Rigidity for an integral factor with only constant functions

Let $Y_n=\mathrm{Spec}({\mathcal{O}}_{Y,y_0}\mathrm{/}{\mathfrak{m}}_{y_0}^{n+1})$ and define $Z_n$ similarly at $z_0$. Constancy on the original fibre says the ideal of $z_0$ pulls back into the ideal of $X\times \{y_0\}$; its $(n+1)$st power pulls back into the corresponding power. Thus $f$ restricts to $f_n:X\times Y_n\to Z_n$. These finite infinitesimal neighbourhoods are affine. By [F1] and the constant-function hypothesis, $\mathrm{\Gamma }(X\times Y_n,\mathcal{O})=\mathrm{\Gamma }(Y_n,\mathcal{O})$. Hence $f_n$ factors through $Y_n$, and evaluation at $x_0$ identifies the factor with $f(x_0,-){\mathrm{\mid }}_{Y_n}$.

Let $E$ be the closed equalizer of $f$ and $f(x_0,-)\circ {\mathrm{pr}}_Y$, using the closed diagonal of $Z$. Step 1.1 says $E$ contains $X\times Y_n$ for every $n$. In the Noetherian local ring of $X\times Y$ at $(x_0,y_0)$, the equalizer ideal is therefore contained in every power of the ideal generated by ${\mathfrak{m}}_{y_0}$. That ideal is contained in the local maximal ideal, so [F2] says the equalizer ideal is zero there. Since the ideal sheaf is coherent, $E$ contains an open neighbourhood of $(x_0,y_0)$. The product of integral varieties over the algebraically closed field is integral; an ideal on it vanishing on a nonempty open is zero, since it injects into the rational function field on every affine chart. Thus $E=X\times Y$, giving the claimed identity. AC is inherited from [F2] and the integral-variety suppliers.