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authorAdrian Kummerlaender2018-03-01 20:30:48 +0100
committerAdrian Kummerlaender2018-03-01 20:30:48 +0100
commitd8181f09290e69043185ed63e4ad6ab8e74869ab (patch)
treef54e024f0148278ab14ccb2e2420b499ed835614 /content/eaz.tex
parentb69625a5fb01e835a754aad84709ae1de3135b69 (diff)
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Update non-inline math environment syntax
Diffstat (limited to 'content/eaz.tex')
-rw-r--r--content/eaz.tex10
1 files changed, 5 insertions, 5 deletions
diff --git a/content/eaz.tex b/content/eaz.tex
index 010c4a8..ba08f7d 100644
--- a/content/eaz.tex
+++ b/content/eaz.tex
@@ -488,11 +488,11 @@ Das Bild von $F^\times \to F^\times, b \mapsto b^2$ ist Quadratmenge.
Sei $p \geq 3$ Primzahl. Für $a \in \Z$ ist def.:
\vspace*{-2mm}
-$$\legendre{a}{p} = \begin{cases}
+\[ \legendre{a}{p} = \begin{cases}
0 & p | a \\
1 & \exists x \in \Z \setminus p\Z : a \equiv x^2 \ (mod \ p) \\
-1 & \text{sonst}
-\end{cases}$$
+\end{cases} \]
$\legendre{a}{p}$ ist das \emph{Legendre-Symbol} von $a$ modulo $p$.
@@ -503,10 +503,10 @@ $\legendre{a}{p}$ ist das \emph{Legendre-Symbol} von $a$ modulo $p$.
Sei $a \in \Z, m, n \in \Z : a=mn, p \in \Primes$:
\vspace*{-2mm}
-$$\begin{array}{ll}
+\[ \begin{array}{ll}
\legendre{a}{p} = \legendre{a-p}{p} & \legendre{m \cdot n}{p} = \legendre{m}{p}\legendre{n}{p} \\
\legendre{2}{p} = (-1)^{\frac{p^2-1}{8}} & \legendre{-1}{p} = (-1)^{\frac{p-1}{2}}
-\end{array}$$
+\end{array} \]
Sei $l, p \in \Primes$ mit $l, p \neq 2$:
@@ -578,7 +578,7 @@ Sei $R$ ein solcher Hauptidealring. Dann:
Seien $R$ Hauptidealring, $r, s \in R$ teilerfremd (d.h. $1=rx+sy$ für geeignete $x, y \in R$). Dann gilt für Ideale $I = Rr, J = Rs$ der Chinesische Restsatz s.d.:
\vspace*{-2mm}
-$$R/(Rrs) \cong R/(Rr) \times R/(Rs)$$
+\[ R/(Rrs) \cong R/(Rr) \times R/(Rs) \]
$\forall a, b \in R \exists x \in R : x \equiv a \ (mod \ Rr) \land x \equiv b \ (mod \ Rs)$