Notation: If $X$ is a $K$-scheme, then a point $x\in X$ is said $K$-rational if its residue field $k(x)=\frac{\mathcal O_{X,x}}{\mathfrak m_{X,x}}$ is isomorphic to $K$. The set of all $K$-rational point is denoted by $X(K)$.
Now I have problems with the proof of the following proposition:
Proposition: Let $K$ be a field and consider the ring $A:=\frac{K[T_1,\ldots,T_n]}{\mathfrak a}$ where $\mathfrak a$ is a generic ideal of $K[T_1,\ldots,T_n]$. If $Z(\mathfrak a)=\{\alpha\in K^n\,:\, f(\alpha)=0\;\forall f\in\mathfrak a\}$ and $X=\operatorname {Spec} A$, then we have a bijection $\lambda: Z(\mathfrak a)\longrightarrow X(K)$.
Proof: If $\alpha=(\alpha_1,\ldots,\alpha_n)\in Z(\mathfrak a)$ we consider the maximal ideal of $K[T_1,\ldots,T_n]$ $\mathfrak m_\alpha=(T_1-\alpha_1,\ldots,T_n-\alpha_n)$. Clearly $\mathfrak a\subseteq\mathfrak m_{\alpha}$ so we have a closed point $\overline{\mathfrak m_{\alpha}}$ in $X$; we have defined the function $\lambda:\alpha\mapsto\overline{\mathfrak m_{\alpha}} $ and we must show that $\overline{\mathfrak m_{\alpha}}\in X(K)$. But $\mathcal O_{X,\overline{\mathfrak m_{\alpha}}}=A_{\overline{\mathfrak m_{\alpha}}}$ and the maximal ideal of $A_{\overline{\mathfrak m_{\alpha}}}$ is $\overline{\mathfrak m_{\alpha}} A_{\overline{\mathfrak m_{\alpha}}}$, so:
$$k(\overline{\mathfrak m_{\alpha}})\cong\frac{A_{\overline{\mathfrak m_{\alpha}}}}{\overline{\mathfrak m_{\alpha}} A_{\overline{\mathfrak m_{\alpha}}}}\cong \frac{A}{\overline{\mathfrak m_{\alpha}}}\cong K$$
We conclude that the function $\lambda$ is well defined and clearly injective.
I can't find the inverse of $\lambda$ or equivalently I can't show that $\lambda$ is surjective on $X(K)$. Please help me.
Thanks in advance.
Let $\overline{\mathfrak m}\in X(K)$ be a rational maximal ideal of $X$ coming from the maximal ideal $\mathfrak m\subset K[T_1,\ldots,T_n]$ .
The $K$-rationality hypothesis means exactly that the structural map $s:K\to K[T_1,\ldots,T_n]/\mathfrak m=A/\overline {\mathfrak m}$ is bijective (that structural map is implicitly given by saying that $X$ is a $K$-scheme).
Hence there exists exactly one $\alpha_i\in K$ with image $$s(\alpha_i)=\overline {T_i}\in K[T_1,\ldots,T_n]/\mathfrak m= A/\overline {\mathfrak m}$$ And this gives you the required inverse to $\lambda$, namely: $$\lambda ^{-1}(\overline{\mathfrak m})=(\alpha _1, \ldots,\alpha _i,\ldots,\alpha _n) $$