Show that $I(\overline{Y_1 \setminus Y_2}) = I(Y_1):I(Y_2)$

Question

Let $X$ be an affine variaty in a commutative ring $R$ and let $,Y_1,Y_2$ be subvarieties of $X$. Show that in the coordinate ring $A(X)=K[x_1,...,x_n]/I(X)$, where $K$ is an algebraically closed field, holds $I(\overline{Y_1 \setminus Y_2}) = I(Y_1):I(Y_2)$.

Remark: In a ring $R$ the ideal quotient of the ideals $J_1, J_2 \vartriangleleft R$ is defined by $(J_1:J_2)=\{f\in R\mid fJ_2\subset J_1\}$.

My attempt is as follows:

$\subseteq:$ If $f \in I(\overline{Y_1 \setminus Y_2})$, then by the definitions $f(x) = 0$ for all $x \in \overline{Y_1 \setminus Y_2}$, which, by the fact that for any $y \in Y_1$ either $y \in Y_1 \setminus Y_2$ or $y \in Y_2$ holds, implies $fI(Y_2) \subseteq I(Y_1)$.

$\supseteq:$ If $f \in I(J_1):I(J_2)$, then $fI(Y_2) \subseteq I(Y_1)$. We now make a case distinction: If $Y_1 = Y_2$, then $I(\overline{Y_1 \setminus Y_2}) = I(\emptyset) = A(X)$, thus $I(\overline{Y_1 \setminus Y_2})$. If $Y_1 \ne Y_2$, then $fI(Y_2) \subseteq I(Y_1)$ implies $f \in I(\overline{Y_1 \setminus Y_2})$.

I am unsure why this seemingly does not require us to use that we are working in $A(X)$. I also do not get we need to consider $\overline{Y_1 \setminus Y_2}$ instead of $Y_1 \setminus Y_2$. Could you please explain this to me?

Answer 1

I don't quite get why you have to distinguish cases, and how you arrive at the final implication (if $Y_1\neq Y_2$ and $fI(Y_2)\subseteq I(Y_1)$ then $f\in I(\overline{Y_1\setminus Y_2})$). A concise argument would go as follows:

Let $f$ be such that $fI(Y_2)\subseteq I(Y_1)$. We will show first that $f$ vanishes on $Y_1\setminus Y_2$ (so if $Y_1=Y_2$ there is nothing to prove). Assume by contradiction that there exists $y\in Y_1\setminus Y_2$ such that $f(y)\neq 0$. As $y\notin Y_2$, there exists $g\in I(Y_2)$ such that $g(y)\neq 0$ (as $Y_2=\{x\ |\ h(x)=0\ \forall h\in I(Y_2)\}$). But then $fg\in I(Y_1)$ by hypothesis, and $(fg)(y)=f(y)g(y)\neq 0$ by construction, so we have a contradiction.

As the set of points $V(f)$ where $f$ vanishes is closed and contains $Y_1\setminus Y_2$, it must contain $\overline{Y_1\setminus Y_2}$. So we obtain $f\in I(V(f))\subseteq I(\overline{Y_1\setminus Y_2})$.

We are implicitly using that we are working in $A(X)$, because we use various properties (like $V(I(Y))=Y$ for any closed subset etc). Also, although we can associate a vanishing ideal $I(S)$ to every subset $S$ (not necessarily closed), we have in general that $V(I(S))=\overline{S}$. So indeed $I(\overline{Y_1\setminus Y_2})=I(Y_1\setminus Y_2)$, but $V(I(Y_1\setminus Y_2))\neq Y_1\setminus Y_2$ in general. I think this is the conceptual reason for working directly with the closure.