Let $k$ be an algebraically closed field. How to prove that $\dim_k(k[x_1,\dotsc,x_n]/\sqrt{I})=|V(I)|$?
I know that Hilbert Nullstellensatz will be required, but I can't figure out how.
With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as ${\displaystyle {\hbox{I}}({\hbox{V}}(I))={\sqrt {I}}}$ where $ {\displaystyle I\subset k[X_{1},\ldots ,X_{n}]} $ ideal.