3
$\begingroup$

Let $K$ be a complete field with respect to a discrete valuation $v$ and let $O_K$ be its valuation ring, $m$ its maximal ideal. Suppose $K$ has characteristic $0$ and that $O_K/m$ is of characteristic $p$.

Let $A$ be an $O_K$ algebra and suppose that $A$ is a complete regular local ring of dimension $d+1$. In general, I know that there exists a complicated result (Cohen Structure Theorem, unequal characteristic case) that asserts that $A$ is more or less a homomorphic image of a ring of power series in $d$ variables over a suitable ring of Witt vectors.

My question is: assume that $A\otimes_{O_K} K = K[[X_1, \ldots, X_d]]$ and that the square of the maximal ideal $m_A$ of $A$ does not contain the uniformizer $\pi$ of $O_K$ (in Cohen's language: A is unramified). Can we deduce in this case that $A=O_K[[X_1, \ldots, X_d]]$ (possibly without Cohen's theorem)?

$\endgroup$
1
  • $\begingroup$ Note that $O_K[[X_1,…,X_d]]\otimes_{O_K} K \ne K[[X_1, \dots, X_d]]$. The left hand side consists in powers series whose coefficients have lower bounded valuations. $\endgroup$
    – user18119
    Commented Jun 17, 2012 at 21:08

1 Answer 1

1
$\begingroup$

First notice that $A/\mathfrak m_{A}$ is equal to $\mathcal O_{K}/\pi$ and thus that $A$ is equal to $\mathcal O_{K}+\mathfrak m_A$. Indeed, otherwise your first condition on $A$ would not be true. Because $\pi$ is not in the square of $\mathfrak m_{A}$, there is a system of parameters $(\pi,x_{1},\cdots,x_{d})$ of $\mathfrak m_{A}$ which contains $\pi$.Hence, any element of $A$ can be expanded as a power-series in the $x_{i}$ with coefficients in $\mathcal O_{K}$. Hence $A$ is a quotient of $\mathcal O_K[[x_1,\cdots,x_d]]$. Comparing dimensions give what you want.

All this is treated in details for example in Matsumura Commutative Ring Theory section 29.

$\endgroup$

You must log in to answer this question.

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