Example 3.32 - Algebraic Geometry and Atithmetic Curves, of Qing Liu Can anyone explain me why $k(x)=k[T_1,\ldots,T_n]/\mathfrak{m}_\alpha$?

 A: The residue field at a point $x\in X$ is by definition
$$
k(x) := \mathcal{O}_{X,x}/\mathfrak{m}_x,
$$
where $\mathcal{O}_{X,x}$ is the stalk of the structure sheaf of $X$ at $x$, and $\mathfrak{m}_x$ is its maximal ideal. If you choose an affine open $\operatorname{Spec} A\subseteq X$ containing $x$, then $k(x)$ may be computed in the following way: the point $x$ corresponds to a prime ideal $\mathfrak{p}\subseteq A$, and we have
$$
\mathcal{O}_{X,x} \cong\mathcal{O}_{\operatorname{Spec}A,\mathfrak{p}}\cong A_\mathfrak{p},
$$
where $A_\mathfrak{p}$ is the localization of $A$ at $\mathfrak{p}$. This has maximal ideal $\mathfrak{p}A_\mathfrak{p}$.
In your case, $X$ is already affine, so we can take all of $X$ to be the affine open $\operatorname{Spec}A$ containing $x$. Now, $x$ is the point in $X$ corresponding to the maximal ideal $\mathfrak{m}_\alpha = (T_1 - \alpha_1,\dots, T_n - \alpha_n)$, so we have
$$
k(x) = \mathcal{O}_{X,x}/\mathfrak{m}_x\cong \left(k[T_1,\dots, T_n]/I\right)_{\mathfrak{m}_\alpha}/\mathfrak{m}_\alpha\left(k[T_1,\dots, T_n]/I\right)_{\mathfrak{m}_\alpha}.
$$
Localization commutes with taking quotients, so we have
\begin{align*}
k(x) &\cong \left(k[T_1,\dots, T_n]/I\right)_{\mathfrak{m}_\alpha}/\mathfrak{m}_\alpha\left(k[T_1,\dots, T_n]/I\right)_{\mathfrak{m}_\alpha}\\
&\cong \left(k[T_1,\dots, T_n]_{\mathfrak{m}_\alpha}/I_{\mathfrak{m}_\alpha}\right)/\mathfrak{m}_\alpha\left(k[T_1,\dots, T_n]_{\mathfrak{m}_\alpha}/I_{\mathfrak{m}_\alpha}\right)\\
&\cong k[T_1,\dots, T_n]_{\mathfrak{m}_\alpha}/\mathfrak{m}_\alpha k[T_1,\dots, T_n]_{\mathfrak{m}_\alpha}\qquad\textrm{(3rd ring isomorphism theorem)}\\
&\cong \left(k[T_1,\dots, T_n]/\mathfrak{m}_\alpha k[T_1,\dots, T_n]\right)_{\mathfrak{m}_\alpha}\\
&\cong k[T_1,\dots, T_n]/\mathfrak{m}_\alpha\cong k.
\end{align*}
