Affine varieties and their ideals (part2)

Question

On wikipedia, they talk about varieties $V,W$ and the $I(V)$ and $I(W)$ as well as the quotient ideal,

$$I(V):I(W) = I(V - W)$$

Can someone show me a quick proof of the identity?

Answer 1

1. Show

$$I(V):I(W) \supset I(V-W)$$

This is the easy part. By the definition of the ideal quotient, we have to show $I(V-W)I(W) \subset I(V)$ and this immediate: If $f$ vanishes on $V-W$ and $g$ vanishes on $W$, then $fg$ vanishes on all of $V$.

2. Show

$$I(V):I(W) \subset I(V-W)$$

For this we need a crucial ingredient:

Lemma. If $W \subset \mathbb A^n$ is an affine variety and $x \notin W$, there is some $f \in I(W)$, such that $f(x) \neq 0$.

Proof: Assume the opposite. We then have $I(W) \subset I(\{x\})$. Apply $V$ and get $V(I(W)) \supset V(I(\{x\}))$. For varieties $X$, we have $V(I(X))=X$. Since both are varieties, we have $W \supset \{x\}$, contradiction!

Now let us finish the proof:

Let $f \in I(V):I(W)$ and $x \in V-W$. We have to show $f(x)=0$.

By the lemma above, we have some $g \in I(W)$ with $g(x) \neq 0$. We have $fg \in fI(W) \subset I(V)$, hence $0=f(x)g(x)$. Then $g(x) \neq 0$ implies $f(x)=0$ and we are done.