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Let $M_1$ and $M_2$ be commutative monoids, $M_1$ written additively with identity $0$ and $M_2$ multiplicatively with identity $1$. Furthermore, let $M_2$ act on the left on $M_1$ via monoid endomorphisms, that is, we have a homomorphism of monoids $\varphi:M_2\rightarrow\mathrm{End}(M_1)$; as usual, instead of $\varphi(m_2)(m_1)$ for $m_1\in M_1,m_2\in M_2$, we write $m_2\cdot m_1$ or just $m_2m_1$. Then we can define a commutative monoid structure on the cartesian product $M_1\times M_2$ with identity $(0,1)$ and the binary operation $(m_1,m_2)\odot(m_1',m_2'):=(m_2'm_1+m_2m_1',m_2m_2')$. This is a bit like the semidirect product of monoids, but the action is applied to both first entries. If we choose, for instance, $\varphi$ to be the trivial homomorphism, then this is the direct product of $M_1$ and $M_2$.

My question simply is: Has this construction already been studied? And if so, what is known about it? Does it have a name? Thanks in advance!

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  • $\begingroup$ This looks like the two-sided semi direct product and shouldn't be commutative. $\endgroup$ Sep 8, 2014 at 20:21
  • $\begingroup$ @BenjaminSteinberg: Thank you for your comment! It is commutative, though, if both $M_1$ and $M_2$ are assumed to be commutative. $\endgroup$ Sep 8, 2014 at 20:36
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    $\begingroup$ This is a special case of the 2-sided semidirect product of Rhodes and Tilson. There you have a monoid N with commuting left and right actions. In your case they are both the same action and you are writing them both on the same side. This can be done because everything is commutative. $\endgroup$ Sep 8, 2014 at 20:43
  • $\begingroup$ @BenjaminSteinberg: I see. Thank you very much, this was helpful! $\endgroup$ Sep 8, 2014 at 20:48

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