I am studying for an algebra qualifying exam and came across the following problem.

Let $R$ be the ring of Gaussian Integers. Of the three quotient rings $$R/(2),\quad R/(3),\quad R/(5),$$ one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?

I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.

I hope that this can be answered easily and quickly. Thanks.

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    $\begingroup$ A product of fields is not an integral domain. Consider the zero divisors $(1,0)$ and $(0,1)$. $\endgroup$ – Jared May 17 '13 at 1:07
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    $\begingroup$ That's helpful, thanks. Since R/(2) has just 4 elements, if it were isomorphic to a product of fields, then it would necessarily be isomorphic to Z/(2)xZ/(2). But $i^2=1$ in R/(2), and every element of Z/(2)xZ/(2) when squared equals itself. So... From the unusual phrasing of the question, I suppose R/(5) is isomorphic to a product of fields. Is this right? Would that product be Z/(5)/Z/(5)? $\endgroup$ – John Adamski May 17 '13 at 1:34
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    $\begingroup$ You might find this helpful: Splitting of prime ideals in Galois Extensions (Wikipedia). $\endgroup$ – kahen May 17 '13 at 1:51
  • $\begingroup$ Related: math.stackexchange.com/questions/23358 $\endgroup$ – Watson Nov 28 '18 at 9:56

Now is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then

$$\frac{R}{I\cap J}\cong\frac{R}{I}\times\frac{R}{J}.$$

First let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\bf Z}$ iff $\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\cap(m)=({\rm lcm}(m,n))$. As $nm=\gcd(n,m){\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\cap(m)=(nm)$. Thus we have ${\bf Z}/(nm)\cong{\bf Z}/(n)\times{\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\bf Z}/\prod p_i^{e_i}{\bf Z}\cong\prod{\bf Z}/p_i^{e_i}{\bf Z}$.

(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)

Without coprimality, there are counterexamples though. For instance, if $p\in\bf Z$ is prime, then the finite rings ${\bf Z}/p^2{\bf Z}$ and ${\bf F}_p\times{\bf F}_p$ (where ${\bf F}_p:={\bf Z}/p{\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).

Now here's the proof. Define the map $R\to R/I\times R/J$ by $r\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\in I$, $j\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\mapsto(I,1+J)$ and $j\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\times R/J$ as an $R$-module so the image must be the whole codomain.

Now let's work with ${\cal O}={\bf Z}[i]$, the ring of integers of ${\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have

  • ${\bf Z}[i]/(3)$ is a field and
  • ${\bf Z}[i]/(5)\cong{\bf Z}[i]/(1+2i)\times{\bf Z}[i]/(1-2i)$ is a product of fields.

Go ahead and count the number of elements to see which fields they are. However, ${\bf Z}[i]/(2)={\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\bf F}_2[\varepsilon]/(\varepsilon^2)$, which is not a product of fields since $\epsilon\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.


Why don't you first think through the case of $\mathbb Z/n\mathbb Z$. When is this a field? When is it a product of fields? When is it neither? Try to relate your answer to the factorization of $n$ and the Chinese Remainder Theorem. Now see if you can generalize it to $\mathbb Z[i]$ (or to any PID).


I find quotients of polynomial rings easier to reason with algebraically than subfields of $\mathbb{C}$.

Because $\mathbb{Z}[i] \cong \mathbb{Z}[x] / (x^2 + 1)$, we can do arithmetic with rings and ideals. Calculating in more detail than is usually necessary:

$$\begin{align} \mathbb{Z}[i] / (3) &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / (3) \\&\cong \mathbb{Z}[x] / (3, x^2 + 1) \\&\cong \left( \mathbb{Z}[x] / (3) \right) / (x^2 + 1) \\&\cong \left(\mathbb{Z} / (3)\right)[x] / (x^2 + 1) \\&\cong \mathbb{F}_3[x] / (x^2 + 1) \end{align}$$

and so we've reduced the problem to one of quotients of polynomial rings over the finite field $\mathbb{F}_3$.


$\Bbb Z[i]/2\,$ isn't a field or $\color{#c00}{\Pi F}$ (product of fields), since it has a nonzero nilpotent: $\,(1\!+\!i)^2 = 2i = 0$.
$\Bbb Z[i]/5\,$ isn't a field by $\,(2\!-\!i)(2\!+\!i)= 5 = 0,\,$ so it is the $\color{#c00}{\Pi F},\,$ leaving $\,\Bbb Z[i]/3\,$ as the field. $\ $ QED


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