I want to prove that $I_{V}(W_{1} \cup W_{2}) = I_{V}(W_{1}) \cap I_{V}(W_{2})$, where $V$, $W_{1}$ and $W_{2}$ are affine algebraic varieties with $W_{1},W_{2} \subset V$ and $I_{V}$ is the ideal function relative to $V$, i.e. $I_{V}(W) = \frac{I(W)}{I(V)}$.

We obviously have that $\frac{I(W_{1}) \cap I(W_{2})}{I(V)} \subset \frac{I(W_{1})}{I(V)} \cap \frac{I(W_{2})}{I(V)}$, but the other inclusion doesn't seem obvious to me since I can't tell if having elements $f \in I(W_{1})$ and $g \in I(W_{2})$ with $f \sim g$ (where $\sim$ is the equivalence relation induced by $I(V)$) gets us an element $h \in I(W_{1}) \cap I(W_{2})$ with $h \sim f \sim g$.

One thing that would fix this would be to have $\epsilon^{-1}(\frac{I(W)}{I(V)}) = I(W)$, where $\epsilon \colon R:=\mathbb{K}[X_{1},...,X_{n}] \to \frac{R}{I(V)}$ is the canonical epimorphism, which in turn is the same as asking if we have $\epsilon^{-1}(\frac{J}{I})= J$ for $I,J \subset R$ radical ideals, since $I(\bullet)$ gives a bijection between radical ideals and varieties, by the Nullstellensatz.

Any help would be appreciated! Thanks in advance :)

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    $\begingroup$ Try saying it in words: "a function on $V$ vanishes on $W_1\cup W_2$ if and only if it vanishes on $W_1$ and it vanishes on $W_2$". $\endgroup$ – KReiser Feb 23 at 20:07
  • $\begingroup$ Thank you, I was thinking about this in a purely algebraic way. This solves it easily. $\endgroup$ – Daàvid Feb 23 at 21:33

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