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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 Hessian curves ： ウィキペディア英語版 Hessian form of an elliptic curve In geometry, the Hessian curve is a plane curve similar to folium of Descartes. It is named after the German mathematician Otto Hesse. This curve was suggested for application in elliptic curve cryptography, because arithmetic in this curve representation is faster and needs less memory than arithmetic in standard Weierstrass form.〔Cauchy-Desbove's Formulae: ''Hessian-elliptic Curves and Side-Channel Attacks'', Marc Joye and Jean-Jacques Quisquarter〕 == Definition == Let $K$ be a field and consider an elliptic curve $E$ in the following special case of Weierstrass form over $K$: : $Y^2+a\_1\; XY+a\_3\; Y=X^3$ where the curve has discriminant $\backslash Delta\; =\; (a\_3^3(a\_1^3\; -\; 27a\_3))\; =\; a\_3^3\; \backslash delta.$ Then the point $P=(0,0)$ has order 3. To prove that $P=(0,0)$ has order 3, note that the tangent to $E$ at $P$ is the line $Y=0$ which intersects $E$ with multiplicity 3 at $P$. Conversely, given a point $P$ of order 3 on an elliptic curve $E$ both defined over a field $K$ one can put the curve into Weierstrass form with $P=(0,0)$ so that the tangent at $P$ is the line $Y=0$. Then the equation of the curve is $Y^2+a\_1\; XY+a\_3\; Y=X^3$ with $a\_1,a\_3\backslash in\; K$. Now, to obtain the Hessian curve, it is necessary to do the following transformation: First let $\backslash mu$ denote a root of the polynomial : $T^3-\backslash delta\; T^2+T+\; a\_3\backslash delta^2=0.$ Then : $\backslash mu=.$ Note that if $K$ has a finite field of order $q\backslash equiv\; 2$ (mod 3), then every element of $K$ has a unique cube root; in general, $\backslash mu$ lies in an extension field of ''K''. Now by defining the following value $D=\backslash frac$ another curve, C, is obtained, that is birationally equivalent to E: : $C$ : $x^3\; +\; y^3\; +\; z^3=\; Dxyz$ which is called ''cubic Hessian form'' (in projective coordinates) : $C$ : $x^3\; +\; y^3\; +\; 1=\; Dxy$ in the ''affine plane'' ( satisfying $x=\backslash frac$ and $y=\backslash frac$ ). Furthermore, $D^3\backslash ne1$ (otherwise, the curve would be singular). Starting from the Hessian curve, a birationally equivalent Weierstrass equation is given by : $v^2\; =\; u^3\; -\; 27D(D^3\; +\; 8)u\; +\; 54(D^6\; -\; 20\; D^3\; -\; 8),\; \backslash ,$ under the transformations: : $(x,y)\; =\; (\backslash eta\; (u\; +\; 9D^2),\; -\; 1\; +\; \backslash eta(3D^3\; -\; Dx\; -12))\; \backslash ,$ and : $(u,v)\; =\; (-9D^2\; +\; \backslash varepsilon\; x,\; 3\backslash varepsilon(y\; -\; 1))\; \backslash ,$ where: :$\backslash eta\; =\; \backslash frac$ and :$\backslash varepsilon\; =\; \backslash frac$ 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「Hessian form of an elliptic curve」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.