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?

  • 2
    $\begingroup$ This is the maximal ideal in the localization of $S$ at $I$. $\endgroup$ Commented Dec 1, 2010 at 20:09
  • 1
    $\begingroup$ Other than "localization of the module $\left\lbrace f_1,f_2,...,f_s\right\rbrace$ at the complement of the ideal $\left\lbrace f_1,f_2,...,f_s\right\rbrace$"? $\endgroup$ Commented Dec 1, 2010 at 20:10
  • $\begingroup$ Sorry, the brackets should be < > rather than { }. $\endgroup$ Commented Dec 1, 2010 at 20:10

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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