Algebraic extensions of power series rings
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- by Jimmy T. Arnold PDF
- Trans. Amer. Math. Soc. 267 (1981), 95-110 Request permission
Abstract:
Let $D$ and $J$ be integral domains such that $D \subset J$ and $J[[X]]$ is not algebraic over $D[[X]]$. Is it necessarily the case that there exists an integral domain $R$ such that $D[[X]] \subset R \subseteq J[[X]]$ and $R \cong D[[X]][[\{ {Y_i}\} _{i = 1}^\infty ]]$? While the general question remains open, the question is answered affirmatively in a number of cases. For example, if $D$ satisfies any one of the conditions (1) $D$ is Noetherian, (2) $D$ is integrally closed, (3) the quotient field $K$ of $D$ is countably generated as a ring over $D$, or (4) $D$ has Krull dimension one, then an affirmative answer is given. Further, in the Noetherian case it is shown that $J[[X]]$ is algebraic over $D[[X]]$ if and only if it is integral over $D[[X]]$ and necessary and sufficient conditions are given on $D$ and $J$ in order that this occur. Finally if, for every positive integer $n$, $D[[{X_1}, \ldots ,{X_n}]] \subset R \subseteq J[[{X_1}, \ldots ,{X_n}]]$ implies that $R \ncong D[[{X_1}, \ldots ,{X_n}]][[\{ {Y_i}\} _{i = 1}^\infty ]]$, then it is shown that $J[[{X_1}, \ldots ,{X_n}]]$ is algebraic over $D[[{X_1}, \ldots ,{X_n}]]$ for every $n$.References
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Additional Information
- © Copyright 1981 American Mathematical Society
- Journal: Trans. Amer. Math. Soc. 267 (1981), 95-110
- MSC: Primary 13J05; Secondary 13G05
- DOI: https://doi.org/10.1090/S0002-9947-1981-0621975-4
- MathSciNet review: 621975