Hi. From one of the forms of Hilbert's Nullstellensatz we know that all the maximal ideals in a polynomial ring $k[x_1, \dots, x_n]$ where $k$ is an algebraically closed field, are of the form $(x_1  a_1, \dots , x_n  a_n)$. So that any maximal ideal in this case is generated by polynomials $g_j \in k[x_1, \dots, x_n]$ for $j = 1, \dots , n$, where $g_j$ only depends on the variable $x_j$ (Obviously by taking $g_j = x_j  a_j$). Now, I'm interested in the case of a polynomial ring $k[x_1, \dots, x_n]$ where $k$ is an arbitrary field (i.e., I can't make use of the Nullstellensatz). I suppose this may no longer be the case, i.e., I don't expect any maximal ideal to be generated by n polynomials, each of them only dependent on one of the variables $x_j$, but my question is if maybe the maximal ideals can be generated by polynomials $g_j$ that only depend on the first $j$ variables $x_1, \dots , x_j$? If so, does anybody know how to prove this or can anyone suggest me some references that may help me?. Thanks.
The stronger version of the Nullstellensatz states that a maximal ideal $I$ of $R=k[x_1,\ldots,x_n]$ is the kernel of a $k$homomorphism from $R$ to $L$ where $L/k$ is a finite extension. Let $a_1,\ldots,a_n$ be the images of $x_1,\ldots,x_n$ under such a homomorphism. Then $a_1$ has a minimal polynomial $m_1$ over $k$. Let $f_1(x_1,\ldots,x_n)=m_1(x_1)$. Then $f_1\in I$
Now $a_2$ has a minimal polynomial $m_2$ over $k(a_1)$. We can write the coefficients of $m_2$ as polynomials in $a_1$ over $k$. Doing this, and replacing $a_1$ by $x_1$ and the free variable by $x_2$ gives a polynomial $f_2$ in $x_1$ and $x_2$. Also $f_2\in I$.
Keep going. We get a sequence of polynomials $f_i$ in $x_1,\ldots,x_i$ and it's not hard to prove these generate $I$.

$\begingroup$ Isn't the $g_1(x_1)$ you wrote in the first step actually $m_1(x_1)$? $\endgroup$ – Adrian BarqueroSanchez May 30 '10 at 21:42

$\begingroup$ Robin, where can I find that version of the Nullstellensatz? $\endgroup$ – Adrian BarqueroSanchez May 30 '10 at 22:42

2$\begingroup$ This is equivalent to the form given in Lang (and reproduced at mathoverflow.net/questions/15226/… ). $\endgroup$ – Harry Gindi May 31 '10 at 6:18

$\begingroup$ It is an easy exercise to prove Robin's statement from that one. $\endgroup$ – Harry Gindi May 31 '10 at 6:26
In response to the OP's request for an explicit reference:
see Theorems 128 and 130 on pages 8283 of

$\begingroup$ Thanks Pete, I couldn't find the part that says the extension is finite. $\endgroup$ – Adrian BarqueroSanchez May 31 '10 at 3:06

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2$\begingroup$ What would the reference be for your updated notes? $\endgroup$ – Andrew Kelley Mar 9 '16 at 20:49