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Let $(S,+,\cdot)$ be a semiring with or without 0 but necessarily with 1. Let $f: S \rightarrow S$ be defined by $f(k)=k+k$. What is the weakest possible extra assumption I need to make on $S$ so that $f$ is injective.

"Weak" will mean an assumption that achieves injectivity but does not imply the following sufficient condition, multiplicative cancellation: $\forall a,b,c \in S$, $a\cdot b=a\cdot c \Rightarrow b=c$. This is sufficient since

$f(k)=f(k')$ $\Rightarrow k+k=k'+k'$ $\Rightarrow (1+1)k=(1+1)k'$ $\Rightarrow k=k'$

by cancellation.

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    Your question could be restated as: Give examples of "weak" conditions implying the (left) cancellation property for the element 2=1+1 in a semiring.2012-06-25

1 Answers 1

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One sufficient condition is that $2$ is (left) invertible, that is, there is an $h \in S$ such that $h\cdot2=1$. Indeed, $k+k=(1+1)\cdot k= 2\cdot k$ implies $2\cdot k=2\cdot k'$ and so $k = 1\cdot k= (h\cdot 2)\cdot k= h\cdot(2\cdot k) = h\cdot(2\cdot k') = \cdots=k'$

(Of course, asking that $2$ be left-cancellable is just a restatement of the question, so I'm offering a slightly stronger but still natural condition.)

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    I agree, this is a natural condition. It's one that I've considered but didn't mention in the question due to the following. I'm trying to define a function as follows: $f(s)=...$, if $\exists k \in S$ such that $s=k+k$ and $f(s)=...$, otherwise. Unfortunately, if I demand that$2$be invertible $f(s)$ always reduces to the first case since we can just let $k=2^{-1}s$.2012-05-21