Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 2 / 1

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

The commutative Laurent polynomial ring $B$ is an integral domain and $\sigma$ is invertible. Write nonzero elements asThe largest exponent in their product is $m+n$, with coefficientThere is only one pair of exponents contributing at this degree, and the coefficient cannot vanish in the integral domain $B$. Hence $rs\ne 0$. Equivalently, $rs=0$ forces $r=0$ or $s=0$. This is skew Laurent extensions of domains are domains, and the argument works over the original field in any characteristic.

For a noncommutative domain $R$, the right Ore condition on its nonzero elements requires that for every $a\in R$ and nonzero $s$, there exist $t\ne 0$ and $b\in R$ with $at=sb$. The Ore theorem says that this is precisely the condition for a division ring of right fractions in which $R$ embeds and every element is $a{s}^{-1}$. The left-handed condition similarly gives left fractions.

For completeness, a right Noetherian domain satisfies the condition. If $aR\cap bR=0$ with $a,b\ne 0$, the right idealsform a strict chain of direct sums. Indeed, in a relation ${\sum }_{i=0}^n{a}^{i}b{r}_i=0$, the first term lies in $bR$ and the rest in $aR$. Their zero intersection makes $b{r}_0=0$, hence $r_0=0$; cancel $a$ and repeat. Every new summand is nonzero in a noncommutative domain, contradicting the ascending chain condition. Thus $au=bv\ne 0$ exists, with $u,v\ne 0$, giving right common multiples. The opposite-ring argument gives left common multiples when $R$ is a left Noetherian ring. This proves right Noetherian domains satisfy the Ore condition.

A sketch of the fraction construction also explains the theorem. Pairs $(a,s)$ represent $a{s}^{-1}$. They represent the same fraction when a common right denominator $su=tv\ne 0$ also satisfies $au=bv$. Addition uses $(au+bv)(su{)}^{-1}$. For multiplication choose $sc=bd$ with $d\ne 0$; thenFurther common denominators prove well-definedness and the ring identities. The original map is injective because all nonzero denominators are regular; a zero fraction numerator would satisfy $au=0$ for $u\ne 0$. Every nonzero fraction has inverse $s{a}^{-1}$, so the ring is a division ring. This construction has the universal property of inverting all nonzero elements.

Apply this to the two-sided Noetherian noncommutative domain $kG$. Its left Ore property expresses any right fraction as a left one: choose $u\ne 0$ with $ua=bs$, yielding $a{s}^{-1}=u^{-1}b$. Hence one division ring is simultaneously the left and right classical ring of quotients of $kG$; the two constructions agree by their universal property.