Is this necessary that $k$ be algebraic close field in this theorem?

Question

let $k$ be algebraic close field and $A=k\left[x_{1}, \ldots, x_{n}\right]$

for any subset $Y \subseteq \mathbf{A}^{n}_k$,define the ideal of $Y$ in $A$ by

$ I(Y)=\left\{f \in A \mid f(P)=0 \text { for all } P \in Y\right\}$

and $Z(T)=\left\{P \in \mathbf{A}^{n}_k \mid f(P)=0\right.$ for all $\left.f \in T\right\}$

Theorem : For any ideal $a \subseteq A, I(Z(a))=\sqrt{a},$ the radical of $\mathfrak{a} .$

Is this necessary that $k$ be algebraic close field in the above theorem ?

Answer 1

Yes. Take $k=\mathbb R$ and $\mathfrak a=(x^2+1)\subset k[x]$ for example. We have have $Z(\mathfrak a)=\varnothing$ and $I(Z(\mathfrak a)) = k[x]$, which is not the radical of $\mathfrak a$.