Let $f:X \rightarrow Y = \operatorname{Spec}A$ be a morphism and $Y = \bigcup U_\alpha$ where $U_\alpha = \operatorname{Spec}A_\alpha$. Given the ideal $I = \{a\in A: f^*(a) = 0\}$, show that $I \otimes _A A_\alpha \cong I_\alpha$ where $I_\alpha := I|_{U_\alpha}$ = {$a \in A_\alpha : f^*(a) = 0$}
I was proving a different problem and got stuck on this. This seems fine if the covering is a covering of principal opens, but I don't see why it should hold for an arbitrary affine open cover.
Here is a partial answer, provided that $f$ is quasi-compact (which your comment indicates is the case in the situation you are interested in). Your ideal $I$ is the module of global sections of the kernel of $\mathcal O_Y\to f_\ast \mathcal O_X$. Let's denote this sheaf by $\mathcal I$. As $f$ is quasi-compact, this is a quasi-coherent sheaf. Then restricting the morphism $f$ to $f^{-1}U_\alpha$ corresponds to restricting $\mathcal O_Y\to f_\ast \mathcal O_X$ to $U_\alpha$, and the kernel of the restriction is the restriction of the kernel. But on affines pullback of quasi-coherent modules is given by tensoring, that is we have $\mathcal I\vert_{U_\alpha}\cong (I\otimes_A A_\alpha)^\sim$, which proves the claim.