general polynomial f with integer coefficients, f(i) = a + bi where a is determined by the coefficients of even degree, and b is determined by the coefficients of odd degree. Write f(x) = g(x^2) + x h(x^2).

By definition, f is in the kernel of phi if and only if f(i) is an integer multiple of p. This is equivalent to the simultaneous conditions that g(-1) is a multiple of p and h(-1)=0. The latter condition implies that h(x^2), as a polynomial, is a multiple of x^2 + 1. The former implies that there is an integer z such that g(x^2)-zp, as a polynomial, is a multiple of x^2 + 1. Both conditions together imply that f(x), as a polynomial, is a linear combination of the constant p and the binomial x^2+1.

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  • $\begingroup$ Actually my definition of the kernel of phi is incorrect as well. f is in the kernel if and only f(i) is a multiple of p in the ring Z[i] which is slightly broader than saying f(i) is an integer multiple of p. To repair the argument, the "latter" condition implies NOT that h(x^2) is itself a multiple of x^2+1, but that there exists another integer w such that h(x^2)-w.p is a multiple of x^2+1. The rest of the argument stands. $\endgroup$ – Justpassingby Nov 25 '15 at 15:45
  • $\begingroup$ thansk, I like the idea, but I cannot justify that the conditions implies that $h(x^2)-wp$ is a multiple of $x^2+1$. We have $h(-1)=0$ ok but next ? Can you give more details? Please. $\endgroup$ – JeSuis Nov 25 '15 at 16:26
  • $\begingroup$ If h(-1)=0 then h(s) is divisible, as a polynomial, by (s+1). Actually in my revised argument we do not have h(-1)=0 but h(-1)=w.p for some integer w. But that still means that the polynomial h(s)-w.p has a zero at s=-1, so h(s)-w.p is divisible by (s+1), so h(x^2)-w.p is divisible by (x^2+1). $\endgroup$ – Justpassingby Nov 26 '15 at 8:10

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