$J \subset I(V(J))$ 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?

Think this through again. $I(V(J))$ is the set of polynomials that vanish on every point on which every element of $J$ vanishes. Certainly every element of $J$ vanishes on every point on which every element of $J$ vanishes. On the other hand, $x$ vanishes on every point on which $x^2$ vanishes, but isn't in the ideal generated by $x^2$. 
