$J \subset I(V(J))$ where $J$ is an ideal. The textbook says it's by definition, but as I see it the inclusion should be reversed should it not?
I mean $I(V(J))= \{ f\in k[x_1,\dots, x_n]: f(a_1,\dots ,a_n)=0 \text{ for an arbitrary element } (a_1,\dots, a_n)\in V(J) \}$.
Now $V(J)$ is the set: $V(J)=\{ (a_1,\dots ,a_n)\in k^n: \text{ for an arbitrary element } g \in J , \ g(a_1,\dots , a_n)=0 \}$.
It seems to me that it should be $I(V(J)) \subset J$, because it includes the condition that the polynomial in $J$ will vanish at some point in $V(J)$.
Can someone show me the inclusion, explicitly?

We show that $$J \subseteq \sqrt J \subseteq I(V(J)).$$
That $J \subseteq \sqrt J$ is immediate.
That $\sqrt J \subseteq I(V(J))$ follows from: Let $f \in \sqrt J$, then $f^n \in J$ for some $n > 0$. But $V(J) = \{P \in \mathbb A^n : g(P) = 0$ for all $g \in J\}$ and $f^n \in J$, so $f^n(P) = 0$ for all $P \in V(J)$. But $0 = f^n(P) = \big(f(P)\big)^n$ and since $A = k[x_1, \ldots, x_n]$ is an integral domain, then $f(P) = 0$. Hence $f \in I(V(J))$.
Notice that if you had the Nullstellensatz at your disposal, you would have the stronger result: $$J \subseteq \sqrt J = I(V(J)).$$
