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

Question

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.

Answer 1

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 $R/I$.

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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.