Let $\mathcal{o}$ be a Dedekind domain, $K$ its field of fractions, $L$ a finite separable field extension of $K$, and $\mathcal{O}$ the integral closure of $\mathcal{o}$ in $L$. When $\mathfrak{p}$ is a prime ideal in the Dedekind domain $\mathcal{o}$, then one can consider the ideal $\mathfrak{p}\mathcal{O}$ in the Dedekind domain $\mathcal{O}$, and this is sometimes abbreviated as simply $\mathfrak{p}$. What is the relation between $\mathfrak{p}$ and $\mathfrak{p}\mathcal{O}$? Are they in 1-1 correspondence? The latter is usually not a prime ideal in $\mathcal{O}$, right?
Asked
Active
Viewed 96 times
0
-
You are right, the latter is often not a prime ideal. Prime ideals often split in extensions, sometimes they ramify, sometimes they remain prime. You might wish to google "Ramification theory" or some variant thereof to read about such phenomena. – Edward Evans Jul 05 '18 at 16:06
-
Richt, google for "ramification" in algebraic number theory. – Dietrich Burde Jul 05 '18 at 16:06
-
Thank you for the comments. – yhsai Jul 05 '18 at 16:19
-
1I guess $\mathfrak{p}\mathcal{O}\cap\mathcal{o}=\mathfrak{p}$ so $\mathfrak{p}$ and $\mathfrak{p}\mathcal{O}$ are in 1-1 correspondence. Right? – yhsai Jul 05 '18 at 16:26