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.



Maximal Ideals in the ring k[x1,...,xn ]
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 $I$.


