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If $p$ is a prime integer, then $\mathbb{Z}[i]/(p)$ is isomorphic to $\mathbb{Z}_p/(x^2 + 1)$

I suspect that there is a typo, and that the $\mathbb{Z}$ on the r.h.s. should be $\mathbb{Z}[x]$ (polynomials).

Either way, I need help getting started and understanding the notation. $\mathbb{Z}[i]$ is the Gaussian integers, so $\mathbb{Z}[i]/p$ means taking modulus $p$... so any elements that differ by a multiple of $p$ are congruent?

What about the expression on the right?

Thanks.

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The expression on the right is likely meant to be $\mathbb{Z}_p[x]/(x^2+1)$, not $\mathbb{Z}[x]/(x^2+1)$ (the latter is isomorphic to $\mathbb{Z}[i]$, but not to $\mathbb{Z}[i]/(p)$). –  Arturo Magidin Mar 7 '11 at 1:03
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The notation $\mathbb{Z}[i]/(p)$ means $\mathbb{Z}[i]$ modulo the ideal generated by $p$, i.e., the multiples of $p$ (in the Gaussian integers).

The expression on the right is supposed to be $\mathbb{Z}_p[x]/(x^2+1)$. So first, $\mathbb{Z}_p$ is the integers modulo $p$, $\mathbb{Z}/p\mathbb{Z}$. Then you take $\mathbb{Z}_p[x]$, which are the polynomials with coefficients in $\mathbb{Z}_p$. Finally, you take the quotient modulo $x^2+1$, which means two polynomials with coefficients in $\mathbb{Z}_p$ are equivalent if and only if their difference is a multiple of $x^2+1$ (multiple in $\mathbb{Z}_p[x]$).

If you know the Isomorphism Theorem for rings, then the simplest way to do this is to construct a ring homomorphism either from $\mathbb{Z}[i]$ to $\mathbb{Z}_p[x]/(x^2+1)$ that is onto, and has kernel equal to the multiples of $p$; or a ring homomorphism from $\mathbb{Z}_p[x]/(x^2+1)$ to $\mathbb{Z}[i]/(p)$ that is onto and whose kernel are exactly the multiples of $x^2+1$. The latter is probably simpler.

If you don't know the Isomorphism Theorem for rings, you can try to define a homomorphism directly. You want to decide how to map an equivalence class of the form $a+bi + (p)$ ($a,b\in\mathbb{Z}$) to an equivalence class of the form $\overline{r}+\overline{s}x + (x^2+1)$, where $\overline{r}$ and $\overline{s}$ are integers modulo $p$ (you should first prove that every element in $\mathbb{Z}_p[x]$ is equivalent modulo $x^2+1$ to a polynomial of degree at most $1$; try using the division algorithm). There is an obvious choice: map $a+bi + (p)$ to $\overline{a}+\overline{b}x + (x^2+1)$ (the reason you would want to do this is that $x+(x^2+1)$ satisfies $t^2 + 1 = 0$ in $\mathbb{Z}_p[x]/(x^2+1)$; that is, it is a "square root of $-1$", just like $i$). You would need to show this is well-defined, one-to-one (on classes), onto, and a ring homomorphism.

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This is very helpful. I was struggling to even identify what elements of these rings would look like, so I'm glad you clarified that for me. –  The Chaz 2.0 Mar 7 '11 at 1:18
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The expression on the right should be $\mathbb{Z}_p[x]/(x^2+1)$.

As a number theory person, I prefer to reserve the notation $\mathbb{Z}_p$ for the $p$-adic integers, and stick to $\mathbb{Z}/p\mathbb{Z}$ (or $\mathbb{F}_p$, depending on the setting) for "the integers mod $p$", but as Arturo points out $\mathbb{Z}_p$ is quite common outside of algebra / number theory circles.

What's going on here is that both rings are (isomorphic to) the integer polynomials $\mathbb{Z}[x]$, mod the ideal $(p,x^2+1)$. This problem is asking you to show that it doesn't matter if you first

  • quotient $\mathbb{Z}[x]$ by the ideal $(x^2+1)$ (which produces $\mathbb{Z}[i]$), and then quotient the resulting ring $\mathbb{Z}[i]$ by the ideal generated by $p$, or
  • quotient $\mathbb{Z}[x]$ by the ideal $(p)$ (which produces $\mathbb{Z}_p[x]$), and then quotient the resulting ring $\mathbb{Z}_p[x]$ by the ideal generated by $x^2+1$

Arturo's advice about how to approach proving that these rings are isomorphic is spot on. The above is more of an explanation for why you might expect these two rings to be isomorphic. When we "mod out", we are making certain identifications of elements in our ring (specifically, when we mod out the ring $R$ by the ideal $I$, we identify $a\in R$ and $b\in R$ when $a-b\in I$), and this question is asking you to show that the order in which we identify elements of a ring doesn't matter.

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@Zev: It's often written $\mathbb{Z}_p$ (by non-number theorists). –  Arturo Magidin Mar 7 '11 at 1:02
    
Thank you both. Would it be appropriate for me to post my work (as a comment or an "answer") later? –  The Chaz 2.0 Mar 7 '11 at 1:09
    
@Zev: Yes, the notation is not compatible with standard notation in number theory. It is, nonetheless, widely used by non-number-theorists, so it is not more "not correct" than any other of the very numerous clashes of notation and nomenclature among disparate branches of mathematics. –  Arturo Magidin Mar 7 '11 at 1:10
    
@The Chaz: You can post it as an "answer". –  Arturo Magidin Mar 7 '11 at 1:10
    
@Arturo: Fair enough, I will adopt the OP's notation in my answer. –  Zev Chonoles Mar 7 '11 at 1:11
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