I am having trouble solving the following problem.
Let $\mathcal B$ be a small filtered subcategory of the category of commutative $k$-algebras, and suppose that all objects in $\mathcal B$ are integral domains. Show that the colimit of this inclusion diagram is an integral domain.
I think I can show that, if $L$ is the colimit of this diagram with maps $\psi_B:B\to L$ for all $B\in\operatorname{ob}(\mathcal B)$, then $L\cong\cup_{B\in\operatorname{ob}(\mathcal B)} \operatorname{Im}(\psi_B)$. Now I suppose that $ab=0$ in $L$, then there exists $x\in B_1$ and $y\in B_2$ in $\mathcal B$ such that $\psi_{B_1}(x)=a$ and $\psi_{B_2}(y)=b$. Then, using the filtered condition, we get maps to a common object $B$ in $\mathcal B$ so $x\mapsto x'\in B$ and $y\mapsto y'\in B$. Now we have $\psi_B(x')=\psi_{B_1}(x)$ and $\psi_B(y')=\psi_{B_2}(y)$, so $0=ab=\psi_B(x')\psi_B(y')=\psi_B(x'y')$.
Now I am stuck, because I don't know that $x'y'=0$ in $B$. If I did, then I could use that $B$ is an integral domain to get the $x'=0$ or $y'=0$, and if $x'=0$ then $a=0$ and if $y'=0$ then $b=0$.
Any help on where I am stuck would be appreciated, or if anyone has an alternative solution then I would also like to see that.