# Examples of commutative semirings satisfying $kb = hb + r \ \; \text{ iff } \; k = h \text{ and } r = 0$.

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 $$\Bbb Z$$?

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 $$kb=hb+r$$.

• 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 – CopyPasteIt Jun 4 at 12:27