I would like to understand the symmetric product of a union of two varieties. Assume that $X=X_1\cup X_2$, where $X_1,X_2$ are irreducible. What can we say about $Sym^2(X):=\frac{X\times X}{S_2}$, where $S_2$ acts on $X\times X$ by permuting the factors?
EDIT: My guess is that it should be something like $Sym^2(X_1)\times Sym^2(X_2)$. If I write $X_i=V(f_i), i=1,2$, then $$(X_1\cup X_2)\times (X_1\cup X_2)= (V(f_1)\cup V(f_2))\times (V(f_1)\cup V(f_2))=V(f_1f_2)\times V(f_1f_2)=V(f_1f_2,f_1f_2), $$
where $V(f_1f_2,f_1f_2)$ is considered as a subset the product. Then the step I am unsure about is $$ V(f_1f_2,f_1f_2)=V(f_1,f_1)\cup V(f_2,f_2)$$ which should then give that $X\times X=(X_1\times X_1)\cup (X_2\times X_2)$.
So is the step I mentioned above correct? If not, how can I describe $Sym^2(X)$ in terms of $Sym^2(X_1)$ and $Sym^2(X_2)$?
Any comments/references are very welcome!
If $X_1 \cap X_2 = \varnothing$ then $$ Sym^2(X_1 \sqcup X_2) = Sym^2(X_1) \sqcup (X_1 \times X_2) \sqcup Sym^2(X_2). $$ If the intersection is a subvariety $X_{12} = X_1 \cap X_2$ then one should glue $Sym^2(X_1)$ and $(X_1 \times X_2)$ along $X_1 \times X_{12}$, $Sym^2(X_2)$ and $(X_1 \times X_2)$ along $X_{12} \times X_2$ and $Sym^2(X_1)$ and $Sym^2(X_2)$ along $Sym^2(X_{12})$.