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In Görtz/Wedhorn it is stated that if $Z$ is a locally closed subset of a scheme $X$, then there exists a unique reduced subscheme $Z_{\operatorname{red}}$ with the same topological space as $Z$.

My questions: why is this unqie? (Basically I would like to understand the argument (one sentence) they give for this assertion.)

My thoughts: basically we could start with any (sub)scheme structure on $Z$ and apply the functor $\cdot_{\operatorname{red}}$ to obtain a reduced subscheme, but if we start with different scheme structures at the outset, why don't we end up with different reduced scheme structures? Görtz/Wedhorn refers to the Proposition before the statement, which says that $\cdot_{\operatorname{red}}$ is functorial, so my second thought was that if we assume we have a reduced scheme structure $Z_{\operatorname{red}}$ (where now the $_{\operatorname{red}}$ is rather a fixed expression than a functor I guess), we could apply the functor $\cdot_{\operatorname{red}}$ to the closed immersion $i:Z_{\operatorname{red}}\rightarrow X$ and obtain $(Z_{\operatorname{red}})_{\operatorname{red}}=Z_{\operatorname{red}}$, so that $$i=Z_{\operatorname{red}}\stackrel{i_{\operatorname{red}}}{\rightarrow}X_{\operatorname{red}}\hookrightarrow X,$$ and I thought maybe I could argue from here somewhere, but I have no clue how to go on. I'm grateful for any sort of help.

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I think that the uniqueness property comes from the glueing property. If you have $Z\subset X$ you can take two affine neighbourhoods $U=Spec A$ and $V=Spec B$ and take the reduced rings $A_{red}$ and $B_{red}$. Now we have to see whats happening in the intersection. If $x\in V\cap U$, then there is a open neigh $x\in D(f\in A)=D(g\in B)\subset U\cap V$. Then $A_f=B_g$ by definition of scheme and $(A_f)_{red}=(B_g)_{red}$. Which should be the same as $(A_{red})_f=(B_{red})_g$ (check). So in the intersection everything behaves properly and the fibers are the same. So two structures will give you the same reduced scheme.

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