10
$\begingroup$

The chinese remainder theorem in its usual version says that for a finite set of pairwise comaximal ideals $R/\bigcap _jI_j\cong \prod _j R/I_j$.

In the binary case, the following general statement holds without conditions on the ideals $R/(I\cap J)\cong R/I\times _{R/I+J}R/J$. In this question I wanted to generalize the more general version to several ideals, but got stuck and only contrived an ad hoc justification for pairwise comaximality.

A few weeks ago I finally thought of $R/(I\cap J)\cong R/I\times _{R/I+J}R/J$ as a sheaf condition for a cover by two elements. Then I told myself the diagram below must be an equalizer, because pairwise comaximality pops out of it so naturally. $$R/\bigcap _jI_j\rightarrow \prod _j R/I_j \rightrightarrows \prod _{i,j}R/(I_i+I_j)$$ Several satisfied days later I stumbled upon this comment which to my dismay says the diagram above fails to be an equalizer for more than three ideals. But it just seems so perfect...

Can anyone give some counterexamples which show why the diagram above is not an equalizer and explain why things fail geometrically?

$\endgroup$
  • $\begingroup$ It would be weird for that diagram to be an equalizer since the first map is already an isomorphism! And the geometric way of thinking about this is that the vanishing sets of comaximal ideals are disjoint in Spec R, and disconnected Spec's are the Spec of the products of the coordinate rings of the components. $\endgroup$ – Dylan Wilson Jan 4 '16 at 12:40
  • $\begingroup$ @DylanWilson I'm asking about the general case, in which the ideals are not assumed comaximal. $\endgroup$ – Arrow Jan 4 '16 at 17:58
8
$\begingroup$

Each ideal $I\subset R$ corresponds to a closed subscheme of $\mathrm{Spec}(R)$, intersection of ideals corresponds to union of subschemes, and sum of ideals corresponds to intersection of subschemes. If a finite collection of closed subschemes is an open cover of their union (which is the case if and only if the union is disjoint), then indeed your diagram is an equalizer, precisely because of the sheaf condition. But in general the sheaf condition won't hold for coverings by closed sets.

A counterexample: let $k$ be a field, $R=k[x,y]$, $I_1=(x)$, $I_2=(y)$, $I_3=(x-y)$. Then $\mathrm{Spec}(R)$ is the affine plane and the ideals $I_1$, $I_2$, and $I_3$ correspond to the lines $L_1:x=0$, $L_2:y=0$, and $L_3:x=y$. The statement that your diagram is an equalizer is equivalent to the statement

the data of a regular function on $L_1\cup L_2\cup L_3$ is the same as the data of a triple of regular function $f_1$, $f_2$ , $f_3$ on $L_1$, $L_2$, $L_3$ resp., such that all three functions take the same value at the origin.

This is false because we need an extra condition on the functions $f_1$, $f_2$, $f_3$, namely for these functions to determine a function on $L_1\cup L_2\cup L_3$, the derivative of $f_3$ in the direction $(1,1)$ needs to equal the sum of the derivative of $f_1$ in the direction $(0,1)$ and the derivative of $f_2$ in the direction $(1,0)$.


The data of $\prod R/I_j$ and the maps to $\prod R/(I_i+I_j)$ corresponds to data of a bunch of closed subschemes and their pairwise intersections. This is not enough to determine a scheme. For instance, the union $L_1\cup L_2\cup L_3$ is not isomorphic to the union of the coordinate axes in $\mathbb{A}^3$, because the tangent space to the former at the singular point is $2$-dimensional, while the tangent space to the latter is $3$-dimensional. But both schemes are the union of three lines, such that the pairwise intersection is a single point. For the example $I_1,I_2,I_3\subset R$ above, the equalizer of your diagram is the coordinate ring of the union of the coordinate axes in $\mathbb{A}^3$, rather than of $L_1\cup L_2\cup L_3$.

$\endgroup$
  • 3
    $\begingroup$ Thanks for the instructive answer! I have asked about the equalizer of $\prod _j R/I_j \rightrightarrows \prod _{i,j}R/(I_i+I_j)$ here - perhaps you could answer that question as well :) $\endgroup$ – Arrow Jan 6 '16 at 17:12

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.