Injectivity of $I\leadsto V_I(-)$ and relation to Hilbert's Nullstellensatz The functor $\mathbb A^n_k(-):\mathbf {Alg}_k\to \mathbf{Set}$ is represented by $\operatorname{Hom}_k(k[x_1,\dots ,x_n],-)$; for every ideal $I\subseteq k[x_1,\dots ,x_n]$, there is a subfunctor $V_I(-)\subseteq \ A^n_k(-)$ that sends a $k$-algebra $K$ to the set (in $K^n$) of the solutions of (all the polynomials in) $I$.
Clearly $V_I(-)$ is represented by $\operatorname{Hom}_k(k[x_1,\dots ,x_n]/I,-)$, meaning that the association $I\leadsto V_I(-)$ is injective. However if we restrict the domain of our functors to $\mathbf {Alg}^{\mathbf{red}}_k\subseteq \mathbf {Alg}_k$, i.e. the reduced $k$-algebras, we lose the injectivity: for example, $V_{(x)}(-)$ and $V_{(x^2)}(-)$ become the same functor.
My first question is: if I only consider radical ideals, then the association $I\leadsto V_I(-)$ returns injective? I would say yes because if $I=\sqrt{I}$, then $k[x_1,\dots ,x_n]/I$ is reduced and so $V_I(-)$ is representable also as a functor on $\mathbf{Alg}^{\mathbf{red}}_k$.
My second question is: does Hilbert's Nullstellensatz link in any way with these observations? They remind me of the bijection between radical ideals and algebraic varieties, but I can't see clearly the connection, if it exists. Thanks for any clarification.
 A: Question: "My second question is: does Hilbert's Nullstellensatz link in any way with these observations? They remind me of the bijection between radical ideals and algebraic varieties, but I can't see clearly the connection, if it exists. Thanks for any clarification."
Answer: It seems your claims follow from the nullstellensatz in the following way: If $k \subseteq K$ is the algebraic closure of a field $k$ and if $I,J \subseteq R:=k[x_1,..,x_n]$ let
$$V(I):=\{(a_i)\in K^n: f(a_i)=0\text{ for all $f\in I$}\}.$$
It follows by the HNS that $I(V(J)) = \sqrt{J}$ is the radical of $J$. Hence if $I,J$ are radical ideals with $V(I)=V(J)$ it follows $I=J$. Hence if there is an equality of functors $V_I(-)=V_J(-)$ it follows $V(I)=V(J)$ and hence $I=\sqrt{I}=\sqrt{J}=J$ and hence the correspondence $I \leadsto V_I(-)$ is injective on ideals.
Question: "My first question is: if I only consider radical ideals, then the association $I\leadsto V_I(-)$ returns injective? I would say yes because if $I=\sqrt{I}$, then $k[x_1,\dots ,x_n]/I$ is reduced and so $V_I(-)$ is representable also as a functor on $\mathbf{Alg}^{\mathbf{red}}_k$."
Answer: A similar argument proves this claim. If there is an equality of functors $V_I(-)=V_J(-)$, it follows $V_I(K)=V_J(K)$ hence $V(I)=V(J)$ and hence $I=\sqrt{I}=\sqrt{J}=J$.
