Can anybody suggest how to prove or disprove the statement that every commutative right primitive ring is a field?
Definition 1: A ring is defined to be a set $(R,+, \cdot)$ with two binary operations, satisfying the following three axioms:
$(R, +)$ is an abelian group with non-trivial $0$ element,
$(R, \cdot)$ is a semigroup with unity,
the distributive identities are satisfied: for all $a, b, c \in R$, $a(b + c) = ab + ac$, $(a + b)c = ac + bc$.
Definition 2: A ring $R$ is called (right) primitive if there is a faithful simple right $R$-module.
Definition 3: A right $R$-module $M$ is called a faithful module if its annihilator in the ring $R$ is equal to zero; that is, for all $r \in R$, $M r = 0$ implies $r = 0$.
Let $S$ be a faithful simple right $R$ module.
By the first isomorphism theorem and the correspondence theorem, $S\cong R/T$ as a right $R$ module for some maximal ideal $T$.
But $T$ annihilates $R/T$, so $T=\{0\}$.
Since zero is a maximal ideal iff $R$ is a field, we’re done.