Matrix rings over polynomial identity rings
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- by Elizabeth Berman
- Trans. Amer. Math. Soc. 172 (1972), 231-239
- DOI: https://doi.org/10.1090/S0002-9947-1972-0308187-3
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Abstract:
We prove that if $A$ is an algebra over a field with at least $k$ elements, and $A$ satisfies ${x^k} = 0$, then ${A_n}$, the ring of $n$-by-$n$ matrices over $A$, satisfies ${x^q} = 0$, where $q = k{n^2} + 1$. Theorem 1.3 generalizes this result to rings: If $A$ is a ring satisfying ${x^k} = 0$, then for all $n$, there exists $q$ such that ${A_n}$ satisfies ${x^q} = 0$. Definitions. A checkered permutation of the first $n$ positive integers is a permutation of them sending even integers into even integers. The docile polynomial of degree $n$ is \[ \prod \limits _{i = 1}^p {D({x_{i1}}, \cdots ,{x_{ik}}){u_i},} \]athewhere the sum is over all checkered permutations $f$ of the first $k$ positive integers. The docile product polynomial of degree $k,p$is \[ \prod \limits _{i = 1}^p {D({x_{i1}}, \cdots ,{x_{ik}}){u_i},} \] where the $x$’s and $u$’s are noncommuting variables. Theorem 2.1. Any polynomial identity algebra over a field of characteristic 0 satisfies a docile product polynomial identity. Theorem 2.2. If $A$ is a ring satisfying the docile product polynomial identity of degree $2k,p$, and $n$ is a positive integer, and $q = 2{k^2}{n^2} + 1$; then ${A_n}$ satisfies a product of $p$ standard identities, each of degree $q$.References
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Bibliographic Information
- © Copyright 1972 American Mathematical Society
- Journal: Trans. Amer. Math. Soc. 172 (1972), 231-239
- MSC: Primary 16A42
- DOI: https://doi.org/10.1090/S0002-9947-1972-0308187-3
- MathSciNet review: 0308187