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If $c$ is a regular element of a ring in a right Noetherian ring and a nonzero right ideal $B$ had $B\cap cR=0$, cancellation would make $B+cB+c^{2}B+\cdots$ an infinite direct sum of nonzero right ideals. Its finite partial sums contradict the ascending chain condition. Thus $cR$ is an essential right ideal.

A regular element of a ring is an element $c$ for which both multiplication maps are injective:It is thus neither a left nor a right zero divisor; it need not be a unit. Here “ideals” means two-sided ideals. Put $A=A_n(\mathbb{C})$, and let $T_2(A)$ denote the specified upper triangular ring.

We first establish the additional facts about the Weyl algebra needed below. Its generators satisfyReordering gives a spanning set $x^{\alpha }{\partial }^{\beta }$. These monomials are independent. Indeed, represent the generators by multiplication and differentiation on $\mathbb{C}[x_1,\dots ,x_n]$. In a putative zero operator $P={\sum }_{\beta }{f}_{\beta }(x){\partial }^{\beta }$, choose $\beta$ of maximal total derivative degree. Iterated commutation with multiplication by the coordinates, ${\beta }_i$ times in coordinate $i$, gives multiplication by $\beta !f_{\beta }(x)$. All other derivative terms of that total degree disappear unless they have exactly the same multiindex, and lower-degree terms disappear. Thus $f_{\beta }=0$, and descending induction eliminates all coefficients. This proves the ordered monomial basis of a Weyl algebra and faithfulness of its polynomial action.

The Weyl algebra is a simple ring. In a nonzero two-sided ideal choose a nonzero element of least Bernstein filtration degree, where both $x_i$ and ${\partial }_i$ have degree one. Commuting it with any generator stays in the ideal and lowers that degree. Minimality therefore makes all these commutators zero. Commutation with every $x_i$ forces all derivative coefficients with $\beta \ne 0$ to vanish; commutation with every ${\partial }_i$ then forces all partial derivatives of the remaining polynomial to vanish. In characteristic zero the element is a nonzero scalar. The ideal consequently contains $1$. This proves simplicity of a Weyl algebra in characteristic zero. Its Jacobson radical is zero, since the radical is a proper two-sided ideal of a nonzero unital simple ring.

Use the two diagonal idempotents to isolate the entries of any ideal of $T_2(A)$. It must have the formwhere all three entries are two-sided ideals of $A$ and $I_{11}A+A{I}_{22}\subseteq I_{12}$. Conversely this condition makes the indicated set an ideal. Simplicity says that each $I_{ij}$ is $0$ or $A$, and either nonzero diagonal entry forces $I_{12}=A$. Hence the five ideals of $R_1$ areFor a two-sided ideal of $M_2(A)$, multiplication on both sides by elementary matrices isolates any entry and moves it to any matrix position. The entries obtained form one two-sided ideal $I$ of $A$, and the original ideal is $M_2(I)$. Thus the only ideals of $R_2$ are $0$ and $R_2$.

Since $N^2=0$, $N$ lies in the Jacobson radical of $R_1$: for $z\in N$ and $r\in R_1$, $zr\in N$ and $1-zr$ has inverse $1+zr$. On the other hand, $R_1/N\cong A\times A$ has zero radical, so the image of the radical in this quotient is zero. The elementary unit criterion for the Jacobson radical justifies both assertions. ThereforeThe second equality follows from simplicity of $R_2$.

We now construct the denominators used to describe regular matrices. A left Noetherian noncommutative domain satisfies the left Ore condition. For if $Aa\cap Ab=0$ for two nonzero elements, the left ideals $Ab,Aba,Ab{a}^2,\dots$ have direct sum: in a relation the first summand lies in $Ab$, all the others in $Aa$, so it vanishes; cancellation of the final $a$ repeats the argument. Their finite partial sums form a strictly ascending chain, contradicting left Noetherianity. Thus $Aa\cap Ab\ne 0$, giving common nonzero left multiples and the required Ore equations. The analogous argument with right ideals gives the right Ore condition under right Noetherianity. Both hypotheses for $A$ are allowed in the question.

The Ore localization therefore embeds $A$ in a division ring $Q$. Briefly, its elements are left fractions $a^{-1}b$, with nonzero $a$, and two fractions are identified after passing to a common left denominator. Common multiples supply addition and multiplication, while cancellation in the domain makes the embedding injective. A nonzero fraction has an inverse, obtained by interchanging its numerator and denominator in the appropriate Ore fraction calculation. The right fraction construction gives the same division ring, so finite lists have common left and common right denominators.

A triangular matrix $c=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ with $r,t\ne 0$ is invertible over $T_2(Q)$, with inverseIt is consequently regular over $T_2(A)$. If $r=0$, then $c{E}_{11}=0$; if $t=0$, then $E_{22}c=0$. Hence

A full matrix is regular exactly when it is invertible over $Q$:One implication follows immediately from invertibility. Conversely, a singular matrix over a division ring has a nonzero right kernel column and a nonzero left kernel row, by row reduction. A common right denominator turns the column into a nonzero column over $A$, providing a right annihilator matrix over $A$. A common left denominator does the same for the row and gives a left annihilator. Thus a singular matrix is not regular. Explicitly, for $c=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ with $r\ne 0$, row elimination says that regularity is equivalent to $u-t{r}^{-1}s\ne 0$ in $Q$. If $r=0$, it is equivalent to both $s\ne 0$ and $t\ne 0$. A commutative determinant formula must not be substituted here.

Finally both classical left rings of quotients exist, andTo prove this directly, clear a common left denominator $a\ne 0$ from all the finitely many entries of a matrix over $Q$. It becomes $(a{I}_2{)}^{-1}b$ with $b$ in the original triangular or full ring and $a{I}_2$ regular. Every original regular element becomes a unit in the corresponding ring over $Q$, by the descriptions above. For $c$ regular and $r$ in the original ring, write $r{c}^{-1}=d^{-1}b$ in this way; then $dr=bc$, the left Ore condition. Since the denominators are regular, the additional annihilator condition for localization is automatic. This proves the matrix and triangular localization over an Ore domain assertions. In particular, the nonzero nilpotent radical of the triangular ring does not prevent its classical localization; that localization retains the triangular radical.

An essential right ideal intersects every nonzero right ideal nontrivially. A regular element of a ring is a two-sided non-zero-divisor: both $cr=0$ and $rc=0$ imply $r=0$. Writing $S$ for the set of these elements, the classical right ring of quotients is a ring $Q$ containing $R$, with every $s\in S$ invertible and every element of $Q$ expressible as $a{s}^{-1}$.

Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it is a right Goldie ring: it satisfies the ascending chain condition on right annihilators and has finite uniform dimension. In particular, every semiprime ring that is a right Noetherian ring satisfies the theorem. We prove the requested case directly using the supplied assumption about essential right ideals.

First let $c\in S$. If a nonzero right ideal $B$ had $B\cap cR=0$, the sumwould be direct. Indeed, a finite relation $b_0+c{b}_1+\cdots +c^m{b}_m=0$ gives $b_0\in B\cap cR$, hence $b_0=0$; cancellation of $c$ repeats the argument. Each $c^{j}B$ is a nonzero right ideal, so the partial sums form a strictly ascending chain. This contradicts right Noetherianity. Thus a regular principal right ideal in a right Noetherian ring is essential, and every right ideal containing $c$ is essential.

For $r\in R$ and $c\in S$, putThis is an essential right ideal. To check this, take a nonzero right ideal $B$. If $rB=0$, then $B\subseteq E$. Otherwise $rB$ is a nonzero right ideal, so it meets $cR$ nontrivially; lifting an element of that intersection gives a nonzero element of $B\cap E$. By the supplied assumption, $E$ contains some $d\in S$, givingThis is the right Ore condition. The set $S$ is multiplicatively closed, and the denominator cancellation condition holds because its elements are non-zero-divisors. Therefore Ore theorem constructs $Q$ and embeds $R$ in it.

For every right ideal $J$ of $Q$,Indeed, if $x=a{s}^{-1}\in J$, then $a=xs\in J\cap R$. Contracting an ascending chain of right ideals of $Q$ to $R$ therefore proves that $Q$ is a right Noetherian ring.

Suppose $J$ is an essential right ideal of $Q$. For a nonzero right ideal $B$ of $R$, choose $0\ne x\in J\cap BQ$. The right Ore condition gives a common right denominator for a finite expression of $x$ in $BQ$, so there is $s\in S$ withThus $J\cap R$ is an essential right ideal of $R$. It contains a regular element of a ring, which becomes a unit in $Q$, and consequently $J=Q$.

Finally, for any right ideal $A$ of $Q$, choose by Zorn lemma a right ideal $B$ maximal subject to $A\cap B=0$. Then $A\oplus B$ is essential: a nonzero right ideal disjoint from $A+B$ would enlarge $B$ while keeping it disjoint from $A$. Hence $A\oplus B=Q$. Every submodule of the right regular module has a complement, making it a semisimple module. Since it is also Noetherian, it is a finite direct sum of simple modules. It is therefore Artinian, and $Q$ is a semisimple Artinian ring. This is semisimplicity of a classical quotient from regular elements in essential ideals.

Suppose $R$ is a right Noetherian ring and every essential right ideal contains a regular element of a ring. Preimages of essential regular principal right ideals under left multiplication are essential, so they contain regular elements and prove the right Ore condition. In the classical right ring of quotients $Q$, every right ideal is generated by its contraction to $R$, making $Q$ right Noetherian. An essential right ideal of $Q$ contracts to an essential right ideal of $R$, hence contains a regular element and is all of $Q$. A maximal disjoint complement to any right ideal has essential sum with it, so every right ideal is complemented. The right regular module is therefore a finite direct sum of simple modules, making $Q$ semisimple Artinian.