10
Suppose you have a commutative ring $R$, a square zero ideal $I\subset R$, a $R$-module $M$ and an endomorphism $u$ of $M$ which is the identity modulo $IM$.
Then $v:= 1_M-u$ maps $M$ to $IM$, hence $v^2=0$. Therefore $u=1_M-v$ is invertible — its inverse is $1_M+v$.
4
Since $\vec{u},\vec{v}$ both generate $\mathfrak{m}$, there exist matrices $U,V \in \mathrm{Mat}_{n \times n}(R)$ such that $\vec{u} = U\vec{v}$ and $\vec{v} = V\vec{u}$. Then $(UV-I_{n})\vec{u} = \vec{0}$ so $UV-I_{n}$ has entries contained in $\mathfrak{m}$ since $\vec{u}$ is a regular sequence; thus $UV$ is invertible (because it is $I_{n}$ modulo $\...
4
Here is a counterexample.
Let $S$ be a smooth surface (irreducible). Let $C_1, C_2 \subset S$ be two disjoint smooth curves in $S$ which happen to be isomorphic. Let $X$ be the result of glueing $C_1$ and $C_2$ by this isomorphism. The singular locus of $X$ is a curve $C$ which is mapped onto isomorphically by $C_1$ and $C_2$ via the morphism $S \to X$. On ...
2
The corrected formulation of the K"unneth formula in the Stacks project can also be used to given an answer this question. Let me explain.
Observe that the differentials in the de Rham complex $E \otimes \Omega^\bullet_{X/S}$ of $(E, \nabla_E)$ are differential operators of finite order and similarly for $(F, \nabla_F)$. The construction in Section 0G4A ...
Only top voted, non community-wiki answers of a minimum length are eligible
