# Completion w.r.t. ideal generated by a part of regular system of parameter

Let $$R$$ be a $$d$$-dimensional Noetherian regular local $$k$$-algbera ($$k$$ any field of char($$k$$) = 0, $$d \geq$$ 2). Let $$x, y$$ be a part of regular system of parameter for $$R$$. Let $$I = (x, y)$$ be a ideal generated by $$x$$ and $$y$$ and $$\hat{R}$$ denote the completion of $$R$$ with respect to $$I$$.

Is it true that $$\hat{R} = \frac{R}{I}[[x, y]]$$?

I know the very special case of it is true when $$d = 2$$ and $$I$$ is the maximal ideal. The above statement seems correct but I am not entirely sure. Any help would be great.

• Is the "$n=1$" case known, i.e. whether the $(x)$-adic completion of $R$ is isomorphic to $R/(x)[[s]]$? Commented Aug 27, 2020 at 18:50
• I am not entirely sure. But isn't it follows from $n =1$. Since if you go mod one parameter, we still get regular local ring with less number of regular parameter and then again apply your result which will imply required result. So basically this first case is key step. Commented Aug 28, 2020 at 1:29

Suppose additionally that $$R$$ is essentially of finite type over $$k$$. Then $$\hat{R}$$ contains a subring that maps isomorphically on to $$R/I$$.

Proof: Note that $$R/I$$ is regular and essentially of finite type over $$k$$. Hence it is $$0$$-smooth over $$k$$ (Matsumura, Commutative Ring Theory, Theorem 30.3). We can therefore lift the identity map of $$R/I$$ to a $$k$$-algebra map $$f_2 : R/I \to R/I^2$$. Repeating the argument, we get a $$k$$-algebra map $$f_j : R/I \to R/I^j$$ lifting $$f_{j-1}$$, for every $$j \geq 3$$. Thus we get a map $$R/I \to \hat{R}$$ such that the composite $$R/I \to \hat{R} \to R/I = \hat{R}/I\hat{R}$$ is the identity map of $$R/I$$.

Older answer, left here so that the comments below this make sense.

$$\hat{R}$$ is a flat $$R$$-module (Matsumura, Commutative Ring Theory, Theorem 8.8) but $$\frac{R}{I}[[x,y]]$$ is not, since every non-zero element of $$I$$ is a zero-divisor on it.

• I think the question is whether they are isomorphic as rings, not as $R$-modules.
– abx
Commented Aug 27, 2020 at 16:29
• @Manoj. Yes I am ask whether they are isomorphic as rings. Commented Aug 27, 2020 at 17:39
• Sorry, I misunderstood your question. Are you asking whether $\hat{R}$ contains a subring which maps isomorphically to $R/I$? (As one would have, if $I$ were the maximal ideal.) Commented Aug 27, 2020 at 18:35
• I am asking whether these two rings are isomorphic rings or not and the weaker question is whether $\hat{R}$ contains a subring isomorphic to R/I. Commented Aug 28, 2020 at 1:19