I'm reading a paper about Gröbner basis. Suppose $P$ is a set of multivariate polynomials in $n$ variables and $V(P)$ is the set of all the points $\alpha$ such that for every polynomial in $P$ like $q$, $q(\alpha )=0$, it means $V(P)$ is affine variety of $P$ and we define $I(V(P))$ be the set of all polynomials such that they are zero on whole $V(P)$. It is said that if $P$ is a Gröbner basis then $\left<P\right>=I(V(P))$ but I don't understand why $I(V(P))\subset \left <P\right>$. Could you please give me a hint?
Here $\left <P\right >$ is the ideal generated by $P$.
This is not true in general unless you are working over an algebraically closed field. In that case, the result you need is Hilbert's Nullstellensatz.
For example, if you were working over the field of real numbers, the polynomial $f(T)=T^{2}+1$ does not have any roots, so its vanishing locus is empty. But the polynomial constantly equal to 1 vanishes on every point of $V(f)$. Hence $1\in I(V(f))$, so we get that $I(V(f))=\mathbb{R}[T] \nsubseteq (T^{2}+1)$.