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Let $N_1$, $N_2$ be $Q_1$, $Q_2$-subsemimodules of an $R$-semimodule $M$, with $N_1 \cap N_2 = \{0\}$. Is $N_1 + N_2$ a $Q$-subsemimodule of $M$?

I think the answer of the question is yes, but I cannot find the subset $Q$ of $M$. For more information about $Q$-subsemimodules, refer to my paper:

J. N. Chaudhari and D. R. Bonde, On Partitioning and Subtractive Subsemimodules of Semimodules over Semirings; Kyungpook Math. J. 50(2010), 329-336.

1 Answers 1

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I am going to say no. Consider the $\mathbb{R}^{\geq 0}$-semimodule $M=\{(x,y)\mid x+y\geq 0\}$, and the subsemimodules $N_1=\{(x,0)\mid x\geq 0\}$ and $N_2=\{(0,y)\mid y\geq 0\}$. Let $Q_1=Q_2=\{(x,y)\mid x+y=0\}$. Observe that $N_1$ is a $Q_1$-subsemimodule of $M$, and $N_2$ is a $Q_2$-subsemimodule of $M$. But $N_1+N_2$ is the positive orthant, $\{(x,y)\mid x,y\geq 0\}$, so for any $(x,y),(x',y')\in M$ we have: $(\max(x,x'),\max(y,y'))\in((x,y)+N_1+N_2)\cap ((x',y')+N_1+N_2)$.