Let $n \in ℕ$. Show that the ring $ℤ/nℤ$ is a field if and only if $n$ is prime.

Let $n$ prime. I need to show that if $\bar{a} \neq 0$ then $∃\bar b: \bar{a} \cdot \bar{b} = \bar{1}$. Any hints for this ?

Suppose $ℤ/nℤ$ is a field. Therefore: for every $\bar{a} \neq 0$ $∃\bar b: \bar{a}\cdot \bar{b}=1$. How can I show that $n$ must be prime ?


If $n$ is prime and $\overline a\ne0$ so $a$ isn't a multiple of $n$ and then $a$ and $n$ are coprime so by the Bezout theorem there's $b,c\in\mathbb Z$ such that $$ba+cn=1$$ hence by passing to the class we find $\overline a\overline b=\overline1$.

Conversely if $n$ isn't prime then we write $n=ab$ so $\overline0=\overline a\overline b$ where $\overline a\neq \overline0 $ and $\overline b\neq 0$ hence $\mathbb Z/n\mathbb Z$ isn't an integral domain and then it isn't a field.

  • $\begingroup$ Suppose $ℤ/nℤ$ is a field. For every $a : 0 < a < n$ there exists a $b$ with $\bar{a}\bar{b} = \bar{1}$. Therefore $ab+kn = 1$. Therefore $\gcd(a,n)=1$. Therefore $n$ must be prime. $\endgroup$ – 90intuition Oct 11 '13 at 20:37
  • $\begingroup$ @90intuition Yes this is correct. $\endgroup$ – user63181 Oct 11 '13 at 20:39
  • $\begingroup$ Nice work, Sami! (Be careful on speedy upvoting in rapid succession: I've been losing points for "serial upvotes." I usually try to pause just a tad in between my upvotes.) $\endgroup$ – amWhy Mar 3 '14 at 13:18

Hints: If $p \not\mid a$ then $ak \operatorname{mod} p$ are different for $k=0, \ldots, p-1$. If $n$ is not prime, then $n=mk$ with $2\le m,k \le n-1$.

  • $\begingroup$ If $n$ is not prime. Then $n$ could be $1$ and $m,k=1$. Right ? $\endgroup$ – 90intuition Oct 11 '13 at 20:23
  • $\begingroup$ The case $n=1$ is trivial, because then $\mathbb{Z}/n\mathbb{Z}$ contains just one element. $\endgroup$ – njguliyev Oct 11 '13 at 20:27

Hint: For prime implies field, use the fact that there are no zero divisors and the pigeonhole principle to argue there must be such a $\overline b$

  • $\begingroup$ I'm not introducced to both the term pigeon hole and the term zero divisor. $\endgroup$ – 90intuition Oct 11 '13 at 20:24
  • 2
    $\begingroup$ In Z_6, both "2" and "3" are said to be "zero divisors", as they are not zero yet their product $2 \cdot 3 = 6 = 0$ $\endgroup$ – The Chaz 2.0 Oct 11 '13 at 20:28
  • 1
    $\begingroup$ The [pigeonhole principle)[en.wikipedia.org/wiki/Pigeonhole_principle) says if you put $n+1$ pigeons into $n$ holes, one of the holes contains two pigeons. In this case, you want to claim that there are not two distinct elements $c$ and $d$, such that $ac=ad$ (because then $a(d-c)=0$), so all the products of $a$ with elements are distinct, so one of the products must be $1$. $\endgroup$ – Ross Millikan Oct 11 '13 at 20:37

Just for basic idea :

See that , In $\mathbb{Z}_4$, element $\bar{2}$ does not have inverse.

See that , In $\mathbb{Z}_6$ the element $\bar{2}$ and $\bar{3}$ does not have inverse.

See that , In $\mathbb{Z}_8$ the element $\bar{2}$ and $\bar{4}$ does not have inverse.

In general In $\mathbb{Z}_{pq}$ elements $\bar{p}$ and $\bar{q}$ does not have inverse.

Suppose $n$ is prime and let $\bar{a} \in \mathbb{Z}_n$

Consider $\{ \bar{a}.\bar{b} : \bar{b}\in \mathbb{Z}_n\}$.

please check that this can not be a proper subset of $\mathbb{Z}_n$

(You are supposed to use that $n$ is prime to prove above result).

As $\{ \bar{a}.\bar{b} : \bar{b}\in \mathbb{Z}_n\}= \mathbb{Z}_n$, we see that :

for some $\bar{b}\in \mathbb{Z}_n$ we have $\bar{a}. \bar{b}=\bar{1}$ and thus we are done.


I abbreviate your ring to "$R$".

Hint for 1)

Suppose $p$ is prime and first prove that $\bar{a},\bar{b}\neq0$ implies $\bar{ab}\neq \bar{0}$. If to the contrary $a,b$ were not multiples of $p$, but $ab$ is a multiple of $p$, can you see the contradiction? Use this to show that $\{\bar{a}r\mid r\in R\}=R$, proving that there is $b$ such that $ab=1\in R$.

Hint for 2) There exists $1 <j,k<n$ such that $jk=n$. What does that look like in the ring? (I mean $\bar{j}\bar{k}\in R$

  • $\begingroup$ I'm not introduced to ideals yet. $\endgroup$ – 90intuition Oct 11 '13 at 20:22
  • $\begingroup$ For $2)$. $\bar{a} \bar{b} = \overline{n+1}$ right ? $\endgroup$ – 90intuition Oct 11 '13 at 20:25
  • $\begingroup$ @90intuition OK, I worked around it. $\endgroup$ – rschwieb Oct 11 '13 at 20:26
  • $\begingroup$ @90intuition OK, I added bars to be more clear about which things are in the quotient and which aren't. $\endgroup$ – rschwieb Oct 11 '13 at 20:28

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