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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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It's possible you mean "given $s\in S$ with $s=k+k$ for some $k$, what extra conditions on $S$ ensure there is no $k'\ne k$ for which $k'+k'=s$," or you could mean "what conditions on $S$ ensure that $f:S\to S:k\mapsto k+k$ is injective?" The former is anchored to some specific $s$: perhaps it could be that $f$ is not injective yet the preimage $f^{-1}(s)$ is a singleton set for some specific $s\in S$. While your question is explicitly the former, I figure it's also possible you want the latter. –  anon May 21 '12 at 2:01
    
Thanks for pointing that out, I want the latter and that's a nice way to state it. –  Jackson Walters May 21 '12 at 2:09
    
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. –  A. De Luca Jun 25 '12 at 13:00

1 Answer 1

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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This is stronger than the left-canellability that the OP did not want. –  Alex Youcis May 21 '12 at 2:46
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@AlexYoucis, no, it is not. You need to cancel $2$, not every element of $S$. –  lhf May 21 '12 at 2:48
    
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 $k=2^{-1}s$. –  Jackson Walters May 21 '12 at 2:52

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