Let $(R,\mathfrak{m})$ be a Noetherian local domain, let $K(R)$ be the total fraction field of $R$ and let $R^{+}$ be the absolute integral closure of $R$. Consider the $R^{+}$-module $K(R)\otimes_{R} R^{+}$ via extension of scalars. Let $J$ be the ideal $\mathfrak{m}R^{+}$ of $R^{+}$. I don't know if $J(K(R)\otimes_{R} R^{+})=K(R)\otimes_{R} R^{+}$.
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1$\begingroup$ Is the integral closure in the algebraic closure of the field K(R). $\endgroup$– CARLOMay 19, 2023 at 2:46
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3$\begingroup$ Isn't $K(R)\otimes_RR^+$ simply the algebraic closure of $K(R)$? It being a field then trivially implies the equality you ask about. $\endgroup$– WojowuMay 19, 2023 at 7:33
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1$\begingroup$ I don't believe that $K(R)\otimes_{R}R^{+}$ be a field. $R^{+}$ is a local ring with maximal ideal $\mathfrak{m}R^{+}$. $\endgroup$– CARLOMay 19, 2023 at 16:04
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3$\begingroup$ If $f \in R^+$, then from an integral eqn $f^s + \sum_ja_jf^{s-j} = 0$ of $f$ over $R$, you have $f(f^{s-1}+a_{s-1}f^{s-2}+\cdots+a_1) = -a_0$, which is a unit in $K(R) \otimes_R R^+$ $\endgroup$– pinakiMay 19, 2023 at 16:42
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2$\begingroup$ $K(R)\otimes_R R^+$ is the localization of $R^+$ at nonzero elements of $R$. Any element of the algebraic closure has an $R$-multiple in $R^+$ by "clearing denominators" from its minimal polynomial, so will lie in that localization. $\endgroup$– WojowuMay 19, 2023 at 18:13
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