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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 Composition ring ： ウィキペディア英語版 Composition ring In mathematics, a composition ring, introduced in , is a commutative ring (''R'', 0, +, −, ·), possibly without an identity 1 (see non-unital ring), together with an operation : $\backslash circ:\; R\; \backslash times\; R\; \backslash rightarrow\; R$ such that, for any three elements $f,g,h\backslash in\; R$ one has # $(f+g)\backslash circ\; h=(f\backslash circ\; h)+(g\backslash circ\; h)$ # $(f\backslash cdot\; g)\backslash circ\; h\; =\; (f\backslash circ\; h)\backslash cdot\; (g\backslash circ\; h)$ # $(f\backslash circ\; g)\backslash circ\; h\; =\; f\backslash circ\; (g\backslash circ\; h).$ It is ''not'' generally the case that $f\backslash circ\; g=g\backslash circ\; f$, ''nor'' is it generally the case that $f\backslash circ\; (g+h)$ (or $f\backslash circ\; (g\backslash cdot\; h)$) has any algebraic relationship to $f\backslash circ\; g$ and $f\backslash circ\; h$. ==Examples== There are a few ways to make a commutative ring ''R'' into a composition ring without introducing anything new. *Composition may be defined by $f\backslash circ\; g=0$ for all ''f'',''g''. The resulting composition ring is a rather uninteresting. *Composition may be defined by $f\backslash circ\; g=f$ for all ''f'',''g''. This is the composition rule for constant functions. *If ''R'' is a boolean ring, then multiplication may double as composition: $f\backslash circ\; g=fg$ for all ''f'',''g''. More interesting examples can be formed by defining a composition on another ring constructed from ''R''. *The polynomial ring ''R''() is a composition ring where $(f\backslash circ\; g)\; (x)=f(g(x))$ for all $f,\; g\; \backslash in\; R$. *The formal power series ring ''R'' also has a substitution operation, but it is only defined if the series ''g'' being substituted has zero constant term (if not, the constant term of the result would be given by an infinite series with arbitrary coefficients). Therefore the subset of ''R'' formed by power series with zero constant coefficient can be made into a composition ring with composition given by the same substitution rule as for polynomials. Since nonzero constant series are absent, this composition ring does not have a multiplicative unit. *If ''R'' is an integral domain, the field ''R''(''X'') of rational functions also has a substitution operation derived from that of polynomials: substituting a fraction ''g''1/''g''2 for ''X'' into a polynomial of degree ''n'' gives a rational function with denominator $g\_2^n$, and substituting into a fraction is given by ::$\backslash frac\backslash circ\; g=\backslash frac.$ :However, as for formal power series, the composition cannot always be defined when the right operand ''g'' is a constant: in the formula given the denominator $f\_2\backslash circ\; g$ should not be identically zero. One must therefore restrict to a subring of ''R''(''X'') to have a well-defined composition operation; a suitable subring is given by the rational functions of which the numerator has zero constant term, but the denominator has nonzero constant term. Again this composition ring has no multiplicative unit; if ''R'' is a field, it is in fact a subring of the formal power series example. *The set of all functions from ''R'' to ''R'' under pointwise addition and multiplication, and with $\backslash circ$ given by composition of functions, is a composition ring. There are numerous variations of this idea, such as the ring of continuous, smooth, holomorphic, or polynomial functions from a ring to itself, when these concepts makes sense. For a concrete example take the ring $()$, considered as the ring of polynomial maps from the integers to itself. A ring endomorphism : $F:()\backslash rightarrow()$ of $()$ is determined by the image under $F$ of the variable $x$, which we denote by : $f=F(x)$ and this image $f$ can be any element of $()$. Therefore, one may consider the elements $f\backslash in()$ as endomorphisms and assign $\backslash circ:()\backslash times()\backslash rightarrow()$, accordingly. One easily verifies that $()$ satisfies the above axioms. For example, one has : $(x^2+3x+5)\backslash circ(x-2)=(x-2)^2+3(x-2)+5=x^2-x+3.$ This example is isomorphic to the given example for ''R''() with ''R'' equal to $\backslash mathbb\; Z$, and also to the subring of all functions $\backslash mathbb\; Z\backslash to\backslash mathbb\; Z$ formed by the polynomial functions. 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「Composition ring」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.