Lifting an idempotent recovers a product decomposition

For every $n\ge 1$ one has $I^n=(t^n)k\mathrm{⟦}t\mathrm{⟧}\times (t^n)k\mathrm{⟦}t\mathrm{⟧},$ so $A\mathrm{/}{I}^n\cong k\mathrm{⟦}t\mathrm{⟧}\mathrm{/}(t^n)\times k\mathrm{⟦}t\mathrm{⟧}\mathrm{/}(t^n)\mathrm{.}$ Thus $A$ is the inverse limit of the quotients $A\mathrm{/}{I}^n$ coordinatewise, and ${\bigcap }_{n\ge 1}{I}^n=0\times 0,$ again coordinatewise. Hence $A$ is $I$-adically complete and separated, so [L1] shows that $(A,I)$ is Henselian.

The residue ring is $A\mathrm{/}I\cong k\times k$, and $(1,0)$ is an idempotent there. The obvious lift in $A$ is $e=(1,0)$, which indeed satisfies $e^2=e$. By [L2], that lift is the unique idempotent of $A$ reducing to $(1,0)$. Its complement is $1-e=(0,1)$.

Multiplication by $e$ and $1-e$ gives $A=eA\times (1-e)A=(k\mathrm{⟦}t\mathrm{⟧}\times 0)\times (0\times k\mathrm{⟦}t\mathrm{⟧})\mathrm{.}$ Thus the lifted idempotent exactly recovers the original product decomposition.