# How do I prove that $I_{V}(W_{1} \cup W_{2}) = I_{V}(W_{1}) \cap I_{V}(W_{2})$

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 :)

• 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 $W_2$". – KReiser Feb 23 at 20:07
• Thank you, I was thinking about this in a purely algebraic way. This solves it easily. – Daàvid Feb 23 at 21:33