$\mathbb{Z}_n[i]$的精确定义问询：环扩张定义的适配疑问

Great question! This is a common confusion—since $\mathbb{Z}_n$ (the integers modulo $n$) can't be embedded into $\mathbb{C}$ as a subring, the original "subring plus evaluated polynomials" definition doesn't apply directly. Instead, we use a more general, ring-theoretic construction that works for any base ring, regardless of embeddability: quotients of polynomial rings.

Here are two equivalent, precise definitions that align with the practical operations you already know:

1. Quotient of Gaussian Integers by $n\mathbb{Z}[i]$ #

First, recall that the Gaussian integers $\mathbb{Z}[i]$ are defined as $\mathbb{Z}[x]/(x^2 + 1)$ (polynomials with integer coefficients, modulo the ideal generated by $x^2 + 1$). We can define $\mathbb{Z}_n[i]$ as:
$$
\mathbb{Z}_n[i] = \mathbb{Z}[i]/n\mathbb{Z}[i]
$$
In plain terms, this takes all Gaussian integers $a + bi$ (where $a,b \in \mathbb{Z}$) and identifies two elements if their difference is a multiple of $n$ (i.e., $a \equiv c \mod n$ and $b \equiv d \mod n$). The resulting elements are exactly pairs $(a,b) \in \mathbb{Z}_n \times \mathbb{Z}_n$, written as $a + bi$, with addition and multiplication working just like you expect:

• $(a+bi) + (c+di) = (a+c) + (b+d)i$ (coefficients added modulo $n$)
• $(a+bi)(c+di) = (ac - bd) + (ad + bc)i$ (all coefficient operations done modulo $n$)

2. Direct Quotient of $\mathbb{Z}_n[x]$ by $(x^2 + 1)$ #

We can skip the Gaussian integers entirely and build $\mathbb{Z}_n[i]$ directly from $\mathbb{Z}_n$:
$$
\mathbb{Z}_n[i] = \mathbb{Z}_n[x]/(x^2 + 1)
$$
Here, we take all polynomials with coefficients in $\mathbb{Z}_n$, then quotient out by the ideal generated by $x^2 + 1$. The "i" in this context is just the equivalence class of $x$ (written $x + (x^2 + 1)$), which satisfies $i^2 = -1$ because $x^2 \equiv -1 \mod (x^2 + 1)$.

This definition is equivalent to the first one—by the Chinese Remainder Theorem for rings, we have:
$$
\mathbb{Z}[x]/(n, x^2 + 1) \cong (\mathbb{Z}[x]/n\mathbb{Z}[x])/(x^2 + 1) \cong \mathbb{Z}_n[x]/(x^2 + 1)
$$
And this is also isomorphic to $\mathbb{Z}[i]/n\mathbb{Z}[i]$, so both definitions give the same ring.

Why This Works for Your Practical Use Case #

Both definitions produce a ring where elements are $a + bi$ with $a,b \in \mathbb{Z}_n$, and operations behave exactly like the "complex arithmetic modulo $n$" you're already familiar with. The key is that we don't need $\mathbb{Z}_n$ to sit inside $\mathbb{C}$—we're constructing a new ring from scratch that mimics the structure of complex numbers, but with coefficients in $\mathbb{Z}_n$.

This polynomial-quotient approach is the standard way to define "adjoining an element satisfying an equation" to any ring, not just subrings of $\mathbb{C}$. It generalizes the original $R[a]$ definition: when $R$ can be embedded into a ring $R'$ containing an element $a$ with $f(a) = 0$, then $R[x]/(f(x))$ is isomorphic to the original $R[a]$, so it's a natural extension.

内容的提问来源于stack exchange，提问作者John Doe