The Stacks project

In the first sentence of the statement, we can delete the assumption that the Noetherian local $A$ be a domain or that it have dimension $2$ (any dimension is ok). It would be nice to make this improvement. Here is the argument for it, it's pretty close to what is happening in the proof already. (Also note the typo R in the line -4 of the proof, corrected below.)

Let $x_1 \in \mathfrak m$ map to a uniformizer of $R$. Observe that $\kappa(\mathfrak p) = K$ is the fraction field of $R$. Write $\mathfrak p = (x_2,\ldots,x_r)$ with $r$ minimal, so that $\mathfrak m = (x_1, \dotsc, x_r)$. Let $h$ be the dimension of the regular local ring $A_{\mathfrak p}$, so that the dimension of $A$ is $h + 1$. If $r = h + 1$, then we are already done: then $A$ is a Noetherian local ring of dimension $h + 1$ whose maximal ideal $\mathfrak m$ is generated by $h + 1$ elements, to the effect that $A$ is regular. Assume, therefore, that $r > h + 1$.

Since the images of the $x_2, \dotsc, x_r$ generate the maximal ideal $\mathfrak p A_{\mathfrak p}$ of the regular local ring $A_{\mathfrak p}$ of dimension $h$, after renumbering we may assume that $x_2,\ldots,x_{h+1}$ induce a regular system of parameters of $A_{\mathfrak p}$. Then the finite $R$-module $\frac{\mathfrak p}{\mathfrak p^2+(x_2,\ldots,x_{h+1})}$ is annihilated by a power of $x_1$. Thus, for each $i> h+1$, we can find $n_i\geq 0$ and $a_{il}\in A$ such that $$x_1^{n_i}x_i - \sum_{l=2}^{h+1} a_{il}x_l = Q_i,$$ where $Q_i\in\mathfrak p^2$ is a quadratic form in $x_2,\ldots,x_r$ with coefficients in $A$. If $n_i = 0$ for some $i$, then, by the Nakayama lemma, $x_i$ is a redundant generator of $\mathfrak p$ and we get a contradiction to the minimality of $r$. Similarly, if some $a_{il}$ is a unit, then we can remove $x_l$ from the list of generators of $\mathfrak p$ and get the same contradiction. In conclusion, we have $n_i > 0$ and no $a_{il}$ is a unit.

Every summand $a_{il}x_l$ with $a_{il} \in \mathfrak p$ maybe absorbed into $Q_i$. Similarly, for every summand $a_{il}x_l$ with $a_{il} \not\in \mathfrak p$, we have $a_{il}-u_{il}x_1^{m_{il}} \in \mathfrak p$ for some unit $u_{il}\in A^\times$ and $m_{il} > 0$, and this difference multiplied by $x_l$ may be absorbed into $Q_i$. Thus, we may assume that each nonzero $a_{il}$ has the form $a_{il}=u_{il}x_1^{m_{il}}$ with $u_{il}\in A^\times$ and $m_{il} > 0$. We will prove that after blowing the integers $n_i$, $m_{il}$ decrease, which will finish the proof.

Let us see what happens with the equations above on the affine blowup algebra $A' = A[\mathfrak m/x_1]$, on which $x_1$ is a nonzerodivisor. As $\mathfrak m = (x_1, \dotsc, x_r)$, we see that $A'$ is generated over $A$ by $y_i = x_i/x_1$ for $i \ge 2$. The surjection $A \twoheadrightarrow R$ extends to a surjection $A'\twoheadrightarrow R$ with kernel $(y_2, \dotsc, y_r)$ (which contains $\mathfrak p$). Moreover, $y_2,\ldots,y_{h+1}$ still induce a regular system of parameters of $A_{\mathfrak p}$ because $x_1$ is a unit in $A_{\mathfrak p}$. Since $x_i = x_1y_i$ for $i \ge 2$ and $x_1$ is a nonzerodivisor in $A^\prime$, the equation displayed equation above becomes $$x_1^{n_i - 1}y_i - \sum u_{il}x_1^{m_{il} - 1}y_l = Q'_i$$ with $Q'_i \in (y_2, \dotsc, y_r)^2$. In effect, we may replace $A$ by the localization of $A^\prime$ at its maximal ideal $(x_1, y_2, \dotsc, y_r)$ to decrease the $n_i, m_{il}$ as promised, and the proof is complete.

P.S. There is also no need to assume that the DVR $R$ be complete, this assumption is not relevant in the proof.