2015-12-4: Difference between revisions
RobertRogo (talk | contribs) Pagină nouă: <math>Problema:</math> Fie <math>K</math> un corp cu <math>m \geq 2</math> elemente si <math>f \in K[X]</math>. Aratati ca urmatoarele afirmatii sunt echivalente: <math>\begin{enumerate}[label=\alph*)] \item Exista <math>g \in K[X]</math> astfel incat <math>f(X)=g(X^{m-1}</math>; \item Pentru orice <math>a \in K^*</math> avem <math>f(X)=f(aX)</math>. \end{enumerate}</math> |
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< | '''<big>Enunț</big>''' Fie <math>K</math> un corp cu <math>m \geq 2</math> elemente și <math>f \in K[X]</math>. Arătați că următoarele afirmații sunt echivalente: | ||
<math>(i)</math> Există <math>g \in K[X]</math> astfel încât <math>f(X)=g(X^{m-1})</math>; | |||
<math>(ii)</math> Pentru orice <math>a \in K^*</math> avem <math>f(X)=f(aX)</math>. | |||
'''<big>Soluție.</big>''' | |||
<math>(i) \rightarrow (ii)</math> | |||
Din teorema lui Lagrange aplicată grupului <math>(K^*,\cdot)</math> avem că <math>x^{m-1}=1, \forall x \in K^*</math>, deci <math display="block">f(aX)=g((aX)^{m-1})=g(a^{m-1}X^{m-1})=g(X^{m-1})=f(X)</math><math>(ii) \rightarrow (i)</math> <math>(Robert \ Rogozsan)</math> | |||
Ne folosim de următorul rezultat | |||
'''Lemă:''' Fie <math>F</math> un corp finit cu <math>n</math> elemente. Atunci <math display="block">\sum_{a \in F}a^k= | |||
\begin{cases} | |||
0, & \text{dacă } n \text{ nu divide pe } k \\ | |||
-1, & \text{dacă } n \text{ divide pe } k | |||
\end{cases}</math> | |||
Latest revision as of 11:36, 3 September 2023
Enunț Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} un corp cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m \geq 2} elemente și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \in K[X]} . Arătați că următoarele afirmații sunt echivalente:
Există astfel încât ;
![{\displaystyle g\in K[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1414a4dff7c6064b41a4f7921cec55ef1c78a934)
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (ii)} Pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in K^*} avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(X)=f(aX)} .
Soluție.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (i) \rightarrow (ii)} Din teorema lui Lagrange aplicată grupului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (K^*,\cdot)} avem că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^{m-1}=1, \forall x \in K^*} , deci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(aX)=g((aX)^{m-1})=g(a^{m-1}X^{m-1})=g(X^{m-1})=f(X)} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (ii) \rightarrow (i)} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (Robert \ Rogozsan)}
Ne folosim de următorul rezultat
Lemă: Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} un corp finit cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} elemente. Atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{a \in F}a^k= \begin{cases} 0, & \text{dacă } n \text{ nu divide pe } k \\ -1, & \text{dacă } n \text{ divide pe } k \end{cases}}