Nullstellensatz: If $V(f)=V(g)$ we have that $Rad \langle f \rangle =Rad \langle g \rangle$

Question

In my lecture notes I have the following:

From the Nullstellensatz (NSS for short) we have the following:

$$\text{ If } V(f)=V(g) \Rightarrow V(Rad(\langle f \rangle ))=V(Rad \langle g \rangle ) \overset{ \text{ NSS } }{ \Longrightarrow } Rad \langle f \rangle =Rad \langle g \rangle $$

Can you explain it to me? I haven't understood it.

Answer 1

The Nullstellensatz says that $I(V(f)) = \sqrt{(f)}$, so $V(f) = V(g)$ implies (applying $I$ on both sides) that $\sqrt{(f)} = \sqrt{(g)}$, where I note $\sqrt{J}$ the radical of an ideal $J$.