I read Atiyah MacDonald Introduction to commutative algebra. In this book at proposition 10.22 (ii) says that in its proof
ii) $\mathfrak{a}^n / \mathfrak{a}^{n+1} \cong \widehat{\mathfrak{a}}^n / \widehat{\mathfrak{a}}^{n+1}$ by (10.15) iii)
I know $\mathfrak{a}^n / \mathfrak{a}^{n+1} \cong \widehat{\mathfrak{a}}^n / \widehat{\mathfrak{a}}^{n+1}$. But I don't know why we can prove $G_{\mathfrak{a}}(A) \cong G_{\widehat{\mathfrak{a}}}(\widehat{A})$ (as graded ring) by using $\mathfrak{a}^n / \mathfrak{a}^{n+1} \cong \widehat{\mathfrak{a}}^n / \widehat{\mathfrak{a}}^{n+1}$.
I think the following is true.
Let $A = \bigoplus_{n=0}^{\infty}A_n$ and $B = \bigoplus_{n=0}^{\infty}B_n$ be graded rings. And there exists an isomorphism $\varphi_n: A_n \rightarrow B_n$ each $n=0, 1, 2, \cdots$.But let $\varphi_0$ be a ring isomorphism and, for $n \geq 1$, $\varphi_n$ be a group isomorphism. Then, the following $\varphi$ is a ring isomorphism.
$\varphi : A \rightarrow B : (a_n) \mapsto (\varphi_n(a_n))$
First, I don't know if this is true.
Second, if this is true, because I don't know $A / \widehat{\mathfrak{a}} \cong \widehat{A} / \widehat{\mathfrak{a}}$ ("as ring"), I cannot prove prop 10.22 (ii) to use the above.