# Is a ideal of direct limit of rings is itself a direct limit of ideal?

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.

• 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 $i$), – 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 0.6.1.2 (in the 1971 Springer edition).