Depth two gives Hartogs extension on a punctured affine spectrum

By [F1], both $x$ and $y$ act injectively on $M$. Moreover $y$ acts injectively on $M\mathrm{/}{x}^{r}M$ for every $r\ge 1$: its filtration by the powers of $x$ has successive quotients isomorphic to $M\mathrm{/}xM$, on which $y$ is injective. Thus, inside $M_{xy}$, the intersection $M_x\cap M_y$ is $M$. Indeed if $a\mathrm{/}{x}^r=b\mathrm{/}{y}^t$, injectivity of $xy$ gives $y^{t}a=x^{r}b$ after clearing denominators. Injectivity of $y^t$ on $M\mathrm{/}{x}^{r}M$ gives $a=x^{r}c$, so the common element is $c\in M$.

A section of $\tilde{M}$ on $U$ restricts to elements of $M_x$ and $M_y$ agreeing in $M_{xy}$, hence comes from some $c\in M$ by step 1.1. The difference from the section defined by $c$ vanishes on $D(x)$. On any affine principal open contained in $U$, an element whose localization at $x$ is zero is killed by a power of $x$; injectivity of $x$ on the localized module forces it to vanish. Thus the difference vanishes on all of $U$. Conversely $M\to M_x$ is injective, so restriction is injective. This proves the Hartogs assertion.

Assume $\mathrm{depth}A\ge 2$ for these remaining assertions. Generate $\mathfrak{m}$ by $a_1,\dots ,a_s$, so that $U$ is covered by the finitely many affine opens $D(a_i)$. Sections of $\tilde{M}$ are the kernel of the difference of the two maps ${\prod }_i{M}_{a_i}\rightrightarrows {\prod }_{i,j}{M}_{a_i{a}_j}$. A flat base change commutes with this finite kernel and these localizations. For $M=A$, step 2.1 therefore gives $\mathrm{\Gamma }(U_C,\mathcal{O})=C$. A finite projective $P$ is an idempotent summand of $C^r$, so applying that idempotent to the equality of sections gives $\mathrm{\Gamma }(U_C,\tilde{P})=P$. The module ${\mathrm{Hom}}_C(P,Q)$ for two finite projectives is again finite projective (it is $P^{\ast }{\otimes }_{C}Q$); its sections on $U_C$ are therefore exactly its module elements. This proves full faithfulness.