论文标题
海螺最大子环
Conch Maximal Subrings
论文作者
论文摘要
结果表明，如果$ r $是戒指，$ p $是一个积分域$ d \ leq r $带有$ \ bigcap_ {n = 1}^\ infty p^nd = 0 $和$ p \ in u（r）$的主要要素，则$ r $具有conch的最大subimal subsring（请参阅\ cite {fairs {fairs {fairs {fairs {fairs {fairs}）。我们证明，对于每个子$ s $ of $ r $ $ r $的环$ r $具有CONCH $ r $（s）= s \ cap u（r）$（即，$ r $的每个子$ r $均相对于倒数，请参见\ cite {Invsub}）。尤其是，$ r $具有海类的最大supring或$ u（r）$是$ r $的主要子来源不可或缺的。我们观察到，如果$ r $是一个具有$ | r | = 2^{2^{\ aleph_0}} $的整体域，那么$ r $具有最大subring或$ | max（r）| = 2^{\ alleph_0} $，尤其是尤其是另外的$（R）$ $ dim（r）= 1 $，然后是$ r $ $ r $。 If $R\subseteq T$ be an integral ring extension, $Q\in Spec(T)$, $P:=Q\cap R$, then we prove that whenever $R$ has a conch maximal subring $S$ with $(S:R)=P$, then $T$ has a conch maximal subring $V$ such that $(V:T)=Q$ and $V\cap R=S$.结果表明，如果$ k $是一个代数封闭的字段，它不是代数的素数，而$ r $是$ k $上的仿射环，那么对于每个Prime Ideal $ p $ $ p $ $ r $，带有$ ht（p）\ geq dim（r）-1 $的$ r $ $ r $，有一个最大subimal subimal subring subring $ s $ s $ s $ r $ a $ a $（$ $ $ $ $ $（s：s：s：s：r）= p $ p $ p $。如果$ r $是字段$ k $上的普通仿射积分域，那么我们证明$ r $是一个$ t $ t $的总体封闭的最大suzring，而仅当$ dim（r）= 1 $，尤其是在这种情况下，尤其是$（r：t）= 0 $。
It is shown that if $R$ is a ring, $p$ a prime element of an integral domain $D\leq R$ with $\bigcap_{n=1}^\infty p^nD=0$ and $p\in U(R)$, then $R$ has a conch maximal subring (see \cite{faith}). We prove that either a ring $R$ has a conch maximal subring or $U(S)=S\cap U(R)$ for each subring $S$ of $R$ (i.e., each subring of $R$ is closed with respect to taking inverse, see \cite{invsub}). In particular, either $R$ has a conch maximal subring or $U(R)$ is integral over the prime subring of $R$. We observe that if $R$ is an integral domain with $|R|=2^{2^{\aleph_0}}$, then either $R$ has a maximal subring or $|Max(R)|=2^{\aleph_0}$, and in particular if in addition $dim(R)=1$, then $R$ has a maximal subring. If $R\subseteq T$ be an integral ring extension, $Q\in Spec(T)$, $P:=Q\cap R$, then we prove that whenever $R$ has a conch maximal subring $S$ with $(S:R)=P$, then $T$ has a conch maximal subring $V$ such that $(V:T)=Q$ and $V\cap R=S$. It is shown that if $K$ is an algebraically closed field which is not algebraic over its prime subring and $R$ is affine ring over $K$, then for each prime ideal $P$ of $R$ with $ht(P)\geq dim(R)-1$, there exists a maximal subring $S$ of $R$ with $(S:R)=P$. If $R$ is a normal affine integral domain over a field $K$, then we prove that $R$ is an integrally closed maximal subring of a ring $T$ if and only if $dim(R)=1$ and in particular in this case $(R:T)=0$.
