Non Zero ideals of $\mathbf{Z}_{12}$ How many proper non-zero ideals does the ring $\mathbf{Z}_{12}$ have? How many ideals does the ring $\mathbf{Z}_{12}\bigoplus\mathbf{Z}_{12}$ have?
I think that $\mathbf{Z}_{12}$ should have $4$ proper ideals given by following $$\langle 2 \rangle=\langle10\rangle,\langle3\rangle=\langle9\rangle,\langle4\rangle=\langle8\rangle,\langle6\rangle$$
But I am not sure how to do second part as I cant assume that every ideal will be of the form $\langle(a,b)\rangle$ for some  $a,b\in\mathbf{Z}_{12}$. Because it would mean that the direct product ring is a principal ideal ring.
Any help is appreciated. Thanks
 A: The ideals of $\mathbf{Z}_{12}$ are in bijection with the ideals of $\mathbf{Z}$ containing $12\mathbf{Z}$. Since the divisors of $12$ are $1$, $2$, $3$, $4$, $6$ and $12$, there are six of them. Now the answer depends on the meaning you assign to “proper”.
The ideals in $R\oplus S$ are of the form $I\oplus J$, where $I$ and $J$ are ideals of $R$ and $S$ (proof?).
A: Let $I$ be an ideal of $Z_{12}$, whose elements are denoted $\{0,...,11\}$. Let $x$ be the smallest element in $I$. Of course, $x \neq 1$. The claim is that $I$ is generated by $x$ : For any $y \in I$, $y = qx + r$ for $0 < r < x$ then $r  = y - qx \in I$, a contradiction.
Consequently, $I = \langle x\rangle$. For $I \neq Z_{12}$ we require $nx = 0$ for some $n$. Consequently, $I$ is generated by a factor of $12$, so $I = \langle d\rangle$ for some $d | 12 , d \neq 1,12$. How many such $d$ are there?
Now, if we are looking at $Z_{12} \oplus Z_{12}$, then let $I$  be a non-zero ideal of $Z_{12} \oplus Z_{12} = X$.
We claim that the set $S_1 = \{x \in Z_{12} : \exists y = (x,0) \in I\}$ forms an ideal of $Z_{12}$. Similarly for the set of second components $S_2$, and their direct sum is the ideal itself!
To see this, note that $S_1$ is closed under addition : if $a,b \in S_1$ then there are elements $y = (a,0)$ and $z = (b,0)$ in $I$. Since $I$ is closed under addition, $y + z = (a+b , 0)$ is in $I$, so it follows that $a+b \in S_1$. Similarly you can show that $0 \in S_1$ and $S_1$ is closed under additive inverse. Consequently, $S_1$ is a subgroup under addition.
Next, if $a \in S_1$ and $b \in Z_{12}$, then $(a,0)(b,0) = (ab,0) \in I$, so $ab \in S_1$. Finally, $S_1$ is an ideal. Analogously for $S_2$.
It is clear that $I = S_1 \oplus S_2$ : write $(a,b) = (a,0) + (0,b)$, and it is clear that this is unique.
Therefore, every ideal of $X$ is of the form $S_1 \oplus S_2$ for some ideals $S_1,S_2$ of $Z_{12}$.
Now, $S_1 \oplus S_2$  is non-trivial if either $S_1$ or $S_2$ is non-trivial, and it is not the whole of $X$ if and only if either $S_1$ or $S_2$ is proper. Now, you can figure out how many ideals of $X$ there are.
