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A right Noetherian ring satisfies the ascending chain condition on right ideals; equivalently, every right ideal is a finitely generated module.

Here is a noncommutative Hilbert basis theorem proof adapted to the stated hypothesis. Set $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$, by moving one coefficient past one $x$ at a time. These spaces form an exhaustive filtered algebra structure on $B$, with $F_n{F}_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving automorphism.

For a right ideal $I\subseteq B$, defineEach $L_n$ is a right ideal of $A$. In fact, if $a\in L_n$ and $c\in A$, write $c{x}^n=x^{n}b+u$ with $b\in A$, $u\in F_{n-1}$; then $ac{x}^n=(a{x}^n)b+au\in I+F_{n-1}$. Also $L_n\subseteq L_{n+1}$ by right multiplication by $x$. Since $A$ is a right Noetherian ring, this ascending chain stabilizes at some $L_N$.

Choose finite generators $a_{nj}$ for each $L_n$, $0\le n\le N$, and choose $f_{nj}\in I$ with $f_{nj}-a_{nj}{x}^n\in F_{n-1}$. These finitely many elements generate $I$ as a right ideal. To see this, induct on $m$ for $f\in I\cap F_m$. Write $f=a{x}^m+u$ with $u\in F_{m-1}$; then $a\in L_m$. Put $d=\min (m,N)$ and express $a={\sum }_j{a}_{dj}{c}_j$. Write $c_j{x}^d=x^d{b}_j+u_j$ with $u_j\in F_{d-1}$. The differencelies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence $B$ is right Noetherian.

For the quantum torus, take $q\in k^{\times }$ and the convention $YX=qXY$. Begin with the polynomial ring $k[X]$, which is Noetherian by the Hilbert basis theorem. Adjoining $X^{-1}$ preserves the hypothesis because it commutes with $k[X]$, and gives $k[X^{\pm 1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm 1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm 1}][Y;\sigma ]$, where $\sigma (X)=qX$. On a monomial, $Y^{-1}{X}^i{Y}^j=q^{-i}{X}^i{Y}^{j-1}$; when $j=0$ this is in $A{Y}^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving the quantum torus is right Noetherian. Nonzero $q$ is required for this notation.

For a noncommutative ring, a prime ideal of a noncommutative ring means a proper two-sided ideal $P$ such that $UV\subseteq P$ for two-sided ideals $U,V$ implies $U\subseteq P$ or $V\subseteq P$. Equivalently, $aAb\subseteq P$ implies $a\in P$ or $b\in P$. This definition does not require $A/P$ to be a noncommutative domain.

Retain the right Noetherian ring hypothesis for the last assertion. More generally, the ascending chain condition on two-sided ideals suffices. We claim that every proper two-sided ideal $I$ contains a product of finitely many prime ideals of a noncommutative ring, each containing $I$. If not, choose a maximal counterexample $I$. It cannot be a prime ideal of a noncommutative ring. Thus there are two-sided ideals $U,V$ strictly containing $I$ with $UV\subseteq I$: add $I$ to the two witnesses for failure of the defining condition for a prime ideal of a noncommutative ring. By maximality, both $U$ and $V$ contain products of finitely many prime ideals of a noncommutative ring containing them. Concatenating these products gives a product inside $UV\subseteq I$, a contradiction.

Apply the claim to $I=0$ in a nonzero $A$, obtaining $P_1\cdots P_r=0$. Every prime ideal of a noncommutative ring $Q$ contains one of the $P_i$, by repeated application of the definition of a prime ideal of a noncommutative ring. For the prime radical of a noncommutative ring $N$, it follows thatIndeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and ${\bigcap }_i{P}_i\subseteq Q$ for every $Q$ gives equality of the intersections. If $A=0$, the empty intersection is the whole zero ring and the conclusion is immediate.

The final assertion is false for arbitrary algebras without the preceding chain condition. For example, in the commutative ring $k[z_1,z_2,\dots ]/(z_i^2:i\ge 1)$ the nilradical is $(z_1,z_2,\dots )$, its only prime ideal, but the product of any number of distinct $z_i$ is nonzero. Thus this nilradical is not a nilpotent ideal.

A proper two-sided ideal $P$ is prime when $AB\subseteq P$ for two-sided ideals implies $A\subseteq P$ or $B\subseteq P$. The nilpotent ideal $k{e}_{21}$ lies in every prime of $T_2(k)$, and the quotient is $k\times k$. Hence the two primes areHere $\mathcal{C}(P_a)=\{a\ne 0\}$ and $\mathcal{C}(P_d)=\{d\ne 0\}$.

Direct multiplication of triplesshows that all of $\mathcal{C}(P_a)$ satisfies the left Ore equations. For $P_d$, the largest left Ore subset isIndeed units always form an Ore set. If $d\ne 0$ but $a=0$, applying the Ore equation successively to $e_{21}$ and $e_{22}$ gives incompatible equations, so no left Ore subset can contain that element.

The prime radical is the intersection of all prime ideals of a noncommutative ring. For a nonzero ring satisfying the ascending chain condition on two-sided ideals, there exist prime ideals of a noncommutative ring $P_1,\dots ,P_r$ with $P_1\cdots P_r=0$. Indeed, a maximal counterexample among two-sided ideals could not itself be a prime ideal of a noncommutative ring, and two larger witness ideals would contradict its maximality. Every prime ideal of a noncommutative ring contains one $P_i$, so the intersection equals ${\bigcap }_i{P}_i$, and its $r$th power is zero. This nilpotence assertion fails without a chain condition.

A prime ring is a nonzero ring whose zero ideal is a prime ideal of a noncommutative ring. Equivalently, the product of any two nonzero two-sided ideals is nonzero. This condition is weaker than absence of zero divisors: a full matrix ring over a division ring is prime even when its size is greater than one. For a commutative ring, primeness is exactly being an integral domain.