# Why is the reduced scheme structure on a locally closed subset unique?

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.

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.