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Let $D$ be a Noetherian integral domain and let $S$ be a multiplicatively closed subset. The saturation $\tilde{S}$ of $S$ is the set of all divisors of elements of $S$. Clearly, $S^{-1}D = \tilde{S}^{-1}D$, see page 138.

My question: Is it possible to find all the invertible elements of $S^{-1}D$ for a given $S$ and $D$? Trivially, all the elements of $S \cup S^{-1}$ (more precisely, the group this set generates) are invertible in $S^{-1}D$, but what about other invertible elements? For convenience, I do not mind to assume that $S$ is generated by one element only, denote it by $s$.

A special case: If $D$ is also a UFD, and $S= \{p^j\}_{j \in \mathbb{N}}$, where $p$ is a prime element of $D$, then the invertible elements of $S^{-1}D$ are exactly $\{p^j\}_{j \in \mathbb{Z}}$; indeed, if $d_1/p^{a_1}$ is invertible, then $(d_1/p^{a_1})(d_2/p^{a_2}) =1$, but then $d_1 d_2 = p^{a_1+a_2}$, so $d_1=p^{b}$ for some $0 \leq b \leq a_1+a_2$, hence $d_1/p^{a_1}= p^{b}/p^{a_1} =p^{b-a_1}$.

(Actually, a slightly generalized argument shows that in a Noetherian UFD, for every $\tilde{S}$ which is generated by $\{p_1,\ldots,p_m\}$, where $p_i$ are prime elements of $D$, the invertible elements of $S^{-1}D= \tilde{S}^{-1}D$ are exactly $ \{p_1^{\alpha_1}\cdots p_m^{\alpha_m} \}_{\alpha_1,\ldots,\alpha_m \in \mathbb{Z}}$).

But what if $D$ is not a UFD? The assumption that $D$ is Noetherian guarantees that $s \in S \subset D$ can be written (in at least one way) as a product of irreducible elements, but those irreducible factors may not be prime elements of $D$.

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user237522
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1 Answers1

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Given $d\in D$ and $s\in S$, note that $\frac{1}{s}$ is invertible, so $\frac{d}{s}$ is invertible iff $d$ is invertible. Now $\frac{e}{t}$ is an inverse to $d$ iff $de=t$. So $d$ is invertible iff there exists $e\in D$ such that $de$ is an element of $S$. That is, $d\in D$ is invertible iff $d\in\tilde{S}$. Thus the invertible elements of $S^{-1}D$ are just the fractions whose numerator is in $\tilde{S}$.

Eric Wofsey
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  • Thank you very much! (The argument "note that $1/s$ is invertible" is not necessary?). – user237522 Nov 01 '16 at 02:04
  • Well, to deduce that $ab$ is invertible knowing that $a$ is invertible, you need to know $b$ is invertible; I'm doing this with $a=d$ and $b=1/s$. But you could just say that $d=\frac{d}{s}\cdot s$ to deduce the same thing (of course, this is just explicitly using the inverse of $1/s$). – Eric Wofsey Nov 01 '16 at 02:18
  • Thanks for the clarification. – user237522 Nov 01 '16 at 02:24