denotes the universal hypermultiplet). The $\caln=2$ supersymmetry can be broken to $\caln=1$ by adding a Fayet-Iliopoulos term [@APT], [@TV], [@Ivanov]. We are interested in the case $h^{11}=0$ which describes a theory with $h^{21}$ abelian vector multiplets and the universal hypermultiplet. To illustrate the argument it suffices to consider the simpler case of a theory with only one abelian $\caln=2$ vector multiplet, which has an expansion in terms of $\caln=1$ superfields $$\Phi=\phi^{(1)}(\tilde y,\theta)+{\sqrt 2} \tilde \theta^{\alpha}
\phi^{(2)}_{\alpha}(\tilde y,\theta)+\tilde \theta^{\alpha} \tilde
\theta_{\alpha} \phi ^{(3)}(\tilde y,\theta),$$ where $\tilde y^{\mu}=x^{\mu}+i\theta \sigma^{\mu}\bar \theta +i
\tilde \theta \sigma ^{\mu} \bar{\tilde{\theta}}$, and $\theta$, $\tilde \theta$ and their complex conjugates are superspace coordinates. The action for the $\caln=2$ vector then takes the form $$\int d^2\theta d^2\tilde \theta {\cal F}_0(\Phi),$$ where ${\cal F}_0$ is the $\caln=2$ prepotential. Supersymmetry can be broken by adding an FI term $$\int d^2\theta d^2\tilde \theta {\cal F}_0(\Phi)+\xi D,$$ where $D$ is one of the three auxiliary fields of the $\caln=2$ vector multiplet. Integrating out the auxiliary field exactly reproduces the $\caln=1$ flux superpotential containing $H_{RR}$ $$W=\int H_{RR} \wedge \Omega,$$ as $H_{RR}$ appears as an auxiliary field in the ${\cal N}=2$ vector multiplet. The superpotential is not renormalized neither perturbatively nor non-perturbatively as the $\caln=2$ 