Find $\mathbb Z$-modules $L,K\subseteq M:=\mathbb Z\oplus(\mathbb Z/2019\mathbb Z)$ s.t. $L+K=M$ & $\text{Ass}(L)\cup\text{Ass}(K)\ne\text{Ass}(M)$

Question

Let $M:=\mathbb Z\oplus(\mathbb Z/2019\mathbb Z)$ be a $\mathbb Z$-module. Find submodules $M_1,M_2\subseteq M$ such that $M=M_1+M_2$ and $\text{Ass}(M)\ne\text{Ass}(M_1)\cup\text{Ass}(M_2)$.

Recall:

• $\text{Ass}(M)=\{p\in\text{spec}(\mathbb Z):\exists 0\ne m\in M: \text{ann}(m)=p\}$
• $\forall m\in M,\text{ann}(m)=\{r\in R:rm=0\}$

We notice that the only non trivial divisors of $2019$ are $3$ and $673$.

In addition,

• $\text{Ass}(\mathbb Z)=\{\langle 0\rangle\}$
• $\forall n<2019$ s.t. $1=(n,2019)$, $$\text{Ass}(n(\mathbb Z/2019\mathbb Z))=\text{Ass}(\mathbb Z/2019\mathbb Z)=\{\langle0\rangle,3\mathbb Z,673\mathbb Z\} $$
• $\text{Ass}(3(\mathbb Z/2019\mathbb Z))=\{\langle0\rangle,673\mathbb Z\}$

Hence, I can't choose $M_i=n(\mathbb Z/2019\mathbb Z)$ for $n\ne 3,673$. And also $M_1=3(\mathbb Z/2019\mathbb Z),M_2=673(\mathbb Z/2019\mathbb Z))$ won't work either.

Answer 1

Observe that the sum $M=M_1+M_2$ doesn't need to be direct. Choose $M_1 =<1\oplus \overline 1>$ and therefore $Ass(M_1)=0$. For $M_2$ there is another obvious choice.