2015-12-4: Difference between revisions

Revision as of 16:27, 2 September 2023

$Problema:$ Fie $K$ un corp cu $m\geq 2$ elemente si $f\in K[X]$ . Aratati ca urmatoarele afirmatii sunt echivalente:

(i) Exista $g\in K[X]$ astfel incat $f(X)=g(X^{m-1})$ ;

(ii) Pentru orice $a\in K^{*}$ 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)} .

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 Solutie} 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 aplicata 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 ca 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)} .

@@ Line 7: / Line 7: @@
 <math>Solutie</math>
 <math>(i) \rightarrow (ii)</math>
-Din teorema lui Lagrange aplicata grupului <math>(K^*,\cdot)</math> avem ca
+Din teorema lui Lagrange aplicata grupului <math>(K^*,\cdot)</math> avem ca <math>x^{m-1}=1, \forall x \in K^*</math>, deci <math>f(aX)=g((aX)^{m-1}=g(a^{m-1}X^{m-1})=g(X^{m-1})=f(X)</math>.