Given a unital ring $R$ and its unital subring $P$ (with the same unit). Also, given a maximal left ideal $L$ of $R$. Must $L\cap P$ be a maximal left ideal of $P$? I don't think so, but are there any conditions on $R$ or $P$ which ensure that?
Intersection of a subring and an ideal
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$\begingroup$
abstract-algebra
ring-theory
ideals
noncommutative-algebra
2 Answers
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No, $L\cap P$ is not necessarily a maximal left ideal of $P$.
Example:If $R=\mathbb{R}$, $P=\mathbb{Z}$, then $L=(0)$ is a maximal ideal of $R$ because $R$ is a field. So $L\cap P=(0)$ but $(0)$ is not a maximal ideal in $\mathbb{Z}$, because $(0) \subset 2\mathbb{Z} \subset \mathbb{Z}$
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The answer in general is no (see francis-jamet's answer) but there is at least one important case where the answer is affirmative.
If $S\subset R$ is an integral extension and $\frak m$ is a maximal ideal in $R$ then ${\frak m}\cap S$ is a maximal ideal in $S$.
This situation is very relevant for instance in algebraic number theory where $R$ and $S$ can be taken as the ring of integers in algebraic extensions of $\Bbb Q$.