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Let $\mathsf{CRing}_{\mathsf{red}}$ denote the category of reduced commutative rings, and $\mathsf{Sch}_{\mathsf{red}}$ the category of reduced schemes. Let $L : [\mathsf{CRing}_{\mathsf{red}},\mathsf{Set}] \to [\mathsf{CRing},\mathsf{Set}]$ be the left Kan extension (for a sufficiently large version of $\mathsf{Set}$ on the right so that it exists), given by $L(X)(A) = \mathrm{colim}_{R \to A,\, R \text{ reduced}} \, X(R)$. We have a diagram of fully faithful functors:

$$\begin{array}{c} \mathsf{Sch}_{\mathsf{red}} & \rightarrow & \mathsf{Sch} \\ \downarrow && \downarrow \\ [\mathsf{CRing}_{\mathsf{red}},\mathsf{Set}] & \rightarrow & [\mathsf{CRing},\mathsf{Set}]. \end{array}$$

Does it commute? In other words, if $X$ is a reduced scheme and $A$ is an arbitrary commutative ring $A$, is the canonical map $$\mathrm{colim}_{R \to A,\, R \text{ reduced}} \, X(R) \longrightarrow X(A)$$ bijective? This is clear when $X$ is affine. It also holds when $A$ is local.

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    $\begingroup$ Very late to the party, but I think this question can be answered using the framework from my thesis. The point is that reducedness is a local property, so the category of reduced schemes is a certain kind of colimit completion of the opposite of the category of reduced rings, in the same way as the category of schemes is a colimit completion of the opposite of the category of rings. The relevant universal properties then guarantee the commutativity of the diagram. $\endgroup$
    – Zhen Lin
    Jan 4, 2023 at 14:23

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This is true. Recall the following lemma:

Lemma. Let $X \stackrel i\hookleftarrow Z \stackrel f\to Y$ be a span of affine schemes, where $i$ is a closed immersion. Then the pushout $P = X \underset Z\amalg Y$ in $\mathbf{Sch}$ exists and is affine. If $X$ and $Y$ are reduced, then so is $P$.

Proof. If $X = \operatorname{Spec} A$, $Y = \operatorname{Spec} B$, and $Z = \operatorname{Spec} C$, then [Stacks, Tag 0ET0] shows that the pushout is represented by the affine scheme $$P = \operatorname{Spec} \big(A \underset C\times B\big).$$ The final statement follows since products and subrings of reduced rings are reduced. $\square$

Thus, it suffices to show the following:

Proposition. Let $f \colon Z \to S$ be a morphism from an affine scheme $Z$ to a reduced scheme $S$. Then there exists a reduced affine scheme $X$, a closed immersion $i \colon Z \hookrightarrow X$, and a morphism $g \colon X \to S$ such that $f = gi$.

Indeed, this immediately settles surjectivity. Moreover, if $Z \hookrightarrow X \to S$ is a factorisation as in the proposition, and $Z \to Y \to S$ is any factorisation with $Y$ affine reduced, then $P = X \amalg_Z Y$ is affine and reduced by the lemma above, giving a third factorisation $Z \to P \to S$ of which both $Z \hookrightarrow X \to S$ and $Z \to Y \to S$ are refinements. The explicit description of colimits in $\mathbf{Set}$ (see e.g. [Tag 002U]) then gives injectivity.

Proof of Proposition. Write $Z = \operatorname{Spec} A$. If the image of $f$ lands in an affine open $V \subseteq S$, say $V = \operatorname{Spec} B$, then we conclude by factoring $B \to A$ as $B \to B[A] \twoheadrightarrow A$, where $B[A]$ is a polynomial algebra with generators $a \in A$ (which is reduced since $B$ is). In general, cover $S$ by affine opens $V_j$ for $j \in J$, and choose a finite standard affine open cover $\coprod_{i=1}^n U_i \twoheadrightarrow Z$ refining the open cover $\coprod_{j \in J}f^{-1}(V_j) \twoheadrightarrow Z$; say $U_i = \operatorname{Spec} A_{f_i}$. Then we know that the result holds for each $U_i$, so pick $U_i \hookrightarrow X_i \to S$ with $X_i = \operatorname{Spec} R_i$ affine and reduced and $U_i \hookrightarrow X_i$ a closed immersion. Consider the diagram $$\coprod_{i \neq j} U_i \cap U_j \rightrightarrows \coprod_i X_i.\tag{1}\label{1}$$ We claim that the colimit of \eqref{1} exists in $\mathbf{Sch}$ as an affine reduced scheme. Indeed, for each $j \in \{0,\ldots,n\}$, consider the diagram $$\coprod_{i \neq j} U_i \cap U_j \rightrightarrows \Big( \coprod_{i=1}^j X_i \amalg \coprod_{i=j+1}^n U_i \Big)\tag{2}\label{2}.$$ We will prove by induction on $j$ that this diagram has an affine colimit in $\mathbf{Sch}$. For $j = 0$, the colimit is simply $Z$. If the diagram \eqref{2} has an affine colimit $P_j$ for some value of $j$, then the diagram for $j+1$ has affine colimit $P_{j+1} = P_j \underset{U_{j+1}}\amalg X_{j+1}$ by the lemma above.

This proves that \eqref{1} has an affine colimit $X = P_n$ in $\mathbf{Sch}$; say $X = \operatorname{Spec} R$. Applying the universal property to $\operatorname{Hom}(-,\mathbf A^1_{\mathbf Z})$ shows that $$R = \lim \Big( \prod_{i=1}^n R_i \rightrightarrows \prod_{i \neq j} A_{f_if_j} \Big),$$ which is reduced since each $R_i$ is. The map $P_0 \to P_n$ is a closed immersion $i \colon Z \hookrightarrow X$, and the universal property of the colimit of \eqref{1} gives a morphism $g \colon X \to S$ such that $f = gi$. $\square$

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  • $\begingroup$ Thank you so much, Remy! In the proof of injectivity, you consider two closed immersions, but actually it should be two arbitrary morphisms right? I think we can still use the pushout in the category of affine schemes, we don't need the universal property for non-affine schemes. $\endgroup$ Dec 31, 2022 at 7:37
  • $\begingroup$ What does B[A] mean? I have seen this notation being used, but here it would not yield a reduced ring. $\endgroup$ Dec 31, 2022 at 7:45
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    $\begingroup$ @MartinBrandenburg thanks, I had realised the same about your first point, and it's indeed an easy fix. As for $B[A]$, I mean the polynomial algebra on generators $A$, not the group algebra; I realise the notation is ambiguous. $\endgroup$ Dec 31, 2022 at 11:55
  • $\begingroup$ Cool! I think the section starting "Indeed..." still needs some correction. $\endgroup$ Dec 31, 2022 at 12:00
  • $\begingroup$ Hmm, I have exhibited an element on the left hand side and showed that every other element mapping to the same thing on the right is already identified in the colimit, so that should be enough, right? $\endgroup$ Dec 31, 2022 at 13:09

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