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Let $k$ be an algebraically closed field of characteristic $p$ and $A=k[x_1,\cdots,x_n]$ the polynomial ring over $k$ in $n$ variables.

Given a prime ideal $\mathfrak{p}$ in $A$, denote by $A_\mathfrak{p}$ the localization in $\mathfrak{p}$ and also denote by $A(\mathfrak{p})$ the residue field of the local ring $A_\mathfrak{p}$, i.e. $A(\mathfrak{p})=A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$.

Is there a way to describe explicitly $A(\mathfrak{p})$? What about the case where $\mathfrak p$ is maximal?

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If $\mathfrak{p}$ is maximal then the residue field is $A/\mathfrak{p}$ which is $k$ by the (weak) Nullstellensatz. Have you tried any examples, e.g. $A = k[x, y], \mathfrak{p} = (x)$? –  Qiaochu Yuan Aug 6 '12 at 19:23
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What form of explicit description do you have in mind?

If you let $n$ vary, any finitely generated field of extension of $k$ can occur as an $A(\mathfrak p)$, as is easily seen, and so classifying the possible $A(\mathfrak p)$ is the same as classifying the possible f.g. field extensions of $k$, which is essentially the same thing as classifying algebraic varieties over $k$ up to birational equivalence, a fairly difficult problem.

One thing you can say is that the transcendence degree of $A(\mathfrak p)$ over $k$ is the equal to the Krull dimension of $A/\mathfrak p$, which in turn is equal to $n - \mathrm{height} \, \mathfrak p$.


Let's consider the case $n = 2$, as an example. Then $\mathfrak p$ is either maximal (equivalently, $A(\mathfrak p) = k$), the zero ideal (equivalently, $A(\mathfrak p) = k(x_1,x_2)$), or height one (equivalently, $\mathfrak p$ is principal). In the height one case, the field $A(\mathfrak p)$ is the function field of an algebraic curve over $k$.

Any f.g. field extension of $k$ of transcendence dim'n $1$ can arise as an $A(\mathfrak p)$, and for a fixed such extension $K$, the classification of the possible $\mathfrak p$ for which $A(\mathfrak p) \cong K$ is equivalent to the classification of the (possibly singular) plane models of the curve $C$ corresponding to $K$.

The classifiation of the different possible $K$ is the problem of classifying all curves over $k$, which is an elaborate theory (involving the concept of genus of a curve, the moduli space of curves of a given genus, and so on).

Just to be even more specific, if $\mathfrak p = (f)$ with $f$ of degree one or two, then $A(\mathfrak p) \cong k(x).$ But in fact there are irreducible $f$ of every degree for which $A\bigl((f)\bigr) \cong k(x)$, although for most $f$ of degree $> 2$, the field $A\bigl( (f)\bigr)$ will not be isomorphic to $k(x)$. (E.g. if $f$ is of degree three, then the condition for $A\bigl((f)\bigr)$ to be isomorphic to $k(x)$ is that the homogenization of $f$ describe a singular cubic.)

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For instance $k[t]_{(t-a)}/(t-a)k[t]_{(t-a)} \cong k$ and I refer to this as an explicit description (maybe I am using a wrong word here). –  Matt Elly Aug 6 '12 at 19:40
    
Suppose that $n$ is fixed. If $\mathfrak{p}$ and $\mathfrak{p'}$ are prime ideals (or maximal) of $A$, is there any relation between $A(\mathfrak{p})$ and $A(\mathfrak{p'})$? –  Matt Elly Aug 6 '12 at 19:43
    
@MattElly: Dear Matt, As Qiaochu noted in a comment above, as as follows from my more general remark about dimensions, $\mathfrak p$ is maximal if and only if $A(\mathfrak p) = k$. Surely that answers the question in your comment? Regards, –  Matt E Aug 6 '12 at 19:47
    
Ah, ok. So there is an equivalence in the case of maximal ideals which is: $\mathfrak{p}$ is maximal if, and only if, $A(\mathfrak{p})=A/\mathfrak{p}=k$? In this case I understood that for maximal ideals my question is answered. However, my question in the second comment above for prime ideals is also answered? –  Matt Elly Aug 6 '12 at 19:55
    
@MattElly: Dear Matt, What is the question for prime ideals that are not maximal? Regards, –  Matt E Aug 6 '12 at 23:08
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