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Let $K$ be an algebraically closed field. Then let $A = K[x_1, ... , x_n]$.

If $Y \subseteq \mathbb A^n$, then the ideal of $Y$ is to defined to be $I(Y) = \{f \in A | f(P) = 0 \ \forall P \in Y \}$. What is $I(\emptyset)$?

For $T \subseteq A$, define $Z(Y) = \{P \in \mathbb A^n | f(P) = 0 \ \forall f \in A\}$. Am I correct in thinking that $Z(A) = \emptyset$?

Thanks

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You are correct that $Z(A)=0$. Correspondingly, $I(\emptyset)=A$, as all elements of $A$ vanish on every point in $\emptyset$ (trivially, as there are none). In general, for any ideal $J\subseteq A$ we have $I(Z(J))=\sqrt{J}$, where $\sqrt{J}$ denotes the radical of $J$.

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Note that this last statement is one formulation of the weak Nullstellensatz. –  Alex Becker Jan 20 '12 at 21:01
    
Thank you. I'm aware of the statement; I was checking the details of an example that showed that statement doesn't necessarily hold if $K$ wasn't algebraically closed. –  Matt Jan 20 '12 at 21:05

By definition, $$I(\varnothing) = \{f\in A\mid f(P)=0\forall P\in\varnothing\} = A,$$ since every element of $A$ satisfies the condition by vacuity.

Since $P\in Z(A)$ requires $1=\mathbf{1}(P)=0$ (where $\mathbf{1}$ is the constant polynomial $1$), and this is impossible, this proves $Z(A)=\varnothing$.

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