1
$\begingroup$

Let $R$ be a Noetherian regular local ring of dimension $n$ with maximal ideal $\mathfrak{m}$. Given two systems of regular parameters $\vec{u}=\left<u_1, \dots, u_n\right>$ and $\vec{v}=\left<v_1, \dots, v_n\right>$ of $R$, does there exist an element $M$ in $\mathrm{GL}_n(R)$ which takes $\vec{u}$ to $\vec{v}$?

Any comments on this question are most appreciated!

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $\mathfrak{m}$), hence both $U,V$ are invertible.

$\endgroup$
1
  • $\begingroup$ Thanks. I see that if $\det(UV)\in \mathfrak{m}$, then $\det (UV \mod \mathfrak{m})=0$, but $UV$ is invertible $\mod \mathfrak{m}$. So $\det(UV)$ is a unit in $R$. Thanks again! $\endgroup$ Commented May 6, 2021 at 9:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .