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I have a problem seeing how the original formulation of Hilbert's 14th Problem is "the same" as the one found on wikipedia. Hopefully someone in here can help me with that. Let me quote Hilbert first:

Es sei eine Anzahl $m$ von ganzen rationalen Funktionen $X_1,\dots,X_m$ der $n$ Variablen $x_1,\dots,x_n$ vorgelegt: \begin{equation}\begin{split}X_1&=f_1(x_1,\dots,x_n)\\&\vdots\\X_m&=f_m(x_1,\dots,x_n).\end{split}\end{equation} (He calls this system of substitutions (S)). Jede ganze rationale Verbindung von $X_1,\dots,X_m$ wird offenbar durch Eintragung dieser Ausdrücke notwendig stets eine ganze rationale Funktion von $x_1,\dots,x_n$. Es kann jedoch sehr wohl gebrochene rationale Funktionen von $X_1,\dots,X_m$ geben, die nach Ausführung jener Substitution (S) zu ganzen Funktionen in $x_1,\dots,x_n$ werden. Eine jede solche rationale Funktion von $X_1,\dots,X_m$, die nach Ausführung der Substitution (S) ganz in $x_1,\dots,x_n$ wird, möchte ich eine relativganze Funktion von $X_1,\dots,X_m$ nennen. Jede ganze Funktion von $X_1,\dots,X_m$ ist offenbar auch relativganz; ferner ist die Summe, die Differenz und das Produkt relativganzer Funktionen stets wiederum relativganz.

Das entstehende Problem ist nun: zu entscheiden, ob es stets möglich ist, ein endliches System von relativganzen Funktionen von $X_1,\dots,X_m$ aufzufinden, durch die sich jede andere relativganze Funktion in ganzer rationaler Weise zusammensetzen läßt.

Okay, unfortunately this is in German. Here's what I made out of it, but can't guarantee it is right:

Start with $f_1,\dots,f_m\in k[x_1,...,x_n]$ for some field $k$. For every $g\in k[X_1,\dots,X_m]$, $g(f_1,\dots,f_m)$ is then a polynomial in the $x_i$. But there can be $\varphi\in k(X_1,\dots,X_m)$ such that $\varphi(f_1,\dots,f_m)\in k[x_1,\dots,x_n]$, which Hilbert calls "relativganz". Every polynomial in the $X_i$ is of course already relativganz, as well as sum, difference, and product of functions which are relativganz.

The problem: Is it always possible to find finitely many r-g functions $g_1,\dots,g_s$ in $k(X_1,\dots,X_m)$ such that every r-g function $\varphi$ can be expressed in the $g_i$ in a polynomial way?

Now the formulation as it is mentioned on wikipedia:

Let $k$ be a field, and $K\subseteq k(x_1,\dots,x_n)$ be a subfield. Is the ring $R=K\cap k[x_1,\dots,x_n]$ finitely generated as a ring?

So, as for the original formulation, we can choose $R$ to be the ring of r-g functions. But what is $K$ in this case? I feel like the choice of $f_1,\dots,f_m$ should correspond one-to-one to the choice of the subfield $K$. Is it possible, given the second formulation, to translate this back to the notion of "relativganz", i.e., can I always interpret $R$ as a ring of r-g functions? Are they really equivalent?

Sorry for the many question in this last paragraph, but I want to really understand how to translate between those two seemingly different formulations :-)

Thank you very much in advance!

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I only corrected some umlauts and hyphens; your paraphrase of the German passage is correct. –  joriki Mar 19 '13 at 15:17
    
@joriki Thank you! I must have become confused with the hyphens, since Hilbert uses both ways himself. My concern with this passage was not the actual translation German->English, since I'm German, but rather if I screwed up in translating the 'content' ;-) –  InvisiblePanda Mar 19 '13 at 15:25

1 Answer 1

up vote 4 down vote accepted

The two formulations are indeed equivalent.

The map $f: k(X_1,...,X_m) \rightarrow k(x_1,...,x_n)$ that sends $X_i$ to $f_i$ is an injection of fields. It identifies $k(X_1,...,X_m)$ with a subfield $K$ of $k(x_1,...,x_n)$. The intersection $R=K\cap k[x_1,...,x_n]$ is then all elements of $k(X_1,..,X_m)$ that are mapped to $k[x_1,...,x_n]$ via $f$, i.e. exactly the relatinganz functions (relative to f).

It's a fact that any subfield $K$ of $k(x_1,...,x_n)$ is finitely generated. To recover Hilbert's formulation, we only need to choose generators $f_1,...,f_m$ for $K$ (over $k$). Note that the notion of being relativganz depends only on the field $K$ generated by the $f_i$, not the individual generators.

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Hello @Prometheus, and thanks for the response. I already forgot about this, because I now knew that I just interpreted the German passage in a wrong way, that is, I was thinking about a map $k(y_1,\dots,y_m)\to k(x_1,\dots,x_n)$ sending $y_i\mapsto X_i=f_i$, which would not be well-defined at all. Instead, Hilbert directly talks about the subfield $k(f_1,\dots,f_m)$ of the rational function field. –  InvisiblePanda Aug 13 '13 at 6:48
    
But I still had/have a slight problem with the equivalence which you could maybe explain to me (even though I already accepted your answer for the sake of completeness ;)): I know that every subfield $K$ of $k(x_1\dots,x_n)$ is f.g., that is, $K=k(\varphi_1,\dots,\varphi_m)$ for $\varphi_i\in k(x)$. But Hilbert takes the $\varphi_i$ (his $f_i$) to be polynomials, not rational functions. I'm pretty sure that this doesn't make any difference after intersecting with the polynomial ring, but the argument doesn't seem trivial to me, since not every $K$ is $k(f_1,\dots,f_m)$ with polynomial $f_i$'s. –  InvisiblePanda Aug 13 '13 at 6:52
1  
Suppose the statement is true when the subfield is generated by polynomials. If $K$ is any subfield of $K(x_1,...,x_n)$, let $K′$ be the field generated by $R=K \cap k[x_1,...,x_n]$. Then $K'\cap k[x_1,...,x_n]=R$, so it's enough to show the statement holds for $K′$. I claim $K′$ is generated by finitely many polynomials in $R$. Since $K′$ is finitely generated over $k$, $K′=k(f′_1,...,f'_l)$. $K′$ is generated by $R$, so each $f′_i$ is a rational function in elements of $R$: $f′_i=g_i/h_i$, with $g_i,h_i \in R$. Then since $R \subset K′$, $K′=k(g_1,...,g_l,h_1,...,h_l)$. –  Prometheus Aug 13 '13 at 8:19

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