= Solution
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 $F_{-1}=0$ and
$$
F_n=\sum_{i=0}^n Ax^i=\sum_{i=0}^n x^iA.
$$
The 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 $F_nF_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving <automorphism>.
For a <right ideal> $I\subseteq B$, define
$$
L_n=\{a\in A:ax^n\in I+F_{n-1}\}.
$$
Each $L_n$ is a <right ideal> of $A$. In fact, if $a\in L_n$ and $c\in A$, write $cx^n=x^nb+u$ with $b\in A$, $u\in F_{n-1}$; then $acx^n=(ax^n)b+au\in I+F_{n-1}$. 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 $L_n$, $0\leq n\leq N$, 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 $f=ax^m+u$ 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 $c_jx^d=x^db_j+u_j$ with $u_j\in F_{d-1}$. The difference
$$
f-\sum_j f_{dj}b_jx^{m-d}
$$
lies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence \b[$B$ 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 $k[X^{\pm1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm1}][Y;\sigma]$, where $\sigma(X)=qX$. On a <monomial>, $Y^{-1}X^iY^j=q^{-i}X^iY^{j-1}$; when $j=0$ this is in $AY^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving \b[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 that
$$
\boxed{N=P_1\cap\cdots\cap P_r,\qquad N^r=0.}
$$
Indeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and $\bigcap_iP_i\subseteq Q$ 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> $k[z_1,z_2,\ldots]/(z_i^2:i\geq1)$ the <nilradical> is $(z_1,z_2,\ldots)$, its only <prime ideal>, but the product of any number of distinct $z_i$ is nonzero. Thus this <nilradical> is not a <nilpotent ideal>.
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