In my lecture notes there is the following:
$$I \rightarrow V(I) \rightarrow I(V(I))$$
It stands that in general $I \subsetneq I(V(I))$.
The equality stands if and only if $I$ is a radical Ideal.
Can you explain why this stands???
$$V \rightarrow I(V) \rightarrow V(I(V))$$
The equality $V(I(V))=V$ stands always.
Why does this stand???
According to the Nullstelensatz, we have that $$I(V(I))=\operatorname{Rad}(I)$$
so $$I(V(I))=\{f(x_1, x_2, \dots , x_n) \in K \text{ such that } f(a_1, a_2, \dots , a_n)=0, \\ \forall a=(a_1, a_2, \dots , a_n) \in V(I)\}$$
$$\operatorname{Rad}(I)=\{a \in K \text{ such that } \exists\, n \in \mathbb{N} : a^n \in I\}$$
How do we relate these two definitions???
Think about this example: In $k[x]$, if $I = \langle x^2\rangle$ then $V(I) = \{0\}$. But $0$ is a root of $x$ so $I(V(I)) = \langle x\rangle$.
The key is to realize that $I(V(\mathfrak a))$ means take all the polynomials that vanish whenever $\mathfrak a$ vanishes. The roots of a polynomial $f^n$ are exactly the same as the roots of the polynomial $f$ so if $f^n \in \mathfrak a$ then $f$ is going to vanish on $\mathfrak a$ as well, hence $f \in I(V(\mathfrak a))$.