LSU Historical Dissertations and Theses

1996

Dissertation

Degree Name

Doctor of Philosophy (PhD)

Mathematics

Robert Perlis

Abstract

The main theorem in this dissertation provides a partial answer to the following question: Given a $\doubz\sb{p}$-extension $F{\sb\infty} /F$ and a finite extension $K/F$, where F is a number field and p a prime number, to what extent does the K-splitting behavior the prime ideals of F determine the Iwasawa invariants of the $\doubz\sb{p}$-extension $K{\cdot}F{\sb\infty}/K$. The answer is that if two fields K and L are arithmetically equivalent over F, then $K{\cdot}F{\sb\infty} /K$ and $L{\cdot}F{\sb\infty} /L$ have exactly the same Iwasawa invariants for any $\doubz\sb{p}$-extension $F{\sb\infty} /F$, so long as p is not an exceptional divisor for K and L over F. The exceptional divisors are a subset of the primes dividing the degree $\lbrack N : K\rbrack$, where N is the normal closure of K over F. The definition is found in Chapter 2. This theorem comes as a corollary of a theorem concerning representations of finite groups which has importance in its own right. Its statement is: If H, $H\sp\prime$, and B are subgroups of a finite group G, with B normal in G, and if the $\doubz\sb{q}\lbrack G\rbrack$-modules $\doubz\sb{q}\lbrack G/H\rbrack$ and $\doubz\sb{q}\lbrack G/H\sp\prime\rbrack$ are isomorphic, then the $\doubz\sb{q}\lbrack G\rbrack$-modules $\doubz\sb{q}\lbrack G/H\ \cap\ B\rbrack$ and $\doubz\sb{q}\lbrack G/H\sp\prime\ \cap\ B\rbrack$ are also isomorphic. This result is not constructive. Given a specific $\doubz\sb{q}\lbrack G\rbrack$-map M between $\doubz\sb{q}\lbrack G/H\rbrack$ and $\doubz\sb{q}\lbrack G/H\sp\prime\rbrack$, it does not tell us how to obtain a $\doubz\sb{q}\lbrack G\rbrack$-map between $\doubz\sb{q}\lbrack G/H\ \cap\ B\rbrack$ and $\doubz\sb{q}\lbrack G/H\sp\prime\ \cap\ B\rbrack$ which will be an isomorphism whenever M is. Chapter 4 deals with this question, with some partial results for certain choices of G, in particular for the case when $G/B$ is abelian and HB = $H\sp\prime B$.

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