Size of the quotient of a Gorenstein ring with an ideal generated by a regular sequence

Question

Let $(R,\mathfrak{m})$ be a local Gorenstein domain of dimension $2$ with finite residue field, let $x_1,x_2$ be a regular sequence in $R$.

Is it true that $R/(x_1,x_2)$ is finite?

Answer 1

I post an answer following the comments.

It is known (see e.g. Corollary 11.18 of Atiyah, MacDonald, "Commutative algebra") that the quotient of a local Noetherian ring with a non zero divisor has dimension equal to the dimension of the ring minus one. Therefore, $R/(x_1,x_2)$ has dimension $0$, hence it is an Artinian local ring. This implies that the descending chain of ideals $\mathfrak{m}\supseteq\mathfrak{m}^2\supseteq\dots$ must stabilize. By Nakayama's lemma, this sequence stabilizes to $0$. Also, all quotients of subsequent elements in the chain are finitely generated vector spaces over $R/\mathfrak{m}$, that is finite field. Therefore we have a finite chain \begin{equation*} R\supseteq\mathfrak{m}\supseteq \mathfrak{m}^2\supseteq\dots\supseteq \mathfrak{m}^e=\{0\} \end{equation*} where all quotients of subsequent elements are finite. This implies that $R$ is finite.