Corollary 4.4.6 Let $R$ be a f.g. $k$-algebra, $k$ field, and $I$ and ideal of $R$. Then $\sqrt{I}$ is the intersection of all maximal ideals of $R$ containing $I$.
Proof $\quad$ We may assume that $I=0$. Clearly $\sqrt{0}$ is contained in every maximal ideal of $R$. Conversely, suppose $f \in R - \sqrt{0}$. Then $R_f \neq 0$. Let $\mathfrak{n}$ be a maximal ideal of $R_f$ and $\mathfrak{m} = \mathfrak{n} \cap R$. Now, $R_f \big / \mathfrak{n}$ is a f.g. $k$-algebra and it is a field. Hence it is algebraic over $k$. But $R_f \big / \mathfrak{n} \simeq \left ( R \big / \mathfrak{m} \right )_f$. Hence $R \big / \mathfrak{m}$ is integral over $k$. This implies that $R \big / \mathfrak{m}$ is a field, i.e., $\mathfrak{m}$ is a maximal ideal of $R$. Since $f \not \in \mathfrak{m}$, the intersection of all maximal ideals of $R$ is contained in $\sqrt{0}$. $\quad \square$
Here I am not getting, how does it follow $R_f/n\simeq (R/m)_f ?$ From theory of localisation, I know that, $R_f/m_f\simeq (R/m)_f$, but why $m_f=n ?$
Can anyone help me? Thanks in advance.