Disconnected covers of formal schemes Let $A$ be a Dedekind domain, and $B$ a finite torsion-free generically separable $A[[t]]$-algebra. Suppose the preimage of Spec $B$ over $t = 0$ is disconnected. Must Spec $B$ also be disconnected?
 A: I agree with what has been written. If $\text{Spec}\,B/(t)$ is disconnected then so is $\text{Spec}\,B$, by Hensel's lemma. As has been written, disconnectivity is equivalent to the existence of a solution to$$X^2 - X = 0$$other than $0$ and $1$, and this polynomial has derivative $2x - 1$. If $b$ is a root of that polynomial $\text{mod}$ $t$, then the calculation can be rephrased as$$(2b - 1)^2 = 4b^2 - 4b + 1,$$which is congruent to $1$ $\text{mod}$ $t$; and so $2b - 1$ is a unit $\text{mod}$ $t$.
But maybe it's being asked why this is obvious geometrically. Here's an explanation. The ideals $(b)$ and $(1 - b)$ are relatively prime in $B/(t)$, and their intersection in $B/(t)$ is $(0)$. Geometrically, this says that these ideals define disjoint closed subsets $S$, $T$ of the inverse image of $t = 0$ in $\text{Spec}\,B$, and their union is that full inverse image (this is of course the reason why having a root of $X^2 - X$ corresponds to being disconnected). On $S$ (i.e. in $B/(t, b)$), we have $b = 0$, and so$$2b - 1 = -1,$$which is a unit. On $T$ (i.e. in $B/(t, 1 - b)$), we have $b = 1$, and so $$2b - 1 = 1,$$which is again a unit. Since it's a unit on $S$ and on $T$ (i.e. nonvanishing), it's a unit on the inverse image of $t = 0$. Or to say this algebraically, if in $B/(t)$ it's a unit $\text{mod}$ $(b)$ and also $\text{mod}$ $(1 - b)$, then it's a unit mod their intersection, which is $(0)$; i.e. it's a unit in $B/(t)$.
This is in fact the same computation one might do if instead $A$ were a field.
