The problem is the following: Let $f:X\to\operatorname{Spec}(A)$ be a quasi-compact morphism. Let $I\subset A$ be an ideal. Show that $f(X)\subset V(I)$ if and only if $f^\#(\operatorname{Spec}(A))(I)$ is nilpotent in $\mathcal{O}_X(X)$. For simplicity, I'll put $\varphi:=f^\#(\operatorname{Spec}(A))$.
Since $X$ is quasi-compact, I'd like to use the fact that the canonical mapping $\mathcal{O}_X(X)_h\to\mathcal{O}_X(X_h)$ is injective (where $X_h = \{x\in X\;:\; h_x\in\mathcal{O}_X(X)^\times\}$). Using this condition, we have that $h$ is nilpotent if and only if $\mathcal{O}_X(X)_h = 0$, so if I can show $\mathcal{O}_X(X_h) = 0$ for all $h\in\varphi(I)$ if and only if $f(X)\subset V(I)$, then I'm all set.
Unfortunately I'm unable to prove their equivalence. Any help would be much appreciated!
$\mathcal{O}_X(X_h)=0$ iff for all affine opens $U=\operatorname{Spec}B$ of $X$ we have that $h$ is nilpotent in $B$. So we can reduce to the case $X=\operatorname{Spec}B$ affine. if $f(X)\subseteq V(I)$ we get $$X=f^{-1}(f(X))\subseteq f^{-1}(V(I))=V(\varphi(I))$$ The other direction is similar.