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Let $ \mathbb{Z}[i]$ denote the ring of the Gaussian intergers. For which of the following value of n is the quotient ring $ \mathbb{Z}[i]/n\mathbb{Z}[i]$ an integral domain?

$ a. 2$

$ b. 13$

$ c. 19$

$ d. 7$

I'm doubtful with the following attempt I made.

  • I think all 4 options are correct: It suffices to show $n\mathbb Z[i]$ is a prime ideal of $\mathbb Z[i]$ if $n$ is prime. Now $(n)=n\mathbb Z[i].$ So $n$ is prime element of $n\mathbb Z[i]\implies(n)$ is a prime ideal of $\mathbb Z[i].$

    Let $n$ be a prime integer. Of course then $n$ is non zero and non unit. Let $n|(a+ib)(c+id).$ That's $n|(ac-bd)+i(ad+bc)\\\implies\dfrac{ac-bd}{n},\dfrac{ad+bc}{n}\in\mathbb Z\\\implies n|ac,bd,ad,bc\\\implies n|\{a~or~c\}~and~\{b~or~d\}~and~\{a~or~d\}~and~\{b~or~c\}\\\implies n\text{ divides at least $3$ of }a,b,c,d.$

    WLG let $n|a,b\implies n|a+ib.$

Is my attempt correct?

Hirakjyoti Das
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Sriti Mallick
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2 Answers2

4

The following theorem is well known (see a book on algebraic number theory)

Theorem: Let $p$ be a rational prime (that is, a prime in $\mathbb Z$). Then, $p$ is a prime in the Gaussian integers $\mathbb Z[i]$ if and only if $p\equiv 3 \pmod 4$.

Since $\mathbb Z[i]/n \mathbb Z[i]$ is an integral domain if and only if $n$ is prime in $\mathbb Z[i]$, it follows that $(3)$ and $(4)$ are correct.

zacarias
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3

You claim that $n$ prime $\implies (n)$ is a prime ideal in $\mathbb{Z}[i]$. Let's consider $(5)$.

Clearly, $5$ is prime. But $(5) = (2+i)(2-i)$, and thus $(5)$ is not a Gaussian prime.

I encourage you to trace your proof over on this example and see where it goes wrong.

davidlowryduda
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