Let $S$ be a multiplicative subset of a commutative ring $R$. Now consider the homomorphism $\phi_S :S^{-1}R \mapsto R$ where $\frac{r}{s} \mapsto r$ for any $s\in S$. Now my question is: Does this homomorphism create an inclusion preserving bijection between the subrings of $R$ and the subrings of $S^{-1}R$? I am inclined to believe that this is true but I am too tired to prove (or disprove) that. Can somebody just confirm if this is true or not (I am not looking for a solution just a yes or no answer)?
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3I would be very surprised by that. I edited out a solution (teaches me to read), but think of $\mathbb Z\subset\mathbb Q$. – Karl Kroningfeld Dec 16 '13 at 03:19
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In general the function $\phi_S$ isn't even well-defined, since for example $\frac12=\frac24$. You could try to do something with lowest terms, but that would probably not work in general and would no longer be a homomorphism. And, as Karl points out, even in the case of $\mathbb Z\subset \mathbb Q$ no such bijection exists.
Alex Becker
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For the record, the OP reversed the direction of the homomorphism in an edit for some reason. – Karl Kroningfeld Dec 16 '13 at 03:34
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Thanks for the answer. You are correct, I don't know what was I thinking with that map. – user53970 Dec 16 '13 at 03:34
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@KarlKronenfeld, I changed it because I was a little confused. The first case didn't seem to work. I should look harder to find an inclusion preserving map between subrings of $R$ and $S^{-1}R$. – user53970 Dec 16 '13 at 03:37
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1@user53970 First, study the given example. A bijection simply does not exist. – Karl Kroningfeld Dec 16 '13 at 03:37
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@KarlKronenfeld, I know I edited it immediately after posting it but you were very quick. – user53970 Dec 16 '13 at 03:39
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@user53970 I know it's hard to tell across the monitor, but I have no issues with your edit; it helps to show where you are confused, if anything. – Karl Kroningfeld Dec 16 '13 at 03:40