Centralizers in free associative algebras
Centralizers in free associative algebras
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DOI：
10.1090/s0002-9947-1969-0236208-5
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发表时间：
1969-03
影响因子：
1.3
通讯作者：
G. Bergman
中科院分区：
文献类型：
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作者：
G. Bergman
设R是域k上某个生成元集合上的自由结合代数（等价于：某个k-向量空间上的张量代数）。虽然R一般来说是“非常”非交换的，但很容易找到交换元素对：如果我们取任意z E1 R，以及多项式P和Q在k上的一个不定式中，那么P（z）和Q（z）将交换。这里我们将证明任何非标量元素u E1 R的中心化子C（即C={x E R 1 xu=ux}）对于某个非标量z E R具有k[z]的形式。特别地，R的任何一对交换元素都可以写成P（z），Q（z）的形式。(This被P. M.科恩[8，第348页]。让我们首先概述我们的证明，说明结果，因为它们适用于这个问题，而不一定是在最大的一般性，他们将证明：在建立一些一般环理论工具？1、我们将在？2，通过相当初等的论证，我们的中心化子环C是交换的，并且实际上是k[u]的有限整数扩张。为了完成我们的证明，只要证明C是整闭的，并且可以嵌入多项式环k[x]中就足够了，因为多项式代数k[x]的任何整闭的子代数（在它自己的分数域中）和/k都是k[y]的形式[9，命题2.1]。(The证明使用Liuroth定理。在哪？3.我们发现存在一个非零元素eE 1 R，以及C的整闭包C'到R的同态f，使得对所有的xEC，xe=ef（x）.换句话说，f（C）与C“共轭”，并且这个共轭，如果不是C本身，“可以在R内整体闭合”。“进去？4，证明了f（C '）可以“拉回到e上”，从而C本身在R中有积分闭包。这个积分闭包是可交换的并且包含u，它必须与我们的中心化子C重合。所以C是全闭的。在哪？5，我们表明，任何生成的非平凡（即，#Ak）子代数可以映射到多项式环k[x]上，从而具有非平凡像.我们把它应用到C。现在图像将具有对k的超越度I，并且C具有相同的超越度，因为它是k[u]的有限扩展。因此，地图将是1-1（见[16，第二章，？12，Theorem 29，p. 101]），给出了完成我们的证明所需的嵌入。剩下的？？6-9，研究了微分算子环（？）9），前几节结果的推广，和
Let R be the free associative algebra on some set of generators over a field k (equivalently: the tensor algebra on some k-vector-space). Though R is in general " very" noncommutative, it is easy to find pairs of commuting elements: If we take any z E1 R, and polynomials P and Q in one indeterminate over k, then P(z) and Q(z) will commute. We shall here show that the centralizer, C, of any nonscalar element u E1 R (i.e. C={x E R 1 xu=ux}) is of the form k[z] for some nonscalar z E R. In particular, any pair of commuting elements of R can be written in the form P(z), Q(z). (This was conjectured by P. M. Cohn [8, p. 348].) Let us first outline our proof, stating results as they apply to this problem, and not necessarily in the greatest generality in which they will be proved: After setting up some general ring-theoretic tools in ?1, we shall show in ?2, by rather elementary arguments that our centralizer ring C is commutative, and in fact is a finite integral extension of k[u]. To complete our proof, it suffices to show that C is integrally closed, and is embeddable in a polynomial ring k[x], because any subalgebra of a polynomial algebra k[x] that is integrally closed (in its own field of fractions) and /k is of the form k[y] [9, Proposition 2.1]. (The proof uses Liuroth's theorem.) In ?3 we find that there exists a nonzero element e E1 R, and a homomorphismf of the integral closure C' of C, into R, such that for all x E C, xe=ef(x). In other words, f(C) is "conjugate" to C, and this conjugate, if not C itself, "can be integrally closed within R." In ?4, we show thatf(C') can be "'pulled back across e," thus C itself has integral closure in R. This integral closure, being commutative and containing u, must coincide with our centralizer C. So C is integrally closed. In ?5, we show that any finitely generated nontrivial (i.e., #Ak) subalgebra of R can be mapped into a polynomial ring k[x] so as to have nontrivial image. We apply this to C. Now the image will have transcendence degree I over k, and C has the same transcendence degree because it is a finite extension of k[u]. Hence the map will be 1-1 (see [16, Chapter II, ?12, Theorem 29, p. 101]), giving the embedding required to complete our proof. The remaining ??6-9, examine some related points-centralizers in some rings of differential operators (?9), generalizations of results of the preceding sections, and