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In algebraic geometry, the reducedness is a "stalk-local" property: a scheme is reduced (the ring $\mathcal O_X(U)$ is reduced for evert open subset $U$) iff the stalks are reduced at any point.

I wonder if this is true for general commutative rings. Suppose we have a family of commutative rings $\{A_i:i\in \mathcal I\}$, where $\mathcal I$ is a filtered set (the set that makes the colimit exist). I wonder if

$\operatorname{colimit}_{i\in \mathcal I}(A_i)$ is reduced $\Leftrightarrow$ each $A_i$ is reduced.

It seems to me at least $\Rightarrow$ isn't necessarily true, but I can't come up with a counterexample. Also I don't know how to prove/disprove $\Leftarrow$.

No One
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1 Answers1

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$\Rightarrow$ is false. Consider a diagram of the form $A \to B$ where $A$ is not reduced and $B$ is reduced. The colimit of this diagram is $B$ because it satisfies the universal property.

$\Leftarrow$ is proven as follows. Because $A_i$ is a filtered set, an element of $\operatorname{colimit}_{i\in \mathcal I}(A_i)$ is represented by an element $(a, i)$ for some $i \in \mathcal I$ and $a \in A_i$. Let $a$ represent some nilpotent element $b$, so that $b^n = 0$. As multiplication in the colimit is defined by multiplication in $A_i$, and because $a^n = 0$ in the colimit, there exists $A_j$ in the diagram such that $f \colon A_i \to A_j$ and $f(a^n) = f(a)^n = 0$. Thus, as $A_j$ is reduced, $f(a) = 0$, so $a$ is in the same equivalence class as 0 in the colimit. As $a$ is a representative of $b$, this is equivalent to $b=0$, showing that the colimit is reduced.

user477805
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