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Commit ae93a456 by rrueger

### 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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