For ideals $J_1,J_2$ of a commutative ring $R$, the ideal quotient is defined by $(J_1:J_2)=\{f\in R\mid fJ_2\subset J_1\}$. Suppose $X$ is an affine variety. I am trying to understand why if $J_1$ and $J_2$ are radical ideal in the coordinate ring $A(X)=k[x_1,...,x_n]/I(X)$ (where $k$ is an algebraically closed field), then $\overline{V(J_1)-V(J_2)}=V(J_1:J_2)$.

The only tools I can think of to use are the fact that $\overline{S}=V(S)$ and the Nullstellensatz. But I still can't see why it's true. What am I missing?


1 Answer 1


Suppose $f(x)=0$ for all $x\in W:=V(J_1)-V(J_2)$. Then for all $g\in J_2$ we have $fg(x)=0$ for all $x\in V(J_1)$, so $fg\in J_1$, and hence $f\in(J_1:J_2)$. Thus $I(W)\subseteq(J_1:J_2)$, and so $\overline W\supseteq V(J_1:J_2)$.

Conversely, if $f\in(J_1:J_2)$ and $x\in W$, then $fg(x)=0$ for any $g\in J_2$. In particular, since $x\not\in V(J_2)$, we know that there exists $g\in J_2$ with $g(x)\neq0$, and hence $f(x)=0$. Thus $W\subseteq V(J_1:J_2)$, so $\overline W\subseteq V(J_1:J_2)$.


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