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Unverified Commit ae93a456 authored by rrueger's avatar rrueger
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Algebra II: Lecture 11. Update definition of field characteristic

parent a9214d8a
......@@ -21,12 +21,12 @@ which situations such an $\hat{E}$ exists.
\begin{siderules}
\begin{definition}[Reminder Algebra I, Characteristic]
The characteristic of a field, $\chi$, is given by the number of times the
The characteristic, $p$, of a field is given by the number of times the
multiplicative neutral element must be added to itself before we arrive at
the additive neutral element. If no such $\chi$ exists, we say the field has
the additive neutral element. If no such $p$ exists, we say the field has
characteristic 0.
\begin{align*}
\underset{\chi \text{~times}}{\underbrace{1+1+1+\cdots}} = 0
\underset{p \text{~times}}{\underbrace{1+1+1+\cdots}} = 0
\end{align*}
\end{definition}
......@@ -36,6 +36,26 @@ which situations such an $\hat{E}$ exists.
\item The characteristics of $\R, \C, \Q$ are $0$.
\end{itemize}
\end{example}
This definition becomes more natural when considering it in the following
manner: let $\chi \colon \Z \to k$ be a homomorphism, mapping $n \mapsto nl$
for $l$ some unit of $k$. Then $\ker (\chi) = (p)$ for $p$ zero or a prime
(This is a result from Algebra I. It is a direct consequence of $\Z$ being a
PID and $k$ an integral domain). If $p \neq 0$, then $p$ is clearly the
characteristic as defined before, since if $x \in (p) = \ker (\chi)$, then $x
= pl$ as $(p)$ is a prime ideal. Then
%
\begin{align*}
pl
= 0
= pll^{-1}
= p1
= \underset{p \text{~times}}{\underbrace{1+1+1+\cdots}}
\end{align*}
%
Assigning a field characteristic zero, when no such $p$ exists at first seemed
somewhat arbitrary, but now is quite canonical, as it simply is the $p$ that
generates the ideal that is $\ker (\chi)$.
\end{siderules}
\begin{theorem}
......
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