I want to find all prime ideals in $\mathbb{Z}/n\mathbb{Z}$ where $n>1$.

I think I have to use the following theorem (because they asked me to prove it right before this exercise, which wasn't too complicated):

Let R be a ring and I ⊂ R an ideal. Let φ : R → R/I be the natural residue class homomorphism. Let I ⊂ J be another ideal. Then J is a prime ideal in R if and only if φ(J) is a prime ideal in R/I.


Let R be a principal ideal domain and $I\subset R, I\neq(0)$ an ideal. Then every ideal generated by a irreducible element is a prime ideal

An irreducible element in $\mathbb{Z}$ are exactly the prime numbers. I also know that $\mathbb{Z}$ is a principal ideal domain.

I tried supposing that $J=(p)$ is an ideal of $\mathbb{Z}$, and when $p$ prime, it is a prime ideal. Then if $J$ is a superset of $n\mathbb{Z}$, we have that $\phi(J)$ is a prime ideal of $\mathbb{Z}/n\mathbb{Z}$. Is this correct? For what $p$ values does this hold? What if $J$ is not a superset of $n\mathbb{Z}$ as the theorem requires?


Another theorem says the prime ideals of $R/I$ correspond bijectively to the prime ideals of $R$ containing $I$.

In the case of $\mathbf Z/n\mathbf Z$, this means its prime ideals are generated by the congruence classes of the prime divisors of $n$.

If $J=(m)$ does not contain $n$, i.e. if $m$ does not divide $m$, the image of $J$ in $\mathbf Z/n\mathbf Z$ is the ideal $$J\cdot \mathbf Z/n\mathbf Z=(m)\cdot \mathbf Z/n\mathbf Z=(m,n)/(n)=(\gcd(m,n))/(n).$$

  • $\begingroup$ What is $b$? And do the last line and equation go into how to find the prime ideals generated by the congruence classes of the prime divisers of n? So for $\mathbb{Z}/15\mathbb{Z}$ we have that its only prime ideals are $(3)$ and $(5)$? $\endgroup$ – The Coding Wombat Mar 14 at 23:26
  • $\begingroup$ $b$ was a typo, sorry. I meant$ n$. The last equation is an answer to your last paragraph ($J$ not necessary generated by a prime). For $\mathbf Z/15$, the only prime ideals are, more exactly, $(3)/(15)$ and $(5)/(15)$. Any quotient of $\mathbf Z$ has a finite number of prime ideals (they're semi-local rings). $\endgroup$ – Bernard Mar 14 at 23:41
  • $\begingroup$ So $m$ is not prime? Then $J=(m)$ cannot be a prime ideal in $\mathbb{Z}$ right? So is the image of $J$ in $\mathbb{Z}/n\mathbb{Z}$ you wrote out a prime ideal there? $\endgroup$ – The Coding Wombat Mar 15 at 9:49
  • $\begingroup$ Not exactly: $m$ is not necessarily prime, and the image of $I$ is not necessarily a prime ideal. This depends solely on the $\gcd$ being a prime or not. $\endgroup$ – Bernard Mar 15 at 10:23

An ideal $I$ in a commutative ring $A$ is prime if and only if $A/I$ is an integral domain; $\mathbb Z/n \mathbb Z$ is a finite ring, so a the quotient by a prime ideal has to be a finite integral domain, which is a field.

Now it is not hard to see that (since the subgroups of $\mathbb Z/n \mathbb Z$ are all isomorphic to $\mathbb Z/ d \mathbb Z$ with $d|n$, and these subgroups are also ideals), this is possible if and only if $(\mathbb Z/ n \mathbb Z)/ I \simeq \mathbb Z / p \mathbb Z $ with $p$ prime; you can easily deduce what are the ideals you are looking for.


Hint: to solve this exercise, it is enough to use the following facts:

  • an ideal $I\unlhd R$ is prime if and only if $R/I$ is a domain,
  • the characteristic of a domain is prime.

Using these, it is enough to find all quotients of ${\mathbf Z}/n{\mathbf Z}$ of prime characteristic. It should be straightforward to check that they are, in fact, domains.


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