A very basic ring theory question, which I am not able to solve. How does one show that

  • $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.

  • Extending the result: $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$, if $a,b$ are relatively prime.

My attempt was to define a map, $\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}$ and show that the kernel is the ideal generated by $\langle{3-i\rangle}$. But I couldn't think of such a map. Anyhow, any ideas would be helpful.

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    The document http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf might be useful – Watson Jun 19 '18 at 10:06

7 Answers7


Gaussian integers modulo $3-i$

This diagram shows the Gaussian integers modulo $3-i$.

The red points shown are all considered to be $0$ but their locations in $\mathbb Z[i]$ are $0$, $3-i$, $i(3-i)$ and $3-i + i(3-i)$. Every congruence class must be inside that box once and you can see there are $10$ of them.

The arrows show adding by $1$ each time. Doing that takes you through every equivalence class and then back to the start.

So $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.

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Define $$\phi: \mathbb{Z} \rightarrow \mathbb{Z}[i]/(3-i) \text{ where } \phi(z) = z + (3-i)\mathbb{Z}[i].$$ It follows simply that $\ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z}$. So for any such $z \in \ker \phi$, we have $z = (3-i)(a+bi)$ for some $a,b \in \mathbb{Z}$. But $(3-i)(a+bi) \in \mathbb{Z}$ happens if and only if $3b-a=0$. So $$\begin{align*} \ker \phi = (3-i)\mathbb{Z}[i] \cap \mathbb{Z} &= \{(3-i)(3b+bi)\mid b \in \mathbb{Z}\}\\ &= \{(9b + b) + i(3b-3b)\mid b \in \mathbb{Z}\}\\ &= \{10b\mid b \in \mathbb{Z}\}\\ &= 10\mathbb{Z}. \end{align*}$$

To see $\phi$ is surjective, let $(a+bi) + (3-i)\mathbb{Z}[i] \in \mathbb{Z}[i]/(3-i)$. Then $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, so $\phi(a+3b) = (a+bi) + (3-i)\mathbb{Z}[i]$.

Hence $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$.

Brandon Carter
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Firstly: it is not true in general that $\mathbb Z[i]/(a - ib) \cong \mathbb Z/(a^2 + b^2).$ (Consider the case of $3 - 0\cdot i$.)

The claimed isomorphism does hold if $a$ and $b$ are coprime.

Here is a sketch of how to see this:

To begin with, note that it is much easier to consider maps from $\mathbb Z$ to other rings, rather than maps in the opposite direction (as you suggested in your answer), because $\mathbb Z$ maps to any ring with unity in a canonical way, by sending $1$ to $1$.

So consider the canonical map $\mathbb Z \to \mathbb Z[i]/(a - i b).$

The target is finite of order $a^2 + b^2$, and so this map factors to give an injection $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - i b)$ for some $n$ dividing $a^2 + b^2$.

Now if $a$ and $b$ are coprime then $b$ is coprime to $a^2 + b^2$, hence coprime to $n$, and so $b$ is invertible in $\mathbb Z/(n)$. Combining this observation with the equation $a - i b = 0$ (which holds in $\mathbb Z[i]/(a - i b)$) one finds (and I leave this as an exercise!) that the map $\mathbb Z/(n) \hookrightarrow \mathbb Z[i]/(a - ib)$ contains $i$ in its image, and hence is surjective as well as injective, and so we are done.

Matt E
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    Thanks for this generalization. It is very helpful (to *this* third party, at least!) – The Chaz 2.0 Mar 21 '11 at 16:42
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    Hello Matt, something I've been trying to wrap my head around recently is why $\mathbb{Z}[i]/(a−ib)$ has order $a^2+b^2$. Is there a simple algebraic explanation or reference for this? Thank you. – yunone Oct 30 '11 at 01:27
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    @yunone: Dear yunone, It follows e.g. from the division algorithm in $\mathbb Z[i]$. Regards, – Matt E Oct 30 '11 at 02:12
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    @yunone: Think about the chain of ideals $(a^2+b^2) \subset (a+bi) \subset {\mathbf Z}[i]$. For a positive integer $n$, the index $[{\mathbf Z}[i]:(n)]$ is $n^2$. Thus $[{\mathbf Z}[i]:(a^2+b^2)] = N^2$, where $N = a^2+b^2$. From group theory, $[{\mathbf Z}[i]:(a^2+b^2)] = [{\mathbf Z}[i]:(a+bi)][(a+bi):(a^2+b^2)]$. The quotient group $(a+bi)/(a^2+b^2)$ is isom. to ${\mathbf Z}[i]/(a-bi)$. Using cpx. conjugation, ${\mathbf Z}[i]/(a+bi) \cong {\mathbf Z}[i]/(a-bi)$ as additive groups. Hence $N^2 = [{\mathbf Z}[i]:(a+bi)]^2$. Now take square roots. – KCd Apr 07 '12 at 21:25
  • Dear @KCd, thanks for the explanation. – yunone Apr 27 '12 at 21:20
  • @MattE I really liked your argument. +1 solution. How about the convers: If the two rings are isomorphic, does it mean that $a$ and $b$ are coprime ? – Sara.T Dec 04 '18 at 05:55
  • @Sara.T If $d=\gcd(a,b)>1$, then in the first ring $dx=0$ for all $x$. However this doesn't happen in the second ring. – user26857 Jun 20 '19 at 13:10

Go back one step and add the defining equation for $i$ to the ideal. In other words, consider your ring as a quotient of the ring of polynomials $\mathbb Z[x]$:

$$ \mathbb Z[i] / (3-i) = \mathbb Z [x] / (3-x,x^2+1) $$

Manipulating the ideal $(3-x,x^2+1)$ a bit, you will find that the quotient is indeed equal to $\mathbb Z/10\mathbb Z$.

This makes sense because setting $i^2=-1$ and $i=3$ implies that $9 = -1$, which is true in $\mathbb Z/10\mathbb Z$.

Greg Graviton
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In this post, I am quoting the solution to a related question from Artin, with the preceding explanation (Section 11.4, p. 337-338). This is really the same as Greg Graviton's answer, but I found the different point of view and the elaborate explanation very useful. (The impatient may skip to Example 11.4.5 directly.)

Adding Relations

We reinterpret the quotient ring construction when the ideal $I$ is principal, say $I = (a)$. In this situation, we think of $\overline R = R / I$ as the ring obtained by imposing the relation $a = 0$ on $R$, or by killing the element $a$. For instance, the field $\mathbb F_7$ will be thought of as the ring obtained by killing $7$ in the ring $\mathbb Z$ of integers.

Let's examine the collapsing that takes place in the map $\pi: R \to \overline R$. Its kernel is the ideal $I$, so $a$ is in the kernel: $\pi(a) = 0$. If $b$ is any element of $R$, the elements that have the same image in $\overline R$ as $b$ are those in the coset $b + I$ and since $I = (a)$ those elements have the form $b+ra$. We see that imposing the relation $a =0$ in the ring $R$ forces us to set $b = b + ra$ for all $b$ and all $r$ in $R$, and that these are the only consequences of killing $a$.

Any number of relations $a_1 = 0, \ldots, a_n = 0$ can be introduced, by working modulo the ideal $I$ generated by $a_1, \ldots, a_n$, the set of linear combinations $r_1 a_1 + \cdots + r_n a_n$, with coefficients $r_i$ in $R$. The quotient ring $\overline R = R/I$ is viewed as the ring obtained by killing the $n$ elements. Two elements $b$ and $b'$ of $R$ have the same image in $\overline R$ if and only if $b'$ has the form $b + r_1 a_1 + \cdots +r_n a_n$ with $r_i$ in $R$.

The more relations we add, the more collapsing takes place in the map $\pi$. If we add relations carelessly, the worst that can happen is that we may end up with $I = R$ and $\overline R = 0$. All relations $a = 0$ become true when we collapse $R$ to the zero ring.

Here the Correspondence Theorem asserts something that is intuitively clear: Introducing relations one at a time or all together leads to isomorphic results. To spell this out, let $a$ and $b$ be elements of a ring $R$, and let $\overline R = R / (a)$ be the result of killing $a$ in $R$. Let $\overline b$ be the residue of $b$ in $\overline R$. The Correspondence Theorem tells us that the principal ideal $(\overline b)$ of $\overline R$ corresponds to the ideal $(a,b)$ of $R$, and that $R/(a,b)$ is isomorphic to $\overline R / (\overline b)$. Killing $a$ and $b$ in $R$ at the same time gives the same result as killing $\overline b$ in the ring $\overline R$ that is obtained by killing $a$ first.

Example 11.4.5. We ask to identify the quotient ring $\overline R = \mathbb Z[i]/(i-2)$, the ring obtained from the Gauss integers by introducing the relation $i-2=0$. Instead of analyzing this directly, we note that the kernel of the map $\mathbb Z[x] \to \mathbb Z[i]$ sending $x \mapsto i$ is the principal ideal of $\mathbb Z[x]$ generated by $f = x^2 + 1$. The First Isomorphism Theorem tells us that $\mathbb Z[x]/(f) \approx \mathbb Z[i]$. The image of $g = x-2$ is $i-2$, so $\overline R$ can also be obtained by introducing the two relations $f = 0$ and $g = 0$ into the integer polynomial ring. Let $I = (f,g)$ be the ideal of $\mathbb Z[x]$ generated by the two polynomials $f$ and $g$. Then $\overline R =\mathbb Z[x]/I$.

To form $\overline R$, we may introduce the two relations in the opposite order, first killing $g$ and then $f$. The principal ideal $(g)$ of $\mathbb Z[x]$ is the kernel of the homomorphism $\mathbb Z[x] \to \mathbb Z$ that sends $x \mapsto 2$. So when we kill $x-2$ in $\mathbb Z[x]$, we obtain a ring isomorphic to $\mathbb Z$, in which the residue of $x$ is $2$. Then the residue of $f = x^2+1$ becomes $5$. So we can also obtain $\overline R$ by killing $5$ in $\mathbb Z$, and therefore $\overline R \approx \mathbb F_5$.

Alex M.
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In general, one knows that if $\alpha$ is an integer in the number field $K$, then $$ {\rm N}_{K/{\Bbb Q}}(\alpha)=\left|\frac{A}{A\alpha}\right| $$ Here $\rm N$ is the norm and $A$ denotes the ring of integers. In the question's situation, $A={\Bbb Z}[i]$ is the ring of Gaussian integer and since $K$ is quadratic imaginary, ${\rm N}_{K/{\Bbb Q}}(\alpha)=\alpha\bar\alpha$ where the bar denotes complex conjugation.

When $\alpha=3-i$, ${\rm N}_{K/{\Bbb Q}}(\alpha)=(3-i)(3+i)=10$, thus ${\Bbb Z}[i]/(3-i)$ is a ring with 10 elements, whose representatives are $$ \left\{0,1,2,i,i+1,i+2,2i,2i+1,2i+2,-1 \right\}. $$ A minute of thought and a brief inspection of these representatives should convince that the ring is indeed isomorphic to the ring of classes modulo 10.

Andrea Mori
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    Regarding your last sentence: this is true for instance because there is up to isomorphism exactly one (commutative, unital) ring with $10$ elements. The same holds with $10$ replaced by any squarefree positive integer $N$... – Pete L. Clark Feb 23 '11 at 14:33
  • Yes, Pete, I know. My idea was to suggest that the isomorphism could be checked simply by inspection of the representatives without invoking general results. Also, the first part of my answer isn't really necessary to get the representatives, but it's there to put things in a more general context. – Andrea Mori Feb 23 '11 at 19:25
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    my comment was also to put things in a more general context. I didn't mean to imply that you didn't know it -- only that you chose not to mention it, but others who read the question might not know it and be interested. (A lot of comments on this site are like that...) – Pete L. Clark Feb 23 '11 at 19:56

Question: "Extending the result: $Z[i]/(a−ib)≅Z/(a^2+b^2)Z$, if $a,b$ are relatively prime.

Answer: Let $a-ib:=1-(-3)i=1+3i$ and let $B:=\mathbb{Z}/(1+3^2)\cong\mathbb{Z}/(10)$ and let $A:=\mathbb{Z}[i]/(1+3i)$. This approach uses the chinese remainder lemma and it illustrates the "unique factorization of ideals" into products of powers of maximal ideals in Dedekind domains: It follows $-1 \cong 10-1 \cong 9$ hence you get a well defined map

$$\phi: \mathbb{Z}[i] \rightarrow B$$

by definining $\phi(a+bi):=a+3b$. It follows $$-1=\phi(-1)=\phi(i^2)=\phi(i)^2=3^2=9=10-1 =-1.$$

hence the map is well defined. By definition $\phi(1+3i)=1+3^2=10=0$, hence you get a well defined surjective map $\phi: A \rightarrow B$.

In $A$ there are two ideals $I:=(1+i),J:=(2+i)$ and


The ideals $I,J$ are coprime, hence

$$A \cong A/IJ \cong A/I \oplus A/J \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$


$$\mathbb{Z}/(10) \cong \mathbb{Z}/(2) \oplus \mathbb{Z}/(5)$$

by the chinese remainder lemma. There are explicit isomorphisms

$$f:A/I\cong \mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/(2)$$

defined by sending $i$ to $1$.

There is an isomorphism

$$g:A/J \cong \mathbb{Z}[i]/(2+i) \cong \mathbb{Z}/(5)$$

defined by sending $i$ to $3$. It follows




hence $g$ is well defined. You may check that $f,g$ are isomorphisms and you should do this as an exercise in "commutative ring theory/abstract algebra".

Note: This is a general fact: In the ring of integers $\mathcal{O}_K$ in a number field $K$, any ideal $\mathfrak{a}$ may be written (uniquely up to the order) as a product of powers of distinct maximal ideals

$$(*) \mathfrak{a}=\mathfrak{m}_1^{p_1} \cdots \mathfrak{m}_l^{p_l}.$$

The ideals above $I,J$ are maximal since the quotients $A/I,A/J$ are (finite) fields. The equality $(*)$ is proved in Theorem I.3.3 in Neukirch "Algebraic number theory" (this property was in fact one of the reason for the introduction of ideals in algebraic number theory and commutative algebra). In the above case the multiplicities $l_I,l_J=1$ and you get the decomposition


in $\mathcal{O}_K\cong \mathbb{Z}[i]$ where $K:=\mathbb{Q}(i)$.

Question: "A very basic ring theory question, which I am not able to solve. How does one show that $\mathbb{Z}[i]/(3−i)≅\mathbb{Z}/10\mathbb{Z}$?"

Example: We may use the "same method": There a factorization into products of ideals


and the ideals $\mathfrak{m},\mathfrak{n}$ are coprime maximal ideals with finite residue field. There is by the CRL isomorphisms

$$\mathbb{Z}[i]/(3-i)\cong \mathbb{Z}[i]/\mathfrak{m}\oplus \mathbb{Z}[i]/\mathfrak{n} \cong \mathbb{Z}/(2)\oplus \mathbb{Z}/(5) \cong \mathbb{Z}/(10).$$

As above you may write down explicit maps and verify they are isomorphisms.

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  • How did you get the idea of defining $φ(α+bi)=a+3b$? It seems to me like it comes out of the blue. @hm2020 – Νικολέτα Σεβαστού Oct 22 '21 at 13:11
  • @ΝικολέταΣεβαστού - In this case with "small numbers" you may find such a map using the "trial and error"-method. – hm2020 Oct 26 '21 at 12:05
  • My teacher, in a similar exercise, suggested this: $i^2=-1$ and $φ(i^2)=φ(i)^2=-1$ Suppose that $x=φ(i)$ then the equation $x^2=-1 \mod 10$ has solutions $x=[3]$ and $x=[-3]$. One of the solutions doesn't give the right result (the kernel we wanted determined which was the right choice). @hm2020 – Νικολέτα Σεβαστού Oct 27 '21 at 11:47