Consider a Noetherian local ring $(R, \mathfrak m, k).$ We will denote by $\widehat -$ the completion of $-$ with respect to the $\mathfrak m$-adic topology, i.e., the topology on $R$ in which a subset $U$ of $R$ is open if and only if for every element $x$ of $U,$ there exists an integer $n \gg 0$ such that $x + \mathfrak m^n \subseteq U.$ Ultimately, I would like to prove that $R$ is Gorenstein if and only if $\widehat R$ is Gorenstein. I am not aware of the standard proof of this, but I would like to use the facts that $$\widehat{\operatorname{Ext}_R^i (k, R)} \cong \operatorname{Ext}_{\widehat R}^i(k, \widehat R)$$ and $\operatorname{Ext}_R^i(k, R)$ is a finite-dimensional $k$-vector space for each integer $i$ whenever $R$ is Gorenstein to show (if possible) that $\dim_k \operatorname{Ext}_R^i (k, R) = \dim_k \operatorname{Ext}_{\widehat R}^i (k, \widehat R)$ for each integer $i,$ from which the statement follows.
Considering that $\dim_k \mathfrak m / \mathfrak m^2 = \dim_k \widehat{\mathfrak m} / \widehat{\mathfrak m}^2,$ I cannot help but wonder if it is true that $k$-vector space dimension is invariant under completion; however, I am not convinced that this is true in general. Indeed, the equality at the beginning of this paragraph is due to the fact that $\mathfrak m / \mathfrak m^2 \cong \widehat{\mathfrak m} / \widehat{\mathfrak m}^2.$ Given a finite dimensional $k$-vector space $V,$ we have that $\widehat V \cong \widehat k \otimes_k V,$ from which it follows that $$\dim_k \widehat V = \dim_k \widehat k \cdot \dim_k V.$$ But understanding the quantity $\dim_k \widehat k$ eludes me. Considering that $k = R / \mathfrak m,$ we have that $\widehat k = \widehat R / \widehat{\mathfrak m},$ but I am not sure what more can be said from this. I have done a few examples, and it appears to be true that $\dim_k \widehat V = \dim_k V,$ but I would appreciate any assistance in proving that this is the case. Thanks in advance.
As $\mathfrak{m}^nk=0$ for all $n>0$, the $\mathfrak{m}$-adic topology on $k$ is discrete, so $k$ is complete. I.e., $k \cong \hat{k}$. This argument extends to show that any module of finite length is complete. As Chris H mentions, one can also appeal to an inverse limit argument.