Since being a G-ring is a property of the local rings it is clear that a localization of a G-ring is a G-ring. Conversely, if every localization at a prime is a G-ring, then the ring is a G-ring. Thus it suffices to show that $S_\mathfrak q$ is a G-ring for every finite type $R$-algebra $S$ and every prime $\mathfrak q$ of $S$. Writing $S$ as a quotient of $R[x_1, \ldots, x_n]$ we see from Lemma \ref{lemma-G-ring-goes-up-quasi-finite} that it suffices to prove that $R[x_1, \ldots, x_n]$ is a G-ring. By induction on $n$ it suffices to prove that $R[x]$ is a G-ring. Let $\mathfrak q \subset R[x]$ be a maximal ideal. By Lemma \ref{lemma-check-G-ring-maximal-ideals} it suffices to show that $$ R[x]_\mathfrak q \longrightarrow R[x]_\mathfrak q^\wedge $$ is regular. If $\mathfrak q$ lies over $\mathfrak p \subset R$, then we may replace $R$ by $R_\mathfrak p$. Hence we may assume that $R$ is a Noetherian local G-ring with maximal ideal $\mathfrak m$ and that $\mathfrak q \subset R[x]$ lies over $\mathfrak m$. Note that there is a unique prime $\mathfrak q' \subset R^\wedge[x]$ lying over $\mathfrak q$. Consider the diagram $$ \xymatrix{ R[x]_\mathfrak q^\wedge \ar[r] & (R^\wedge[x]_{\mathfrak q'})^\wedge \\ R[x]_\mathfrak q \ar[r] \ar[u] & R^\wedge[x]_{\mathfrak q'} \ar[u] } $$ Since $R$ is a G-ring the lower horizontal arrow is regular (as a localization of a base change of the regular ring map $R \to R^\wedge$). Suppose we can prove the right vertical arrow is regular. Then it follows that the composition $R[x]_\mathfrak q \to (R^\wedge[x]_{\mathfrak q'})^\wedge$ is regular, and hence the left vertical arrow is regular by Lemma \ref{lemma-regular-permanence}. Hence we see that we may assume $R$ is a Noetherian complete local ring and $\mathfrak q$ a prime lying over the maximal ideal of $R$. \medskip\noindent Let $R$ be a Noetherian complete local ring and let $\mathfrak q \subset R[x]$ be a maximal ideal lying over the maximal ideal of $R$. Let $\mathfrak r \subset \mathfrak q$ be a prime ideal. We want to show that $R[x]_\mathfrak q^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is a geometrically regular algebra over $\kappa(\mathfrak r)$. Set $\mathfrak p = R \cap \mathfrak r$. Then we can replace $R$ by $R/\mathfrak p$ and $\mathfrak q$ and $\mathfrak r$ by their images in $R/\mathfrak p[x]$, see Lemma \ref{lemma-check-G-ring-easy}. Hence we may assume that $R$ is a domain and that $\mathfrak r \cap R = (0)$. \medskip\noindent By  Algebra, Lemma \ref{algebra-lemma-complete-local-Noetherian-domain-finite-over-regular} we can find $R_0 \subset R$ which is regular and such that $R$ is finite over $R_0$. Applying Lemma \ref{lemma-G-ring-goes-up-quasi-finite} we see that it suffices to prove $R[x]_\mathfrak q^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is geometrically regular over $\kappa(r)$ when, in addition to the above, $R$ is a regular complete local ring. \medskip\noindent Now $R$ is a regular complete local ring, we have $\mathfrak q \subset \mathfrak r \subset R[x]$, we have $(0) = R \cap \mathfrak r$ and $\mathfrak q$ is a maximal ideal lying over the maximal ideal of $R$. Since $R$ is regular the ring $R[x]$ is regular (Algebra, Lemma \ref{algebra-lemma-regular-goes-up}). Hence the localization $R[x]_\mathfrak q$ is regular. Hence the completions $R[x]_\mathfrak q^\wedge$ are regular, see Lemma \ref{lemma-completion-regular}. Hence the fibre $R[x]_{\mathfrak q}^\wedge \otimes_{R[x]} \kappa(\mathfrak r)$ is, as a localization of $R[x]_\mathfrak q^\wedge$, also regular. Thus we are done if the characteristic of the fraction field of $R$ is $0$. \medskip\noindent If the characteristic of $R$ is positive, then $R = k[[x_1, \ldots, x_n]]$. In this case we split the argument in two subcases: \begin{enumerate} \item The case $\mathfrak r = (0)$. The result is a direct consequence of Lemma \ref{lemma-helper-G-ring}. \item The case $\mathfrak r \not = (0)$. This is Lemma \ref{lemma-another-helper-G-ring}. \end{enumerate}