Suppose R is any commutative noetherian ring with 1 and R is a direct limit of rings rings $R_{i}$.

Now suppose $I$ is any ideal in $R$. Does there direct system of ideals $ \lbrace I_{i} \rbrace$ of $ \lbrace R_{i} \rbrace$ ( where each $I_{i}$ is an ideal of $R_{i}$) so that $ I = colim I_{i}$ and $R/I = colim_{i} R_{i}/I_{i}$?

I was trying to intersect $I$ with $R_{i}$( in case if each $R_{i} \subseteq R$) but that may not be ideal in $R_{i}$. This is usually a trick in case of modules and submodules but the complexity here is that there are various rings. Any help would be great.

  • $\begingroup$ If $I$ is finitely generated (e.g. when $R$ is Noetherian), you can also do this by taking generators $r_1, \ldots, r_m$ of $I$. If the $r_i$ are defined over some $R_{i_0}$, define $I_i$ to be the ideal generated by the $r_i$ in $R_i$ for $i \geq i_0$ (and $0$ if $i \not\geq i_0$, or just throw out those $i$), $\endgroup$ – R. van Dobben de Bruyn Sep 5 '20 at 4:59

Yes. This is done essentially as you tried by taking the inverse images of $I$ in the $R_i$ under the canonical morphisms. For a detailed explanation see EGA (in the 1971 Springer edition).


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