Consider an ideal $I=\langle f_1,f_2,\ldots,f_s\rangle$ in the polynomial ring $\mathbb{Q}[x_1,x_2,\ldots,x_n].$ Build the following set $$ \{ g_1 f_1+g_1 f_2+\cdots+g_n f_n \}, $$ where $g_i$ belongs to the field of fractions $\mathbb{Q}(x_1,x_2,\ldots,x_n)$ and denominators of all $g_i$ does not belong to the ideal $I.$ Is there a special name for this set?
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2$\begingroup$ This is the maximal ideal in the localization of $S$ at $I$. $\endgroup$– Alexander WooCommented Dec 1, 2010 at 20:09
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1$\begingroup$ Other than "localization of the module $\left\lbrace f_1,f_2,...,f_s\right\rbrace$ at the complement of the ideal $\left\lbrace f_1,f_2,...,f_s\right\rbrace$"? $\endgroup$– darij grinbergCommented Dec 1, 2010 at 20:10
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$\begingroup$ Sorry, the brackets should be < > rather than { }. $\endgroup$– darij grinbergCommented Dec 1, 2010 at 20:10
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1 Answer
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This is the image of $I$ in the localization $\mathbb Q[x_1,x_2,\dots,x_n]_{I}$.
There is an issue here though. If $I$ is not a prime ideal, then its complement is not multiplicatively closed, and therefore not a good set to invert in a localization. If $I$ is not a prime ideal, then there are some $f, g$ in $I^c$ such that $fg\in I$. Thus, e.g., $\frac{1}{f}\cdot\frac{1}{g}$ becomes a non-allowed coefficient, while $\frac{1}{f}$ and $\frac{1}{g}$ both are.
The set you describe can still be defined, obviously, but it will lack interesting structure. If $I$ is a prime ideal, then your set is the ideal in the localization ring that I describe above.
answered Dec 1, 2010 at 20:13