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Suppose $A$ and $B$ are two algebras of the same signature, both having maximum condition on sub-algebras. Is it true that $A\times B$ has the same property?

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What is maximum condition? –  The Masked Avenger Feb 19 at 22:38
Take ${\mathbb Z}$ with two unary operations $i:z\mapsto z+1$ and $d:z\mapsto z-1$. Such an algebra has only two subalgebras $\varnothing$ and ${\mathbb Z}$. Whereas in ${\mathbb Z}\times{\mathbb Z}$, a subalgebra is any subset closed with respect to the action of the group $({\mathbb Z},+)$ given by $z\cdot(z_1,z_2):=(z+z_1,z+z_2)$. Any orbit is given by the number $z_1-z_2$. Therefore, a subalgebra is nothing but a collection of integer numbers. Consequently, the maximality condition does not hold. –  Sasha Anan'in Feb 19 at 22:57

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Here is a semigroup example.

Take the semigroup $\mathbb Z_+$ of positive integers. Each of its subsemigroups is finitely generated so it has the maximum condition on subsemigroups. But $\mathbb Z_+\times \mathbb Z_+$ is not finitely generated so it does not have the maximum condition.

Indeed, any element of the form $(m,1)$ with $m>0$ is not a sum of two elements of $\mathbb Z_+\times \mathbb Z_+$. Thus these elements must belong to any generating set.

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Thank you, it seems that determining the type of algebras having this property must be hard. –  M. Shahryari Feb 20 at 3:06

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