Let $R$ be a reduced finite type $\bar{k}$-algebra, a projective morphism $\pi \colon V \rightarrow \mathrm{Spec}(R)$ and ideals $I, J \subseteq R$. Assume there is a split $s_{IJ} \colon \mathrm{Spec}(R/IJ) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(R/IJ)$ of $\pi$ and that the sections $s_{J} \colon \mathrm{Spec}(R/J) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(R/J)$, $s_{I} \colon \mathrm{Spec}(R/I) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(R/I)$ and $s_{I+J} \colon \mathrm{Spec}(R/I+J) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(R/I+J)$ lift to sections $\hat{s}_{J} \colon \mathrm{Spec}(\hat{R}_J) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(\hat{R}_J)$, $\hat{s}_{I} \colon \mathrm{Spec}(\hat{R}_I) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(\hat{R}_I)$ and $\hat{s}_{I+J} \colon \mathrm{Spec}(\hat{R}_{I+J}) \rightarrow V \times_{\mathrm{Spec}(R)} \mathrm{Spec}(\hat{R}_{I+J})$ on the completions. Does then $s_{IJ}$ lift to a morphism $\hat{s}_{IJ}$? Or shorter said: given a morphism such that every section over an irreducible closed affine subscheme lifts to a section on its completion, does then every section over an arbitrary closed affine subscheme lift to a section on its completion?

## 1 Answer

No. Take $R=k[x, y]$ and $I = (x)$ and $J = (y)$. Then let $V$ be the glueing of two copies of $\text{Spec}(R)$ at $0 = (0, 0)$. Let $s_{IJ}$ be the morphism which sends $V(I)$ into the first copy and $V(J)$ into the second. Then $\hat s_{IJ}$ does not exist but $\hat s_I$ and $\hat s_J$ do (and therefore also $\hat s_{I + J}$).

To make a positive statement you need to assume something better than just assuming that $\hat s_{I + J}$ exists (which is automatic from the existence of either $\hat s_I$ or $\hat s_J$), but it seems a bit hard to say exactly what it is.