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

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!

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 $$\{0\}\times \{0\}$$.