Let $R$ be a commutative ring and $I$ and $J$ two ideals. Consider the index category $\mathcal{I}$ with three objects $\{a,b,c\}$ with morphisms $Hom(a,a)=\{id_a\}$, $Hom(b,b)=\{id_b\}$, $Hom(c,c)=\{id_c\}$, $Hom(a,b)$ and $Hom(a,c)$ being of size $1$ as well with morphisms denoted by $f$ and $g$, $Hom(b,c)=\{\varnothing\}$. Show that the colimit of the functor $\mathcal{I} \to Rings$:

-sending $a$ to $R$, $b$ to $R/I$, $c$ to $R/J$ and

-sending the morphisms $f$ and $g$ to the quotient maps

is the ring $R/(I+J)$.

I'm stuck to construct a homomorphism $h: R/(I+J)$ to a any ring $M$ satisfying, for example, $$\begin{array}{cc} R & \xrightarrow{k} & M\\ \downarrow &\nearrow{l} \\R/J \end{array}$$ such that the following diagram commutes $$\begin{array}{cc} R & \xrightarrow{} R/(I+J) \xrightarrow{h} & M\\ \downarrow & \nearrow &\nearrow{l} \\R/J \end{array}$$ My attempt is that for every $\overline{r} \in R/(I+J)$, choose the representative $r \in R$ of $\overline{r}$ and $h(\overline{r})=k(r)$ but it is not well-defined.

P/s: I'm not familiar with drawing diagram here, sorry for any inconvenience.

  • 2
    $\begingroup$ Spoiler: What you are considering is a pushout diagram in the category of rings i.e. the tensor product $R/I \otimes_R R/J = R/(I + J)$. Have a look here for this equality, and here for the fact that the pushout is the tensor product. $\endgroup$ Apr 2 '20 at 19:33

The colimit you are asked to compute is a so called pushout. We can show that the ring $R/(I+J)$ satisfies the universal property, or use previously computed explicit forms of pushouts in rings (as suggested in the comments). Let us do the former.

First denote the morphisms at play like $$ \begin{array}{ccc} R & \overset{\pi_I}{\rightarrow} & R/I\\ \pi_J\downarrow & & \downarrow \tau_J\\ R/J & \overset{\tau_I}{\rightarrow} & R/(I+J)\\ \end{array} $$ where $\tau_J$ denotes the composite $$R/I \rightarrow (R/I)/(J/I) \cong R/(I+J)$$ of quotient map and canonical isomorphism.

We need to check that given a pair of morphisms $t_I:R/I \rightarrow T$ and $t_J:R/J \rightarrow T$ satisfying $$t_I\pi_I = t_J\pi_J \;\;(\star)$$ there is a unique morphism $t:R/(I+J) \rightarrow T$ such that $$t\tau_J = t_I\;\;\text{and}\;\; t\tau_I = t_J$$

The equation $(\star)$ tells us that $J/I\subseteq \ker t_I$ and $I/J \subseteq \ker t_J$. Thus by the isomorphism theorem $t_I$ factors uniquely via $$R/I \overset{\pi_{(I/J)}}{\rightarrow} (R/I)/(J/I) \overset{p_I}{\rightarrow} T$$ and hence uniquely via some $q_I:R/(I+J)\rightarrow T$. The same holds for $t_J$ with some unique $p_J:(R/J)/(I/J) \rightarrow T$ respectively $q_J:R/(I+J) \rightarrow T$.

It is left to show that $q_I = q_J =: t$ holds. But this follows from the calculation $$q_J\tau_I\pi_J = t_J\pi_J = t_I\pi_I = q_I\tau_J\pi_I = q_I\tau_I\pi_J$$ and the fact that $\tau_I\pi_J$ is epic/right-cancellative/surjective as composition of quotient maps and isomorphisms.


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