$R$ is a PID with field of fractions $K$, and $M \subseteq K$ is a finitely generated $R$-submodule. I am trying to show $M$ is in fact generated by one element.
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Consider a finite generating set and find a common denominator for its elements. Now, look at what you've got...
Alternatively, if $M$ is such a submodule, it has no torsion and we know from the structure theorem for f.g. modules over a PID that it must be free. Its rank can be computed by first tensoring with $K$ over $R$ and computing the dimension over $K$ of the resulting $K$-vector space. Because $M$ is contained in $K$, it is very easy to see that that dimension can only be $1$.
Mariano Suárez-Álvarez
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HINT $\: $ Fractions too have $\rm\: gcd\: \bigg(\!\!\dfrac{a}b,\:\dfrac{c}d\!\!\bigg)\ =\ \dfrac{gcd(a,b)}{lcm(b,d)}\ $ $\Rightarrow\:$ Bezoutness $\:\Rightarrow\:$ principality.
Bill Dubuque
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What is this identity meant to say (once presumably corrected)? – darij grinberg Feb 19 '23 at 03:36
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See e.g. here – Bill Dubuque Feb 19 '23 at 05:42
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Ah, I was missing the conditions (and also the $b$ on the top right should be a $c$). Thanks! – darij grinberg Feb 19 '23 at 12:13
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Yes, that's clearly a typo. – Bill Dubuque Feb 19 '23 at 18:06