Proving the following question: Z[3i] isn't PID by use of the quotient map Z/(3i) isn't a field.

Question

I was wondering if this methodology is sound. I've checked that for any element in $\mathbb{Z}[3i]$ if we take the quotient map with ideal $(3i)$, its kernel consists of the elements $ \{ 9a+3bi \}$. Since this essentially sends all of $(3i)$ and $9\mathbb{Z}$ to the 0 coset. I was wondering if this was in any way similar to the quotient map $\mathbb{Z} \, / \, 9\mathbb{Z} $. I realise that the elements of the cosets of $$\mathbb{Z}/(3i)=[x]=\{x: x+ (9a+3bi)\mathbb{Z}\} $$ do not equal the cosets of $$\mathbb{Z} \, / \, 9\mathbb{Z}=[x]=\{x: x+ (9a+3bi)\mathbb{Z}\} $$ but is there any way to relate these two to state they are equivilant, and therefore claim that since 3i is irreducible and prime, if $\mathbb{Z}/(3i)$ maps to domain that is equivalent to a domain that isn't a field, $\mathbb{Z}[3i]$ isn't a PID.

Answer 1

I am assuming you want to show that $\mathbb{Z}[3i]$ is not a PID by showing that quotienting by a prime ideal does not give you field (since prime implies maximal in PID's). Please correct me if I am wrong in this assumption and edit your question to make it clearer what you are asking.

It is true that $\mathbb{Z}[3i]/(3i) \cong \mathbb{Z}/(9)$ but $(3i)$ is not prime in $\mathbb{Z}[3i]$. Let's start with the first claim. Note that we have the following isomorphism. $$\mathbb{Z}[3i] \cong \mathbb{Z}[x]/(x^2+9).$$ The ideal of $\mathbb{Z}[x]/(x^2+9)$ that corresponds to the ideal $(3i)$ is $(x)$ so, $$\mathbb{Z}[3i]/(3i) \cong \mathbb{Z}[x]/(x^2+9,x).$$ Quoienting by $(x)$ first, $$\mathbb{Z}[x]/(x^2+9,x) \cong \mathbb{Z}/(0^2+9) \cong \mathbb{Z}/(9).$$ So yes, $\mathbb{Z}[3i]/(3i) \cong \mathbb{Z}/(9)$. Now for the other claim. The ring $\mathbb{Z}/(9)$ is not an integral domain since $3 \cdot 3 = 0$, so $(3i)$ can not be prime.