Number of generators of the maximal ideals in polynomial rings over a field

Hi I'm trying to prove the following

If $K$ is a field (not necessary algebraically closed) then every maximal ideal of $K[x_{1},\dots,x_{n}]$ is generated by exactly $n$ elements.

I know that if $K$ is algebraically closed by Hilbert basis the problem is done. But if $K$ is not algebraic closed, I know that the Krull dimension of $K[x_{1},\dots,x_{n}]$ is $n$ and by the generalized Krull theorem we have that every ideal has height at most $n$, but I'm stucked in this point.

Thank you for any help.

There's no need to use any dimension theory. Rather, the proof is by induction. The case $n = 1$ is well known, so I won't repeat it here.

Now, suppose $\mathfrak{m}$ is a maximal ideal of $A = K [x_1, \ldots, x_n]$. By a generalised Nullstellensatz, we know that $A / \mathfrak{m}$ is a finite field extension of $K$; in particular, the image of $x_n$ in $A / \mathfrak{m}$ has some minimal polynomial $f_n (t)$ over $K$. Clearly, $f_n (x_n) \in \mathfrak{m}$. Let $L = K[x_n] / (f_n (x_n))$. Then $A / \mathfrak{m} \cong L[x_1, \ldots, x_{n-1}] / \mathfrak{n}$ for some maximal ideal $\mathfrak{n}$ of $L [x_1, \ldots, x_{n-1}]$. Using the induction hypothesis, we know that $\mathfrak{n}$ is generated by $n - 1$ elements, say $f_1 (x_1, \ldots, x_n), f_2 (x_2, \ldots, x_n), \ldots, f_{n-1} (x_{n-1}, x_n)$. Then $\mathfrak{m}$ is generated by $f_1 (x_1, \ldots, x_n), f_2 (x_2, \ldots, x_n), \ldots, f_n (x_n)$, because $K [x_1, \ldots, x_n] / (f_n (x_n)) \cong L [x_2, \ldots, x_n]$.