I want to prove that although $K$ is a field that IS NOT algebraically closed, every maximal ideal in $K[x_1, \ldots, x_n]$ can be generated by $n$ elements.
To prove this, I am following the next steps:
i) Every maximal ideal in $K[x]$ is generated exactly by ONE element.
Induction Hypothesis: Every maximal ideal in $K[x_1, \ldots, x_{n-1}]$ can be generated by $n-1$ elements.
ii) Let $\mathfrak{m}$ be a maximal ideal in $K[x_1, \ldots, x_n]$ and let $\mathfrak{p} = \mathfrak{m} \cap K[x_1, \ldots, x_{n-1}] $. I have considered the extensions:
$$K \subset {K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}} \subset K[x_1, \ldots, x_n] / \mathfrak{m}.$$
I have proved that $K[x_1, \ldots, x_n] / \mathfrak{m}$ is an algebraic extension of $K$ and $\mathfrak{p}$ is a maximal ideal that is generated by $n-1$ elements.
Now I have to consider the extension:
${K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}} \subset ({K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}})[x_n]$ and to find a generator of $\mathfrak{m} / \mathfrak{p} = \overline{\mathfrak{m}} \subset ({K[x_1, \ldots, x_{n-1}]} / {\mathfrak{p}})[x_n] $
but I don't know how to do that. Then I have to conclude that $\mathfrak{m}$ can be generated by $n$ elements. I would appreciate some help.
Thank you very much.