I am studying chapter 1 of Gathmann's notes about algebraic geometry and there is something that I don't understand in the proof of proposition 1.2.4, where Hilbert's Nullstellensatz is proved for $\mathbb{C}$. The part that I don't understand is as follows.
Let $\mathfrak{m}$ be a maximal ideal of $\mathbb{C}[x_1,x_2,\cdots,x_n]$. Since $\mathbb{C}[x_1,x_2,\cdots,x_n]$ is Noetherian, we may assume that $\mathfrak{m}=(f_1,\cdots,f_r)$, where $f_i \in \mathbb{C}[x_1,x_2,\cdots,x_n]$. Let K be the subfield of $\mathbb{C}$ obtained by adjointing to $\mathbb{Q}$ all coefficients of the $f_i$. Let $\mathfrak{m}_{0}=\mathfrak{m} \cap K[x_1,x_2,\cdots,x_n]$. Note that then $\mathfrak{m}=\mathfrak{m}_{0} \cdot \mathbb{C}[x_1,x_2,\cdots,x_n]$.
Next we are going to prove that $\mathfrak{m}_{0}$ is a maximal ideal of $K[x_1,x_2,\cdots,x_n]$. Suppose not, then let $\mathfrak{m}_{0} \subsetneq \mathfrak{m}_{0}^{\prime} \subsetneq K[x_1,x_2,\cdots,x_n]$. Let $\mathfrak{m}^{\prime}=\mathfrak{m}_{0}^{\prime} \cdot \mathbb{C}[x_1,x_2,\cdots,x_n]$. The author then induces a contradiction $\mathfrak{m} \subsetneq \mathfrak{m}^{\prime} \subsetneq \mathbb{C}[x_1,x_2,\cdots,x_n]$ by multiplying $\mathbb{C}[x_1,x_2,\cdots,x_n]$ on $\mathfrak{m}_{0} \subsetneq \mathfrak{m}_{0}^{\prime} \subsetneq K[x_1,x_2,\cdots,x_n]$, which I don't understand.
I can prove $\mathfrak{m} \subsetneq \mathfrak{m}^{\prime}$ by the following way:
It can be verified that $\mathfrak{m} \subseteq \mathfrak{m}^{\prime}$, so it remains to be proved that $\mathfrak{m} \neq \mathfrak{m}^{\prime}$. Suppose not, then $\mathfrak{m} = \mathfrak{m}^{\prime}$. Let $g \in \mathfrak{m}_{0}^{\prime} - \mathfrak{m}_{0}$, by definition $g \in \mathfrak{m}^{\prime}=\mathfrak{m}$ and $g \in K[x_1,x_2,\cdots,x_n]$, so $g \in \mathfrak{m}_{0}$ and we obtain a contradiction.
But the same method becomes invalid when I try to prove $ \mathfrak{m}^{\prime} \subsetneq \mathbb{C}[x_1,x_2,\cdots,x_n]$. Any help?
