Unequal local rings with equal completions

Apply [F2] to $X=\mathrm{Spec}\mathbb{C}[t]$ and its closed point $t=0$. It gives faithful flatness of $A\to B$ and identifies both maximal-ideal completions with $\mathbb{C}\mathrm{⟦}t\mathrm{⟧}$.

Fix any $R>0$ and put $u_n=R^n\mathrm{/}n!$. Choose $N$ with $N+1\ge 2R$. For every $k\ge 0$, $u_{N+k+1}\mathrm{/}{u}_{N+k}=R\mathrm{/}(N+k+1)\le 1\mathrm{/}2$, so induction gives $u_{N+k}\le u_N{2}^{-k}$. The finite identity ${\sum }_{k=0}^K{2}^{-k}=2-2^{-K}\le 2$ bounds every partial sum of the tail by $2u_N$. Define the explicit finite constant $M_R={\sum }_{n=0}^{N-1}{u}_n+2u_N$, with an empty initial sum interpreted as zero. For $n<N$, $u_n$ is one nonnegative summand of $M_R$; for $n\ge N$, the geometric estimate gives $u_n\le u_N\le M_R$. Thus $R^n\mathrm{/}n!\le M_R$ for every $n$, without passing to an infinite sum; hence for any $0<r<R$, $1\mathrm{/}n!\le M_R{R}^{-n}\le M_R{r}^{-n}$. The coefficient family of $e(t)$ therefore satisfies the exact uniform bound required by [F4] on radius $r$, so [F4] makes its sum holomorphic on $\mathrm{\mid }t\mathrm{\mid }<r$. As $R$ is arbitrary, these disks cover a neighborhood of $0$ (indeed all of $\mathbb{C}$), and $e$ defines a holomorphic germ.

Suppose $e(t)=p(t)\mathrm{/}q(t)$ in $A$, with $q(0)\mathrm{\ne }0$. Then $q(t)e(t)=p(t)$ as germs and evaluation at $0$ gives $p(0)=q(0)\mathrm{\ne }0$. Write $q(t)={\sum }_{j=0}^d{q}_j{t}^j$ with $q_d\mathrm{\ne }0$, and let $m=\mathrm{deg}p$. Equality of germs supplies $\epsilon >0$ such that $qe=p$ on $\mathrm{\mid }t\mathrm{\mid }<\epsilon$. Fix $R>0$ and choose $0<r<\min (R,\epsilon )$. By step 1.2 the coefficients $c_n=1\mathrm{/}n!$ of $e$ satisfy $\mathrm{\mid }{c}_n\mathrm{\mid }\le M_R{r}^{-n}$. The coefficient $b_N={\sum }_{j=0}^{\min (d,N)}{q}_j{c}_{N-j}$ of $qe$ satisfies $\mathrm{\mid }{b}_N\mathrm{\mid }\le M_R{r}^{-N}{\sum }_{j=0}^d\mathrm{\mid }{q}_j\mathrm{\mid }{r}^j$. If $p_N$ denotes the coefficient of $t^N$ in $p$, then $\mathrm{\mid }{p}_N\mathrm{\mid }\le M_p{r}^{-N}$ for all $N$, where $M_p={\max }_{0\le N\le m}\mathrm{\mid }{p}_N\mathrm{\mid }{r}^N$. Thus both series satisfy the common-radius bound in [F3], and their sums agree on $\mathrm{\mid }t\mathrm{\mid }<r$, so [F3] gives $b_N=p_N$ for every $N$. For every integer $N>\max (d,m)$ this yields ${\sum }_{j=0}^d{q}_j\mathrm{/}(N-j)!=0$. Multiplying by $N!$ gives $Q(N)=0$, where $Q(z)={\sum }_{j=0}^d{q}_{j}z(z-1)\cdots (z-j+1)$. This is a nonzero polynomial of degree $d$ with leading coefficient $q_d$. To prove the needed root bound locally, a nonzero constant has no roots. If a degree-$d$ polynomial $P$ with $d>0$ has a root $a$, the finite identities $z^j-a^j=(z-a){\sum }_{\mathrm{\ell }=0}^{j-1}{z}^{j-1-\mathrm{\ell }}{a}^{\mathrm{\ell }}$ give $P(z)=P(z)-P(a)=(z-a)S(z)$, where $S$ has degree $d-1$ and the same nonzero leading coefficient. Every other root $b\mathrm{\ne }a$ satisfies $(b-a)S(b)=0$, hence $S(b)=0$ because the nonzero complex number $b-a$ is invertible. Induction gives at most $d-1$ such roots and at most $d$ roots in total. Applying this bound to $Q$ contradicts its vanishing on the infinitely many distinct integers $N>\max (d,m)$. This contradiction proves that $e(t)$ is not in the image of $A$.

Step 1.1 establishes faithful flatness and equality of completions, while step 2.1 establishes failure of surjectivity.