Past exam of the non-commutative rings course of the University of Oxford / 2023 / 1 (Primitive quotients and radicals)

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.

Assume $A\ne 0$, as is implicit in the claimed conclusion. Suppose that a surjective ring homomorphism $f:R\to M_2(A)$ exists. Choose $r_1,r_2\in R$ mapping to the matrix units $E_{12}$ and $E_{21}$. Applying $f$ to the assumed relation gives integers $c_1,\dots ,c_n$ such thatBut $E_{12}^2=0$ and $E_{12}{E}_{21}=E_{11}$, so the right side is $c_1{E}_{11}$. Its $(2,2)$ entry is zero, whereas the $(2,2)$ entry of $E_{22}$ is $1_A\ne 0$. This contradiction proves that no such surjective ring homomorphism exists.

If the convention for rings allows $A$ to be the zero ring, that case must be excluded from the assertion: $M_2(0)=0$, and the zero ring homomorphism is surjective. This is a convention issue in the printed question.

Put $B=R/I$. By the definition of a left-primitive ideal, there is a faithful module $V$ that is a simple module on the left over $B$. The assumed relation passes to the quotient ring $B$, since any pair of elements of $B$ can be lifted to $R$.

By Schur's lemma, the endomorphism ring ${\mathrm{End}}_B(V)$ is a division ring: the kernel and image of a nonzero module endomorphism are submodules, so simplicity makes that module endomorphism both injective and surjective. SetEvaluation makes $V$ a right vector space over a division ring $\Delta$, and the $B$-action is $\Delta$-linear. The opposite ring in the definition ensures the correct order for this right action.

The Jacobson density theorem says that, for any finite $\Delta$-linearly independent family $v_1,\dots ,v_m$ and any $w_1,\dots ,w_m\in V$, some $b\in B$ satisfies $b{v}_i=w_i$ for every $i$.

Suppose that $V$ contains two $\Delta$-linearly independent vectors $v,w$. Applying the Jacobson density theorem twice, choose $a,b\in B$ satisfyingThe inherited relation giveswith integers $c_k$. Acting on $w$, the left side gives $baw=bv=w$, but every term on the right gives $a^{k}bw=0$. This contradicts $w\ne 0$. Thus ${\dim }_{\Delta }V=1$.

The ring homomorphismgiven by the action is injective because $V$ is a faithful module. It is surjective because the Jacobson density theorem, applied to a single nonzero basis vector, realizes every $\Delta$-linear endomorphism of this one-dimensional vector space over a division ring. Every nonzero such endomorphism is invertible. HenceThe use of the Jacobson density theorem here only prescribes actions on finitely many vectors; it does not assume that an arbitrary primitive ring is already a full endomorphism ring.