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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

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  • $\begingroup$ Because it would mean that the direct product ring is a principal ideal ring. What's wrong with that? It is a principal ideal ring. Every finite products of principal ideal rings is principal. $\endgroup$
    – rschwieb
    Jun 29, 2018 at 16:09

2 Answers 2

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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?).

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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.

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  • $\begingroup$ Note that your definition of $S_1$ as those $x$ such that $(x,0)\in I$ is not the set of first components of elements of $I$ a priori. That would be the set of those $x$ such that there is some $z$ with $(x,z)\in I$. $\endgroup$
    – Christoph
    Jun 29, 2018 at 11:56
  • $\begingroup$ Also, for $I=(S_1\times 0)+(0\times S_2)$ you have to show that for $(a,b)\in I$ you have $(a,0),(0,b)\in I$. $\endgroup$
    – Christoph
    Jun 29, 2018 at 12:04
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    $\begingroup$ The first component part I will ignore. The second comment, note that $(a,0) = (a,b) (1,0)$, and similarly for the second component. I think this answers the first part as well, though, since $a \in S_1 \iff (a,0) \in I$ now. $\endgroup$ Jun 29, 2018 at 12:35

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