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Give an example of an $R$-module $M$ such that $M=M_1 \oplus M_2$, $M=\operatorname{span}_R(S)$ and $S \cap M_1 = S \cap M_2 = \emptyset$.

I'm not sure if this is a very well known fact, but I can't see why $\mathbb{Z}_6$ works here. If you take $S=\{1\}$ then the intersection with $\mathbb{Z}_2$ and $\mathbb{Z}_3$ is not empty, is it? I'm very confused.

Michael Albanese
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Lotte
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2 Answers2

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Here $R = \mathbb{Z}$, that is, $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are $\mathbb{Z}$-modules, and $\mathbb{Z}_6 = \mathbb{Z}_2\oplus\mathbb{Z}_3$ as $\mathbb{Z}$-modules (or equivalently, abelian groups). You want $S$ to be a subset of $\mathbb{Z}_2\oplus\mathbb{Z}_3$ (otherwise $\operatorname{span}_{\mathbb{Z}}S$ is not a subset of $\mathbb{Z}_2\oplus\mathbb{Z}_3$), so in order to make sense of the intersections $S\cap \mathbb{Z}_2$ and $S\cap \mathbb{Z}_3$, we identify $\mathbb{Z}_2$ with $\mathbb{Z}_2\oplus\{0\} \subset \mathbb{Z}_2\oplus\mathbb{Z}_3$ and $\mathbb{Z}_3$ with $\{0\}\oplus\mathbb{Z}_3$.

If you take $1 \in S$ to mean $1 + 0 \in \mathbb{Z}_2\oplus\mathbb{Z}_3$, then you do have $S\cap\mathbb{Z}_2 \neq \emptyset$ as claimed, but $\operatorname{span}_{\mathbb{Z}}S = \mathbb{Z}_2\oplus \{0\} \neq \mathbb{Z}_2\oplus\mathbb{Z}_3$. Likewise, if you take $1$ to mean $0 + 1 \in \mathbb{Z}_2\oplus\mathbb{Z}_3$, you do have $S\cap\mathbb{Z}_3 \neq \emptyset$, but again $\operatorname{span}_{\mathbb{Z}}S = \{0\} \oplus\mathbb{Z}_3 \neq \mathbb{Z}_2\oplus\mathbb{Z}_3$.

To see that there is a set $S \subseteq \mathbb{Z}_2\oplus\mathbb{Z}_3$ satisfying $\operatorname{span}_{\mathbb{Z}}S = \mathbb{Z}_2\oplus\mathbb{Z}_3$, and $S\cap\mathbb{Z}_2 = S\cap\mathbb{Z}_3 = \emptyset$, consider the element $a + b \in S$ (clearly $S \neq \emptyset$). If $a = 0$, then the element belongs to the intersection $S\cap\mathbb{Z}_3$; likewise, if $b = 0$, the element belongs to the intersection $S\cap \mathbb{Z}_2$. Therefore all the elements of $S$ must be of the form $a + b$ with $a, b \neq 0$. This gives three possibilities for $S$, all of which have the desired properties.

Michael Albanese
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  • So, according to the last part, the set {1+1,1+2,1+3} spans $\mathbb{Z}_6$? I believe it is true, I'm just making sure... – Lotte Mar 15 '17 at 00:34
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    Note that $3 = 0$ in $\mathbb{Z}_3$, so you don't want $1+3 \in S$. If you are asking about ${1+1, 1+2}$, then yes, this is true. It's true even if you just take one of the two elements. – Michael Albanese Mar 15 '17 at 00:42
  • Your idea of "making sense of the intersections" may be right in this context but it is conjured out of thin air. – Rob Arthan Mar 15 '17 at 02:27
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Hint: the non-trivial subgroups of $\Bbb{Z}_6$ are $\{[0], [3]\}$ and $\{[0], [2], [4]\}$ (where $[i]$ is the equivalence class of $i \in \Bbb{Z}$ modulo $6$). These subgroups are isomorphic to $\Bbb{Z}_2$ and $\Bbb{Z}_3$ respectively and neither of them meets the set $\{[1]\}$.

If you use the standard constructions (as I just have), the intersection $\Bbb{Z}_i \cap \Bbb{Z}_j$ is empty unless $i = \pm j$. So somewhat confusingly, we write $0$ and $1$ for the identity elements in the rings $\Bbb{Z}_i$ and $\Bbb{Z}_j$ but the detailed set-theoretic meaning depends on which ring we are talking about (and on how we have constructed that ring). If we needed to be very precise we would have to write something like $[1]_3$ for $1 \in \Bbb{Z}_3$.

Rob Arthan
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