= Solution
Write $S=\Phi_0$, $u=\Phi_+$, $v=\Phi_-$, $M=M_p$, $P=uv-\zeta$, $x=|S|^2/M^2$ and $q=|u|^2+|v|^2$. Assume $g\ne0$ and, initially, $\zeta\ne0$. The <supergravity F-term potential> is
$$
V_F=e^{K/M^2}\left(K^{i\bar j}D_iW\overline{D_jW}-\frac{3|W|^2}{M^2}\right),\qquad D_iW=W_i+\frac{K_iW}{M^2}.
$$
The canonical <Kähler potential> gives the identity <Kähler metric>. The three <Kähler covariant derivatives of a superpotential> are
$$
D_SW=gP(1+x),\qquad D_uW=gS\left(v+\frac{\bar uP}{M^2}\right),\qquad D_vW=gS\left(u+\frac{\bar vP}{M^2}\right).
$$
Substituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answer
$$
\boxed{V_F=|g|^2e^{x+q/M^2}\left[(1-x+x^2)|uv-\zeta|^2+|S|^2\left(\left|v+\frac{\bar u(uv-\zeta)}{M^2}\right|^2+\left|u+\frac{\bar v(uv-\zeta)}{M^2}\right|^2\right)\right].}
$$
Here $(1+x)^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2)^2+3/4$. Thus this model's <scalar potential> is nonnegative everywhere.
In the stated limit $q/M^2\ll1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the <Kähler covariant derivatives of a superpotential>. Small waterfall fields alone do not justify discarding $\zeta/M^2$ in those derivatives, or expanding in $x$. If also $|\zeta|/M^2\ll1$ and $x\ll1$, the leading result reduces to the global <F-term scalar potential>
$$
V_{\mathrm{global}}=|g|^2\left(|uv-\zeta|^2+|S|^2q\right).
$$
No gauge sector or gauge coupling is supplied, so there is no specified <D-term> contribution to add.
The <supersymmetric vacuum manifold of a bilinear hybrid superpotential> is
$$
\boxed{S=0,\qquad uv=\zeta,\qquad V_F=0.}
$$
Indeed $W=0$ and all three <Kähler covariant derivatives of a superpotential> vanish there, so all <supergravity auxiliary fields> vanish and <supersymmetry> is unbroken. The nonnegative <scalar potential> makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S|^2q=0$; for $\zeta\ne0$ this forces $S=0$. With just the supplied fields this is a complex <vacuum manifold> before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, <D-flatness> typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta|}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.
The other familiar branch of <F-term hybrid inflation> has $u=v=0$. Its nonzero derivative $D_SW=-g\zeta(1+x)$ establishes <supersymmetry breaking>, and its exact <scalar potential> is
$$
\boxed{V_0(S)=|g\zeta|^2e^x(1-x+x^2).}
$$
The <canonical-Kähler correction to an F-term hybrid valley> begins as
$$
V_0=|g\zeta|^2\left(1+\frac{x^2}{2}+O(x^3)\right).
$$
In particular the <inflaton> quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.
To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global <scalar potential> is
$$
|g\zeta|^2+|g|^2\left[|S|^2(|u|^2+|v|^2)-|\zeta|(uv+\bar u\bar v)\right].
$$
The normalized combinations $(u\pm\bar v)/\sqrt2$ diagonalize this quadratic form, with
$$
\boxed{m_\pm^2=|g|^2(|S|^2\pm|\zeta|)\quad\text{in the global approximation}.}
$$
Thus $|S|^2>|\zeta|$ gives a valley of transverse minima, while below the threshold the negative <eigenvalue> produces the <waterfall instability in F-term hybrid inflation>.
The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta|/M^2$. Expansion of the exact <supergravity F-term potential> about $u=v=0$ gives equal diagonal coefficients $|g|^2M^2e^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g|^2M^2e^xb(1+x+x^2)$. Hence
$$
m_\pm^2=|g|^2M^2e^x\left[x+b^2(1+x^2)\pm b(1+x+x^2)\right].
$$
For the usual sub-Planckian symmetry-breaking scale $0<b\ll1$, the lower <eigenvalue> is $|g|^2M^2e^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll1$, and changes sign near $x=b$.
\b[The printed assertion of two full minima requires qualification.] The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In fact
$$
\frac{dV_0}{dx}=|g\zeta|^2e^xx(1+x)>0\qquad(x>0).
$$
Its only stationary point in $S$ is the origin. For $0<|\zeta|<M^2$ that point has a negative waterfall <eigenvalue> $|g|^2(|\zeta|^2/M^2-|\zeta|)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta>0$:
$$
V_F=|g|^2e^{2a^2/M^2}(a^2-\zeta)^2=|g|^2\zeta^2+2|g|^2\left(\frac{\zeta^2}{M^2}-\zeta\right)a^2+O(a^4).
$$
The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is \b[a SUSY-breaking transverse valley and a zero-energy <supersymmetric vacuum> minimum manifold]. The limit on $u,v$ alone does not fix the parameter range $|\zeta|/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.
In the corresponding global theory the exponential prefactor, Kähler corrections to $D_iW$ and universal $-3|W|^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta|^2$ for all $S$, whereas canonical <supergravity> lifts it by the quartic correction above. The <supersymmetric vacuum> $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta=0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.
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