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Let $k$ be an Algebraically closed field. Let $\mathcal C_k$ be the category of integral $k$-schemes of finite type over $k$ (the morphisms between two objects being the morphism of schemes that also commutes with the structure morphism to $k$) . Let $\mathcal D_k$ be the full subcategory (so the morphisms remain the same) of $\mathcal C_k$ whose objects have Krull dimension $0$. So $Z \in \mathcal D_k$ if and only if $Z \cong Spec (k)$.

Now let $X, Y\in \mathcal C_k$ be such that the functors $Hom_k (-,X) $ and $Hom_k (-,Y)$ are isomorphic when considered from $\mathcal D_k \to Set$ . Then is it true that $X \cong Y$ ?

Now I know that $Hom_k (-,X), Hom_k (-,Y) : \mathcal D_k \to Set$ are isomorphic means that $ Hom _k (Spec (k), X)$ and $Hom_k (Spec(k), Y)$ are bijective as sets.

Now if I knew that $ Hom _k (Spec (k), X) \cong Hom_k (Spec(k), Y)$ as locally ringed spaces (where I identify $ Hom _k (Spec (k), X) $ with the set of closed points $X(k)$ of $X$ and the structure sheaf I give is the restriction of that of $X$) , then I would be able to say $X \cong Y$ . Unfortunately, I'm not sure whether I can say that ... Please help

Also note that Yoneda Lemma doesn't apply since $X,Y$ are not necessarily in $\mathcal D_k$

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    $\begingroup$ Let $X,Y$ be, say two curves and let $f:X(k)\to Y(k)$ be any bijection. This induces a map as you desire, since any map from $\mathrm{spec}\ k\to X(k)$ (or $Y(k)$ ) is a morphism. Am I missing something? $\endgroup$
    – Mohan
    Aug 29, 2019 at 1:23
  • $\begingroup$ What happens if $X = Spec(k[t]), Y = Spec(k[t,t^{-1}])$ can you distinguish $Hom_{k-alg}(k[t],k[u]/(f)),Hom_{k-alg}(k[t,t^{-1}],k[u]/(f))$ (is it what your question is about ?). @Mohan also the OP asked a similar question for distinguishing ($\dim \ge 1$) varieties from the curves it contains. $\endgroup$
    – reuns
    Aug 29, 2019 at 2:58
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    $\begingroup$ @reuns If we work in the category of quasi-projective varieties over an algebraically closed field and do the same with the subcategory of one dimensional schemes, I believe the answer is yes. $\endgroup$
    – Mohan
    Aug 29, 2019 at 3:15
  • $\begingroup$ @Mohan: a bijection $X(k) \to Y(k)$ induces what map ? I'm not really getting your point ... $\endgroup$
    – user102248
    Aug 29, 2019 at 3:42
  • $\begingroup$ @reuns: yes in case $X,Y$ are affine-varieties say $X=Spec A$ and $Y=Spec B$ where $A,B$ are finitely generated $k$-algebras, I'm asking whether $Hom_k(A,-); Hom_k(B,-): \mathcal D \to Set$ being isomorphic as functors means $A$ and $B$ are isomorphic as $k$-algebras ? Basically if $Hom_k (A,k) $ and $Hom_k (B,k)$ are isomorphic as $k$-algebras then so are $A$ and $B$ but I'm not sure whether $Hom_k (A,k) $ and $Hom_k (B,k)$ are isomorphic as $k$-algebras follows from the isomorphism as functors ... $\endgroup$
    – user102248
    Aug 29, 2019 at 4:45

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Your $\mathcal D_k$ is equivalent to the category with one object and one morphism, since the only endomorphism of $\mathrm{Spec} k$ over itself is the identity and all objects of $\mathcal D_k$ are isomorphic. Thus an isomorphism of functors on $\mathcal D_k$ is nothing more than a bijection of sets; any two schemes with the same cardinality of their sets of $k$-points are isomorphic in this sense, which is certainly much weaker than an isomorphism of schemes.

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