What are the ideals of $\mathbb{Z}/n\mathbb{Z}$? I am trying to find all the ideals of $\mathbb{Z}/2 \times \mathbb{Z}/4$. I've just proven that the ideals of $R$ × $S$ are precisely the sets of the form $\{(x,y):x \in I,y \in J\}$ for $I\subset R,J \subset S$ ideals.
I can see that I need to find the ideals of $\mathbb{Z}/2$ and $\mathbb{Z}/4$ but I'm struggling how to do this. I think that a proper ideal can't contain 1 otherwise the ideal is the whole ring. Does this mean that the ideal of $\mathbb{Z}/2$ is $\{0\}$ and the ideal of $\mathbb{Z}/4$ is $\{0,2\}$? 
I think this is the case since if 3 was in the ideal of $\mathbb{Z}/4$ we would have 3+2=1 meaning that 1 would need to be in the ideal.
 A: Let $\pi : \mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ be the canonical projection.
Then by the lattice-isomorphism theorem, all ideals of $\mathbb{Z}/n\mathbb{Z}$ are of the form $\pi(I)$ for $I$ ideal of $\mathbb{Z}$ containing $n\mathbb{Z}$. 
Those $I$ are precisely the $d\mathbb{Z}$ for $d\mid n$.
Therefore the ideals of $\mathbb{Z}/n\mathbb{Z}$ are the $d\mathbb{Z}/n\mathbb{Z}$ for $d\mid n$
A: You are right: there are two ideals in $\mathbb{Z}/2\mathbb{Z}$, the zero ideal $(0) = \{0\}$ and the unit ideal $(1) = \{0,1\} = \mathbb{Z}/2\mathbb{Z}$. There are three ideals in $\mathbb{Z}/4\mathbb{Z}$, the zero ideal $(0) = \{0\}$, the ideal $(2) = \{0,2\}$, and the unit ideal $(1) = \{0,1,2,3\} = \mathbb{Z}/4\mathbb{Z}$.
(This is just the same thing that Max said: there are two positive divisors of $2$, $d \mid 2$ for $d=1,2$; and there are three positive divisors of $4$, $d \mid 4$ for $d=1,2,4$.)
So there are $6$ ideals in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$, including the zero and unit ideals.
