# Definition and ideals of $\mathbb{Z}/n\mathbb{Z}$

First, I want to ask what elements are in the rings $$\mathbb{Z}/n\mathbb{Z}$$, the book I have defines the rings $$R/I=\{a+I| a\in R\}$$ where $$a+I=\{x\in R| x-a \in I\}$$ then proceed to give an example I can't understand how it follows the definition above. It says $$\mathbb{Z}/2\mathbb{Z}=\{2\mathbb{Z},1+2\mathbb{Z}\}$$, from what I understand $$\mathbb{Z}/2\mathbb{Z}$$ should have all $$x$$ such that $$x\in \mathbb{Z}$$ and $$a\in \mathbb{Z}$$ and $$x-a \in \mathbb{2Z}$$, which doesn't follow the set. A second example is $$2\mathbb{Z}/4\mathbb{Z}=\{4\mathbb{Z},2+4\mathbb{Z}\}$$ which is also a mystery for me. The last part it confuses me is the set $$\{2\mathbb{Z},1+2\mathbb{Z}\}$$ this set contines all odds and all the even numbers, why is different from $$\mathbb{Z}$$? Can someone explain to me what it's happening above because I am confused.

my second question is about the ideals of $$\mathbb{Z}/n\mathbb{Z}$$,

can I have a simple explanation about why the ideals of $$\mathbb{Z}/n\mathbb{Z}$$ are the rings $$k\mathbb{Z}/n\mathbb{Z}$$ where $$k|n$$

for example, the ideals of $$\mathbb{Z}/6\mathbb{Z}$$ must be : $$\mathbb{Z}/6\mathbb{Z}$$, $$2\mathbb{Z}/6\mathbb{Z}$$ ,$$3\mathbb{Z}/6\mathbb{Z}$$, $$6\mathbb{Z}/6\mathbb{Z}=<0>(?)$$

• Do you agree that $\{1,2,3,4\}$ and $\{\{1,2\},\{3,4\}\}$ are different sets? (Note that one has 4 elements while the other has just two elements.) — The same happens for $\mathbb Z$ and $\{2\mathbb Z,1+2\mathbb Z\}$. (One has infinitely many elements, the other just two.) Jul 4 '21 at 11:16
• Ok yea, I understand what we mean now. Jul 4 '21 at 11:18
• $Z/2Z$ is just 0 and 1 (intuitively). All "modding out" by $2Z$ does is differentiate even/odd integers. $2Z$=evens, $1+2Z$=odds Jul 4 '21 at 11:20
• ok I am starting to get it now thanks a lot Jul 4 '21 at 11:44
• To be precise, $4\mathbb Z/4\mathbb Z = \{4\mathbb Z\}$ which is a ring with one element. It is of course isomorphic to the zero ring, since there is (up to unique isomorphism) only one ring with one element. Jul 4 '21 at 11:48

Here's a very simple answer for your second question, containing some essential rules of thumb to have in mind to understand basic ring theory, and in particular that of PIDs such as $$\mathbb{Z}$$.

In general, if $$A$$ is a commutative ring and $$I$$ an ideal of $$A$$, then the ideals of $$A/I$$ are in natural bijection with those $$A$$ containing $$I$$ (the correspondence is given by the canonical projection). In your case, $$A=\mathbb{Z}$$ and $$I=n\mathbb{Z}$$.

Now, it so happens that $$\mathbb{Z}$$ has the nice property of being a PID, which means in particular that every ideal is generated by a single element, e.g. $$n$$ which yields $$(n)=n\mathbb{Z}$$, and in PIDs, there is an order-reversing relation between inclusion and division : $$(a) \subset(b) \Leftrightarrow b|a$$, so in your case this yields $$(n) \subset (k) \Leftrightarrow k|n$$.

Combining these two facts, you get that the ideals of $$\mathbb{Z}/n\mathbb{Z}$$ are in natural bijection with the set of ideals $$k\mathbb{Z}$$ where $$k|n$$. The correspondence occurring naturally through the quotient map, one immediately concludes that the ideals of $$\mathbb{Z}/n\mathbb{Z}$$ are the $$k\mathbb{Z}/n\mathbb{Z}$$ for $$k|n$$.

Oh and by the way, it is usually customary to define rings such that they contain $$1$$, which is not the case for $$k\mathbb{Z}/n\mathbb{Z}$$ when $$k \neq 1$$, so they are not really rings, they are ideals. Some people call those rng (ring without i, i.e., without identity).

Question: "my second question is about the ideals of Z/nZ, can I have a simple explanation about why the ideals of Z/nZ are the rings kZ/nZ where k|n"

Answer: If $$p,q \neq0$$ are distinct prime numbers, it follows the ideals $$I:=(p),J:=(q)$$ are maximal ideals and their powers $$I^n=(p^n),J^m=(q^m)$$ are coprime for all $$m,n \geq 1$$. If $$n=p_1^{l_1}\cdots l_d^{l_d}$$ is a factorication into a product of powers of distinct prime numbers $$p_i$$, there is by the CRT an isomorphism of rings

$$A:=\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/p_1^{l_1}\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}/p_d^{l_d}\mathbb{Z}:=A_1 \oplus \cdots \oplus A_d.$$

The ring $$A_i$$ is an Artinian ring with maximal ideal $$I_i:=(p_i)$$. Hence the prime ideals in $$A$$ are the ideals on the form

$$J_i:=A_1\oplus \cdots \oplus I_i \oplus \cdots \oplus A_d$$

for $$i=1,..,d$$. If $$m\in \mathbb{Z}$$ is an integer and if $$J \subseteq \mathbb{Z}$$ is any ideal it follows $$J=(m)$$ for some $$m$$ - this is because $$\mathbb{Z}$$ is a "principal ideal domain": Any ideal $$J$$ is the ideal generated by an integer $$m$$. If $$J=(0)$$ in $$A$$ it follows $$m=nd$$. If you write

$$m=p_1^{a_1}\cdots p_d^{a_d} d$$

for some integer $$d$$ that does not contain $$p_i$$ as a factor, it follows $$J$$ gives rise to a non-zero ideal in $$A$$ iff $$a_i for some $$i$$.

Example: Let $$A:=\mathbb{Z}/p^mq^n\mathbb{Z}$$. It follows

$$A\cong \mathbb{Z}/p^m\mathbb{Z} \oplus \mathbb{Z}/q^n\mathbb{Z}\cong A_1 \oplus A_2.$$

There are two prime ideals $$(p)\oplus A_1, A_1\oplus (q)$$ in $$A$$. The ring-structure in $$A_1\oplus A_2$$ is componentwise and the projection map

$$\pi: A \rightarrow A_1\oplus A_2$$

is the map $$\pi(\overline{a}):=(\overline{a},\overline{a})$$.

You should check that the map $$\pi$$ is "well defined". You find a proof of the CRT in Atiyah-Macdonald's book on commutative algebra, Prop.1.10.

• This answer certainly isn't suited for somebody who is still trying to understand the definition of a quotient ring. Jul 4 '21 at 12:00
• I don't dislike your answer at all. Just that, and I could be wrong, OP doesn't seem to be "here". Jul 4 '21 at 12:01
• @Christoph - the question asks for the ideals in the quotient ring and I'm giving a construction of these ideals using the CRT. If the asker is not familiar with this result, this explanation could serve as a motivation for understanding this result. Jul 4 '21 at 12:08
• For the moment It's not very helpful but indeed serves as motivation Jul 4 '21 at 12:10