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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク twists of curves ： ウィキペディア英語版 twists of curves In mathematical field of algebraic geometry, an elliptic curve E over a field K has an associated quadratic twist, that is another elliptic curve which is isomorphic to E over an algebraic closure of K. In particular, an isomorphism between elliptic curves is an isogeny of degree 1, that is an invertible isogeny. Some curves have higher order twists such as cubic and quartic twists. The curve and its twists have the same j-invariant. ==Quadratic twist== First assume K is a field of characteristic different from 2. Let E be an elliptic curve over K of the form: : $y^2\; =\; x^3\; +\; a\_2\; x^2\; +a\_4\; x\; +\; a\_6.\; \backslash ,$ Given $d\backslash in\; K\backslash setminus\; K^2$ and $d\backslash neq\; 0$, the quadratic twist of $E$ is the curve $E^d$, defined by the equation: : $dy^2\; =\; x^3\; +\; a\_2\; x^2\; +\; a\_4\; x\; +\; a\_6.\; \backslash ,$ or equivalently : $y^2\; =\; x^3\; +\; d\; a\_2\; x^2\; +\; d^2\; a\_4\; x\; +\; d^3\; a\_6.\; \backslash ,$ The two elliptic curves $E$ and $E^d$ are not isomorphic over $K$, but over the field extension $K(\backslash sqrt)$. Now assume K is of characteristic 2. Let E be an elliptic curve over K of the form: : $y^2\; +\; a\_1\; x\; y\; +a\_3\; y\; =\; x^3\; +\; a\_2\; x^2\; +a\_4\; x\; +\; a\_6.\; \backslash ,$ Given $d\backslash in\; K$ such that $X^2+X+d$ is an irreducible polynomial over K, the quadratic twist of E is the curve Ed, defined by the equation: : $y^2\; +\; a\_1\; x\; y\; +a\_3\; y\; =\; x^3\; +\; (a\_2\; +\; d\; a\_1^2)\; x^2\; +a\_4\; x\; +\; a\_6\; +\; d\; a\_3^2.\; \backslash ,$ The two elliptic curves $E$ and $E^d$ are not isomorphic over $K$, but over the field extension $K()/(X^2+X+d)$. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「twists of curves」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.