We know that the definition of the Jacobson radical $J(R)$
(a) in a ring $R$ with identity is: $$J(R)=\cap \mbox{ maximal left ideals}.$$
(b) in a ring $R$ without identity is: $$J(R)=\{a\in R\mid aR \mbox{ is right quasi regular}\}$$ $$=\{a\in R\mid \forall r\in R, \exists b\in R, \mbox{ such that }ar+b=arb\}$$
A natural guess is that if $R$ has a maximal left ideal $M$, even $R$ has no multiplicative identity, then the definition (a) coincides with the definition (b). (I am not sure)
If my guess is correct,
then $\boldsymbol{J(R)\subseteq M}$.
But I found out a counterexample.
I use GAP to find a ring of order 8,
the command is: ShowMultiplicationTable(SmallRing(8,37));
The multiplication table of this ring as following figure.
There is a maximal left ideal $M=R(a+b)=\{0,a,b,a+b\}$ and the Jacobson radical of $R$ is $J(R)=\{0,a,c,a+c\}$. Hence $\boldsymbol{J(R)\nsubseteq M=R(a+b)}$.
What did I do wrong? Is there any book discuss this common mistakes?
gap> ShowMultiplicationTable(SmallRing(8,37));
* | 0*a c b b+c a a+c a+b a+b+c
------+------------------------------------------------
0*a | 0*a 0*a 0*a 0*a 0*a 0*a 0*a 0*a
c | 0*a 0*a 0*a 0*a 0*a 0*a 0*a 0*a
b | 0*a 0*a 0*a 0*a b b b b
b+c | 0*a 0*a 0*a 0*a b b b b
a | 0*a c b b+c a+b a+b+c a a+c
a+c | 0*a c b b+c a+b a+b+c a a+c
a+b | 0*a c b b+c a a+c a+b a+b+c
a+b+c | 0*a c b b+c a a+c a+b a+b+c
Consider $R=\mathbb{Z}$ but with zero multiplication. Then every element is quasiregular, so the radical is $R$. On the other hand, the intersection of the maximal ideals of $R$ is $\{0\}$.
The Jacobson radical is the intersection of the maximal regular left (or right) ideals. A left ideal $I$ is regular if there exists $e\in R$ such that $re-r\in I$, for all $r\in R$. It's obvious that no proper left ideal of a zero ring can be regular.