Past exam of the non-commutative rings course of the University of Oxford / 2023 / 1 / a

A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.

The Jacobson radical isHere the second intersection ranges over isomorphism classes of nonzero simple modules. For the equality, every nonzero $v$ in a simple module gives a surjective module homomorphism $R\to V$, $r\mapsto rv$, whose kernel is a maximal left ideal. Conversely, every maximal left ideal $L$ gives the simple module $R/L$. This description by annihilators shows that $J(R)$ is a two-sided ideal.

Let $j\in J(R)$ and put $a=1+j$. If the left ideal $Ra$ were proper, it would lie in a maximal left ideal $L$. Both $a$ and $j$ would belong to $L$, giving $1\in L$, a contradiction. Therefore there is $u\in R$ with $ua=1$.

A left inverse alone need not be a two-sided inverse in a general ring, so one more step is necessary. The equality $u(1+j)=1$ gives $u=1-uj\in 1+J(R)$. Applying the same argument to $u$ supplies $v\in R$ with $vu=1$. Thenand hence $au=vu=1$. ThusFor the zero ring the conclusion holds directly; the maximal left ideal argument concerns the nonzero case.

Let $L$ be any maximal left ideal. If $I⊈L$, maximality gives $L+I=R$, because the two-sided ideal $I$ is in particular a left ideal. Thus $1=\ell +i$ for some $\ell \in L$ and $i\in I$. Butis a unit by hypothesis. A proper left ideal cannot contain a unit, since left multiplication by its inverse would put $1$ in the left ideal. This contradiction proves $I\subseteq L$ for every maximal left ideal $L$, and thereforeTogether with ii, this is the ideal form of the unit criterion for the Jacobson radical.