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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 Ring of polynomial functions ： ウィキペディア英語版 Ring of polynomial functions In mathematics, the ring of polynomial functions on a vector space ''V'' over an infinite field ''k'' gives a coordinate-free analog of a polynomial ring. It is denoted by ''k''(). If ''V'' has finite dimension and is viewed as an algebraic variety, then ''k''() is precisely the coordinate ring of ''V''. The explicit definition of the ring can be given as follows. If $k(\backslash dots,\; t\_n)$ is a polynomial ring, then we can view $t\_i$ as coordinate functions on $k^n$; i.e., $t\_i(x)\; =\; x\_i$ when $x\; =\; (x\_1,\; \backslash dots,\; x\_n).$ This suggests the following: given a vector space ''V'', let ''k''() be the subring generated by the dual space $V^$ * of the ring of all functions $V\; \backslash to\; k$. If we fix a basis for ''V'' and write $t\_i$ for its dual basis, then ''k''() consists of polynomials in $t\_i$; it is a polynomial ring. In applications, one also defines ''k''() when ''V'' is defined over some subfield of ''k'' (e.g., ''k'' is the complex field and ''V'' is a real vector space.) The same definition still applies. == Symmetric multilinear maps == Let $S^q(V)$ denote the vector space of multilinear functionals $\backslash textstyle\; \backslash lambda:\; \backslash prod\_1^q\; V\; \backslash to\; k$ that are symmetric; $\backslash lambda(v\_1,\; \backslash dots,\; v\_q)$ is the same for all permutations of $v\_i$'s. Any λ in $S^q(V)$ gives rise to a homogeneous polynomial function ''f'' of degree ''q'': let $f(v)\; =\; \backslash lambda(v,\; \backslash dots,\; v).$ To see that ''f'' is a polynomial function, choose a basis $e\_i,\; \backslash ,\; 1\; \backslash le\; i\; \backslash le\; n$ of ''V'' and $t\_i$ its dual. Then :$\backslash lambda(v\_1,\; \backslash dots,\; v\_q)\; =\; \backslash sum\_^n\; \backslash lambda(e\_,\; \backslash dots,\; e\_)\; t\_(v\_1)\; \backslash cdots\; t\_(v\_q)$. Thus, there is a well-defined linear map: :$\backslash phi:\; S^q(V)\; \backslash to\; k()\_q,\; \backslash ,\; \backslash phi(\backslash lambda)(v)\; =\; \backslash lambda(v,\; \backslash cdots,\; v).$ It is an isomorphism:〔There is also a more abstract way to see this: to give a multilinear functional on the product of ''q'' copies of ''V'' is the same as to give a linear functional on the ''q''-th tensor power of ''V''. The requirement that the multilinear functional to be symmetric translates to the one that the linear functional on the tensor power factors through the ''q''-th symmetric power of ''V'', which is isomorphic to ''k''()q.〕 choosing a basis as before, any homogeneous polynomial function ''f'' of degree ''q'' can be written as: :$f\; =\; \backslash sum\_^n\; a\_\; t\_\; \backslash cdots\; t\_$ where $a\_$ are symmetric in $i\_1,\; \backslash dots,\; i\_q$. Let :$\backslash psi(f)(v\_1,\; \backslash dots,\; v\_q)\; =\; \backslash sum\_^n\; a\_\; t\_(v\_1)\; \backslash cdots\; t\_(v\_q).$ Then ψ is the inverse of φ. (Note: φ is still independent of a choice of basis; so ψ is also independent of a basis.) Example: A bilinear functional gives rise to a quadratic form in a unique way and any quadratic form arises in this way. 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「Ring of polynomial functions」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.