Define a domain \(\displaystyle{R}_{{{0}}}\) as follows. Take a field K, adjoin an indeterminate \(\displaystyle{x}_{{{0}}}\), and localize at \(\displaystyle{\left({x}_{{{0}}}\right)}\) (that is, adjoin inverses to everything not a multiple of \(\displaystyle{x}_{{{0}}}\)).

\(\displaystyle{R}_{{{0}}}\) has all its ideals principal and linearly ordered: \(\displaystyle{\left({x}_{{{0}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{0}}}^{{{2}}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{0}}}^{{{3}}}}\right)}\)

Now given \(\displaystyle{R}_{{{i}}}\) define \(\displaystyle{R}_{{{i}+{1}}}\) inductively: Adjoin an indeterminate \(\displaystyle{x}_{{{i}+{1}}}\), so we have \(\displaystyle{R}_{{{i}}}{\left[{x}_{{{i}+{1}}}\right]}\). Quotient by \(\displaystyle{\left({{x}_{{{i}+{1}}}^{{{2}}}}-{x}_{{{i}}}\right)}\). Finally, localize at the prime ideal \(\displaystyle{\left({x}_{{{i}+{1}}}\right)}\)

This effectively just gives us one more principal ideal containing all the principal ideals from \(\displaystyle{R}_{{{i}}}:{\left({x}_{{{i}+{1}}}\right)}\) contains \(\displaystyle{\left({{x}_{{{i}+{1}}}^{{{2}}}}={\left({x}_{{{i}}}\right)}\right.}\) contains \(\displaystyle{\left({{x}_{{{i}}}^{{{2}}}}\right)}\)

Now let R be the union of all the \(\displaystyle{R}_{{{i}}}\), and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the \(\displaystyle{x}_{{{i}}}\).