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Define Jacobson radical $\textrm{Rad}(M)$ of a module $M$ as the intersection of all it's maximal submodules.

Let $R$ be an associative ring with unity. It's clear thar $\textrm{Rad}(_RR)$ coincides with the intersection of annihilators of all simple left $R$-modules. Hence, $$\textrm{Rad}(_RR)=\bigcap\limits_{I\textrm{ is a maximal left ideal of }R}\textrm{Ann}_R(R/I)=\bigcap\limits_I\,(I:R)$$ where $(I:R)=\{x\in R\,|\,xR\subseteq I\}$ is the largest (two-sided) ideal of $R$ which lies in $I$. In particular, $\textrm{Rad}(_RR)$ is an ideal of $R$.

Moreover, $\textrm{Rad}(_RR)=\{s\in R\,|\,\forall r\in R: 1+rs\textrm{ has the left inverse in } R\}.$

In fact, this definition is symmetric: $\textrm{Rad}(_RR)=\textrm{Rad}(R_R),$ so, we can talk about Jacobson radical of $R$ which we denote $\textrm{Rad}(R).$

My question is:

Is it true that $\textrm{Rad}(R)=\bigcap I$ where $I$'s are both two-sided ideals in $R$ and maximal left ideals in $R$?

Clearly, $\textrm{Rad}(R)\subseteq\bigcap I.$ I suppose that the inverse inclusion does not hold in general but cannot find a counterexample. Do you know a one? Any help is appreciate!

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Is it true that $\textrm{Rad}(R)=\bigcap I$ where $I$'s are both two-sided ideals in $R$ and maximal left ideals in $R$?

No: consider a field $F$ and $R=F\times M_2(F)$. It has four ideals: $\{0\}\times \{0\}$, $F\times \{0\}$, $\{0\}\times M_2(F)$ and $R$.

The only one which is maximal as a left ideal is $\{0\}\times M_2(F)$, but the Jacobson radical is $\{0\}\times \{0\}$.

But the containment you said is of course correct: a two-sided ideal that is maximal as a left ideal is certainly a maximal two-sided ideal, and the Jacobson radical is contained in such ideals.

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