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Let $\Bbb M$ be a commutative semiring. Setting $b = 1 + 1$, $\, \Bbb M$ also satisfies

P-1: For every $k,h \in \Bbb M$ and $r \in \{0,1\}$

$$\tag 1 kb = hb + r \ \; \text{ iff } \; k = h \text{ and } r = 0$$


I'm trying to prove that $\Bbb M$ is infinite.

Set $k = 0$ and $h = 0$ and $r = 1$ and applying $\text{(1)}$,

$\quad 0b = 0b + 1$

is false. Since $0b = 0$, $\;1 \ne 0$.

Set $k = 0$ and $h = 1$ and $r = 0$ and applying $\text{(1)}$,

$\quad 0b = 1b + 0$

is false. But then $\;1+1 \ne 0$.

Set $k = 0$ and $h = 1$ and $r = 1$ and applying $\text{(1)}$,

$\quad 0b = 1b + 1$

is false. But then $\;1+1+1 \ne 0$.

I then convinced myself that the set

$$\{0,1,1+1,1+1+1,1+1+1+1,1+1+1+1+1\}$$

contains $6$ elements and using induction (not completely thought out) that the commutative semiring $\Bbb M$ is infinite.

Are there examples of these semirings that aren't isomorphic to the $\Bbb N$ or $\Bbb Z$?

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If $\mathbb{M}$ is a finite semiring, then the elements $b^1,b^2...$ cannot all be distinct.

Case 1:

Suppose that we can choose $m,n\ge 1$ so that $b^m=b^n$ and $b^{m-1}\ne b^{n-1}$. Then $k=b^{n-1}, \ h=b^{m-1},\ r=0$ satisfies $kb=hb+r$.

Case 2:

Suppose that for any $m,n\ge 1$ with $b^m=b^n$ the equality $b^{m-1}=b^{n-1}$ also holds. Suppose furthermore that there exists an $m>0$ such that $b^m=b^0=1$ and $b^{m-1} \ne 1$. Then $h=b^{m-1}, \ k=1, \ r=1$ satisfies $kb=hb+r$.

Case 3:

Suppose that any $m,n\ge 1$ with $b^m=b^n$ satisfies $b^{m-1}=b^{n-1}$ and that for any $m>0$ such that $b^m=1$ we also have $b^{m-1}=1$. Then it can be shown that $b=1$. It follows that $h=0, k=1, r=1$ satisfies $kb=hb+r$.

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  • $\begingroup$ So a streamlined proof that $\Bbb M$ is not finite... (+1) Also: Step 1: $b \ne 1$ $\quad$ Step 2: $b^2 \ne b$ $\quad$ Step 3: Apply your Case 1 logic $\endgroup$ – CopyPasteIt Jun 4 at 12:27

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