4
$\begingroup$

Let $R$ be an integral domain, and $K$ its field of fractions. It is well known that for a finitely generated fractional ideal $I$ of $R$, and $S$ a multiplicative set we have $$(R:_KI)_S=(R_S:_KI_S).$$

I suppose that in general the above equation fails, but I don't know such an example.

The usual non-noetherian ring $k[X_1,\dots,X_n,\dots]$ is not an option since it is a Krull domain, or for Krull domains the equality holds for all fractional ideals.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let $R = \mathbb{Z}+X\mathbb{Q}[X]$ and $I = X\mathbb{Q}[X] = (X, X/2, X/3, \ldots)$, and let $S = \{1,2,3,\ldots\}$. Then $K = \mathbb{Q}(X)$, $R_S = \mathbb{Q}[X]$, and $I_S = X \mathbb{Q}[X]$, whence $(R_S:_K I_S) = (1/X)\mathbb{Q}[X]$. However, $(R:_K I) = \mathbb{Q}[X] = R_S$, so that $(R:_K I)_S = (R_S)_S = R_S = \mathbb{Q}[X]$.

To see that $(R:_K I) = \mathbb{Q}[X]$, note that $y \in (R:_K I)$ implies $$y = \frac{f_1}{X} = \frac{f_2}{X/2} = \frac{f_3}{X/3} = \cdots$$ for some $f_1, f_2, f_3, \ldots \in R$, which implies that $2, 3, 4, \ldots$ all divide $f_1$ in $R$, which implies that $f_1(0) = 0$ and therefore $f_1 \in X\mathbb{Q}[X]$ and hence $y \in \mathbb{Q}[X]$. Conversely, if $y \in \mathbb{Q}[X]$, then $Iy \subseteq I \subseteq R$, so $y \in (R:_K I)$.

A similar example that should also work is $R = \operatorname{Int}(\mathbb{Z})$ (the ring of integer-valued polynomials), $I = \{f \in \operatorname{Int}(\mathbb{Z}): f(0) = 0\}$, and $S = \{1,2,3,\ldots\}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.