# Find all prime ideals in $\mathbb{Z}/n\mathbb{Z}$ where $n>1$

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.

And

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 $$J\cdot \mathbf Z/n\mathbf Z=(m)\cdot \mathbf Z/n\mathbf Z=(m,n)/(n)=(\gcd(m,n))/(n).$$

• 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 $(5)$? – The Coding Wombat Mar 14 at 23:26
• $b$ 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). – Bernard Mar 14 at 23:41
• 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? – The Coding Wombat Mar 15 at 9:49
• 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. – 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.