4
$\begingroup$

There's a small theorem in algebra that I've been toying with. Let $R$ be integrally closed in its quotient field, and $E$ a splitting extension of the quotient field, and $S$ be the integral closure of $R$ in $E$. Then for a maximal ideal $p$ of $R$, there are only finitely many maximal ideals of $S$ whose intersection with $R$ is precisely $p$.

I was trying to formulate a small example. Let $f(X)=X^3+X+1$, and $E\supseteq\mathbb{Q}$ a splitting extension for $f$. So taking $R=\mathbb{Z}$ in this case, let $S$ be the integral closure of $\mathbb{Z}$ in $E$. I choose say $(3)$ as a maximal ideal of $\mathbb{Z}$. So there must be finitely many maximal ideals $q$ of $S$ such that $q\cap\mathbb{Z}=(3)$.

This sparked my curiosity, if we know that there are only finitely many such maximal ideals $q$, how many are there actually, at least in this case? I couldn't think of a way to actually enumerate them all to be able to say "There are 9 such maximal ideals!," or something along those lines. Thank you kindly.

  • 0
    For the case of the integers and the rationals, these are well-known. If $r$ is the number of distinct primes lying over $(p)$, and $f$ is the degree of the extension(s) $S/\mathfrak{q}$ over $\mathbb{Z}/(p)$ (where $\mathfrak{q}$ is any prime lying over $(p)$), and $e$ is the largest integer such that $\mathfrak{q}^e$ contains $(p)$ (the "ramification"), then $erf = [E:\mathbb{Q}]$.2012-01-21
  • 0
    Thank you @Arturo, so I would want to solve $e=\frac{[E:\mathbb{Q}]}{rf}=\frac{6}{rf}$? I don't see a way to solve for $r$ and $f$ in general for arbitrary $\mathfrak{q}$.2012-01-21
  • 0
    You would find $f$ by computing the size of $S/\mathbb{q}$ (which will give you the dimension of the extension); the value of $e$ can be found in any number of ways for specific $(p)$ and $E$, but if you have any particular $\mathfrak{q}$ you can simply compute $\mathfrak{q}$, $\mathfrak{q}^2$, etc until you find which one fails to contain $(p)$ (it will be at most $[E:\mathbb{Q}]$). Once you have $e$ and $f$, you can find $r$. I second the advice to take a look at a book on algebraic number theory; Marcus's is very good.2012-01-21

1 Answers 1