$\mathbb{Z}_n$ is not a subring of $\mathbb{Z}$

Question

I come across an example stating that ' $\mathbb{Z}_n$ is not a subring of $\mathbb{Z},n \geq2, n \in \mathbb{Z}$' in the book ' Dummit and Foote , abstract algebra'. Can anyone explain to me why the statement is true?

There are, in fact, no proper sub-rings of $\mathbb Z$. Any sub-ring has to contain $0$ and $1$, so it has to contain all the integers. — Thomas Andrews, Mar 18 '13 at 04:20

score 12 · Answer 1 · answered Mar 18 '13 at 05:08

I'd say there are a few (related) reasons that $\Bbb{Z}/n\Bbb{Z}$ isn't a subring of $\Bbb{Z}$. I feel that while the answers above do give ways of seeing that $\Bbb{Z}/n\Bbb{Z}$ is not a subring of $\Bbb{Z}$, they don't get to the why (the comments in Mariano Suárez-Alvarez's answer do address the issue, though).

First of all, $$ \Bbb{Z}/n\Bbb{Z}\not\subseteq\Bbb{Z}. $$ The former is a set of equivalence classes of integers, not integers themselves. There are alternate definitions of $\Bbb{Z}/n\Bbb{Z} := \Bbb{Z}_n$, but in each, we do not actually deal with integers. It's always classes of integers under an equivalence relation (specifically, equivalence modulo $n$).

Secondly (and this is related to the first reason as well), the operations on $\Bbb{Z}/n\Bbb{Z}$ and $\Bbb{Z}$ are not the same. In $\Bbb{Z}/n\Bbb{Z}$, we're performing addition and multiplication modulo $n$, and in $\Bbb{Z}$, the operations are normal addition and multiplication. The fact that $\underbrace{1 + 1 + \ldots + 1}_{n\,\mathrm{ times}} = 0$ in $\Bbb{Z}/n\Bbb{Z}$ comes from the fact that $n\equiv 0\mod n$. In $\Bbb{Z}$, we have no such equivalence relation restricting our operations: $\underbrace{1 + 1 + \ldots + 1}_{n\,\mathrm{ times}} = n\neq 0$. Hence, $\Bbb{Z}/n\Bbb{Z}$ cannot be a subring of $\Bbb{Z}$, as it is not a subset and the operations on the two rings are not the same.

score 11 · Accepted Answer · answered Mar 18 '13 at 04:14

11

In $\mathbb Z_n$ there is an element such that if you add it to itself $n$ times you get zero. In $\mathbb Z$ there is no such element.

answered Mar 18 '13 at 04:14

Mariano Suárez-Álvarez

135,076

2

I thought when we want to show a subset is a subring, we need to show that it is nonempty ,closed under subtraction and multiplication ? – Idonknow Mar 18 '13 at 04:16
Well, isn't that @Idonknow what Mariano just hinted to you? – DonAntonio Mar 18 '13 at 04:18
@Idonknow Well, $\mathbb Z_n$ is not a subset of $\mathbb Z$, so it can't technically be a subring. Mariano is showing also that it can't be isomorphic to a subring – Thomas Andrews Mar 18 '13 at 04:19
If we manage to show that two sets are not isomorphic, can we conclude that the one is not a subring of another? – Idonknow Mar 18 '13 at 04:22
The question if $\mathbb Z_n$ is a subring of $\mathbb Z$ is not answerable unless you tell us what the elements of the two sets are, and so on. It is also a rather uninteresting question! – Mariano Suárez-Álvarez Mar 18 '13 at 04:23
2

@Idonknow a subring must inherit the operations. However, $(n-1)+1 = 0$ in $Z_n$ and $(n-1)+1=n$ in $Z$. – Mar 18 '13 at 04:27
You should not confuse $ \mathbb{Z}{n} $ with $ n \mathbb{Z} $. In $ \mathbb{Z}{n} $, we are performing arithmetic modulo $ n $, in which adding $ 1_{\mathbb{Z}{n}} $ to itself $ n $ times gives you $ 0{\mathbb{Z}_{n}} $. You cannot have this happening for a subring of $ \mathbb{Z} $. In $ n \mathbb{Z} $, we are adding and multiplying integral multiples of $ n $ to obtain integral multiples of $ n $, so $ n \mathbb{Z} $ is indeed a subring of $ \mathbb{Z} $. – Haskell Curry Mar 18 '13 at 04:32

score 3 · Answer 3 · edited Nov 26 '19 at 22:09

The ring $\mathbb Z$ is torsion-free, meaning that for all $m\in\mathbb Z$ and $x\in\mathbb Z$, if $m\neq 0$ and $x\neq 0$, then $m\cdot x\neq 0$. Any subring of a torsion-free ring is torsion-free (simply because if you have a torsion relation in the subring, then the same torsion relation holds in the original ring, contradicting the torsion-free assumption). Now, for every $n$, the ring $\mathbb Z_n$ is not torsion-free, since $n\cdot x=0$ for all $x\in\mathbb Z_n$. Therefore, $\mathbb Z_n$ is not a subring of $\mathbb Z$.

score 1 · Answer 4 · answered Mar 18 '13 at 04:17

1

To add to Mariano’s answer, the only subrings of $ \mathbb{Z} $ are of the form $ n \mathbb{Z} $, where $ n \in \mathbb{N}_{0} $.

answered Mar 18 '13 at 04:17

Haskell Curry

19,524

score 1 · Answer 5 · answered Jan 21 '20 at 19:54

Let $G = (S,\circ)$ be a group. If $T \subseteq S$ and $H = (T,\circ)$ is a group, then $H$ is called a subgroup of $G$.

Although the underlying set $\mathbb{Z}_n := \{0,1,\dots, n-1\}$ is a subset of $\mathbb{Z}$, the binary operation of $\mathbb{Z}_n$ is addition modulo $n$. Thus, $\mathbb{Z}_n$ can not be a subgroup of $\mathbb{Z}$ because they do not share the same binary operation. Therefore, a fortiori, $\mathbb{Z}_n$ can not be a subring of $\mathbb{Z}$.

$\mathbb{Z}_n$ is not a subring of $\mathbb{Z}$

5 Answers5