Suppose $ D $ is a codimension $ 1 $ irreducible subvariety of a variety $ X, $ and suppose that $ p_{1},p_{2},...,p_{k} $ are irreducible polynomials where $ V(p_{1}), V(p_{2}),\dots,V(p_{k}) $ are also codimension $ 1 $ subvarieties of $ X. $
If $ D \subset \bigcup_{i=1}^{k} V(p_{i}), $ is it true that this implies that $ D = V(p_{i}) $ for some $ i \in \lbrace 1,2,\dots,k \rbrace? $ If so, is this a consequence of the irreducibility of $ D? $
Assume $D$ is not contained in any one of the $V(p_i)$. Then $\bigcup_i (D \cap V(p_i)) = D$. This decomposition of $D$ into (closed) subvarieties must be trivial as $D$ is irreducible, a contradiction.
Assume $D \subset V(p_1)$. Since $D$ has codimension $1$, we can see that $D=V(p_1)$ by considering tangent spaces. I will expand on this later today once I have time.