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Use $q\in k^{\times }$ and the quantum plane convention $yx=qxy$. In this convention $S{L}_q(2)$ is the coordinate Hopf algebra of quantum SL2 with the catalog parameter replaced by $q^{-1}$. It is generated by $a,b,c,d$, subject toEquivalently the last determinant relation is $da-qbc=1$. The quantum determinant convention is essential: these relations use the inverse of the parameter in the catalog's $ab=qba$ convention.

Writing $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the comultiplication is matrix multiplication, $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$. Explicitly,The counit is $\epsilon (a)=\epsilon (d)=1$, $\epsilon (b)=\epsilon (c)=0$. Extend $\Delta$ and $\epsilon$ as unital algebra homomorphisms over a field. Matrix multiplication is associative, giving coassociativity; multiplication by the identity matrix gives the counit laws. The antipode isFor example $(TS(T){)}_{11}=ad-q^{-1}bc=1$ and $(S(T)T{)}_{11}=da-qbc=1$, while the off-diagonal entries vanish by the commutation relations. Thus this is a Hopf algebra, not merely an algebra presentation.

Take the right coaction of the coordinate Hopf algebra of quantum SL2 on the quantum plane to beIt is extended multiplicatively and sends $1$ to $1\otimes 1$. The matrix comultiplication and counit verify the coaction identities on $x,y$. To check that it descends to the quantum plane, expandThe last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.

For the requested calculation, $ca=qac$ and $yx=qxy$ giveMultiply by $\delta (y)$, put all monomials into the order $x^r{y}^s$, and use $bc=cb$. The answer isA left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

Use the quantum plane $A=k⟨x,y⟩/(yx-qxy)$ and the quantum enveloping algebra of sl2 comultiplication from Question 2. The generator actions are $Kx=qx$, $Ky=q^{-1}y$, $Ex=0$, $Ey=x$, $Fx=y$, $Fy=0$. Extend them by the module algebra rulesThese descend to $A$: $E(yx)=q{x}^2=E(qxy)$ and $F(yx)=q{y}^2=F(qxy)$, while $K$ preserves $yx-qxy$. For the ordered monomial basis, induction with $[s+1]=q^{-1}[s]+q^s$ gives the quantum enveloping algebra action on the quantum planeTerms with a negative exponent have zero coefficients and are interpreted as zero.

We now verify the defining relations explicitly on every basis vector, rather than infer them merely from the generator actions. The $K$ weight changes by $q^2$ under $E$ and $q^{-2}$ under $F$, so $KE=q^{2}EK$ and $KF=q^{-2}FK$. AlsoThe middle identity follows by inserting $[j]=(q^j-q^{-j})/(q-q^{-1})$ and cancelling the terms $q^{r+s+1}$ and $q^{-r-s-1}$; the remaining numerator is $(q-q^{-1})(q^{r-s}-q^{s-r})$. Finally $K{K}^{-1}=1$. Thus all the rank-one defining relations hold. The phrase “Serre relations” here refers to these rank-one presentation relations; there are no additional relations involving distinct simple roots for ${\mathfrak{sl}}_2$.

The root-of-unity specialization requires $q\ne 1$: we take $q$ to be a primitive cube root of unity. The displayed presentation contains $(q-q^{-1}{)}^{-1}$, so substituting $q=1$ literally is undefined. Its classical $U({\mathfrak{sl}}_2)$ limit instead has simple finite-dimensional modules of arbitrarily large dimension. This is why the primitive-root qualification matters.

First consider simple modules occurring in the quantum plane. Here $[3]=0$, $K^3=1$, andwith the analogous formula $F^3=0$. Thus the action factors through the finite-dimensional quotient with $E^3=F^3=0$, $K^3=1$, spanned by $F^i{K}^j{E}^l$ with $0\le i,j,l<3$. Any simple module over this quotient is finite-dimensional, since it is generated by one vector. In a nonzero simple submodule choose a $K$ eigenvector $v$ in $\ker E$. This is possible because $E$ is nilpotent and its kernel is $K$-invariant. Then the highest-weight string for quantum sl2 shows thatis invariant under $E,K$, and $F^{3}v=0$ makes it $F$-invariant. Simplicity gives $W=V$, proving $\dim V\le 3$.

The same bound holds even for arbitrary finite-dimensional simple modules over the unrestricted quantum enveloping algebra of sl2. Indeed $E^3,F^3,K^3$ are central: the string commutation identity has a factor $[3]=0$, and the $K$ commutation factors satisfy $q^6=1$. By Schur lemma, they act as scalars $\alpha ,\beta ,\kappa$, with $\kappa \ne 0$. The elementis also central. For example $[E,FE]=(K-K^{-1})E/(q-q^{-1})$, while $[E,\varphi (K)]=[\varphi (q^{-2}K)-\varphi (K)]E=-(K-K^{-1})E/(q-q^{-1})$; the $F$ commutator cancels in the same way. Let $\Omega$ act as $\gamma$.

If $\alpha =0$, choose the highest vector $v$ as above. Its three-vector $F$ string is invariant, now using $F^{3}v=\beta v$, so again its dimension is at most three. If $\alpha \ne 0$, choose a $K$ eigenvector $v$. The space $W=⟨v,Ev,E^{2}v⟩$ is stable under $E,K$ because $E^3=\alpha$, and it is stable under $F$ becauseIt therefore equals the simple module and has dimension at most three.

Over $\mathbb{C}$ this also excludes infinite-dimensional simple modules of the unrestricted algebra. It is a finite module over the central finitely generated commutative algebra over a field $Z_0=\mathbb{C}[E^3,F^3,K^3,K^{-3}]$, using the same 27 ordered spanning elements. A simple module $M$, generated by one vector, is consequently finite over $Z_0$. Choose a maximal ideal $\mathfrak{m}$ in its support. Nakayama lemma makes $M/\mathfrak{m}M\ne 0$; since $\mathfrak{m}M$ is a submodule, simplicity gives $\mathfrak{m}M=0$. The Hilbert Nullstellensatz gives $Z_0/\mathfrak{m}=\mathbb{C}$, so $M$ is finite-dimensional and the preceding argument applies. Thus the usual complex, primitive-root interpretation yields

At a primitive cube root of unity, $[1]=1$, $[2]=-1$, $[3]=0$. In the degree-three quantum plane module, put $z_0=x^3$, $z_1=x^{2}y$, $z_2=x{y}^2$, $z_3=y^3$. The complete actions areHence $W=⟨z_0,z_3⟩$ is a two-dimensional trivial submodule; every line in $W$ is a simple module. The quotient by $W$ is a two-dimensional simple module, with distinct $K$ eigenvalues $q,q^{-1}$ and nonzero maps connecting its two weight spaces.

Let $S$ be a simple module that is a submodule of this cubic component. If $S\cap W\ne 0$, simplicity makes $S\subseteq W$, so $S$ is one of its lines. If $S\cap W=0$, the quotient map injects $S$ into the simple quotient. Its nonzero image is the entire quotient, so $\dim S=2$. Since $K$ is diagonalizable, such an $S$ must contain the unique $q$ weight space $\mathbb{C}{z}_1$. But $E{z}_1=z_0\ne 0$, which then belongs to $S\cap W$, a contradiction. Therefore the cubic-root quantum-plane socle gives the complete answer:All simple submodules lie in the two-dimensional $W$, so their sum cannot equal the four-dimensional cubic component. It is not a direct sum of simple modules. The obstruction is the nonzero extension to the simple two-dimensional quotient, not a failure of that quotient to be simple.

The printed input $x^r{x}^s$ is a genuine typo in the PDF as well as in the TeX: $E$ annihilates every power of $x$, whereas the displayed right-hand side is generally nonzero. The intended input is $x^r{y}^s$, and the calculation below proves that corrected identity.

The given comultiplications make $C_q$ a coalgebra, with counit one on $K,K^{-1},I$ and zero on $E,F$. Write $I_0=p(I)$, $K_0=p(K)$ and $E_0=p(E)$, and similarly for $F$. The measuring coalgebra identities areThe group-like elements act multiplicatively. Although the statement omits an explicit specification of $p(I)$, its identity action follows from the other data and the quantum plane relation. Applying $E_0$ to $yx=qxy$ gives $q{x}^2=q{I}_0(x)x$, so cancellation in the noncommutative domain gives $I_0(x)=x$. Applying $F_0$ to the same relation gives $q{y}^2=qy{I}_0(y)$, so $I_0(y)=y$. Thus $I_0$ is the identity on the entire quantum plane.

Now $E_0(x^r)=0$ by induction. Suppose $E_0(y^{s-1})=[s-1]x{y}^{s-2}$. The measuring rule and $y^{s-1}x=q^{s-1}x{y}^{s-1}$ giveThe induction starts with $E_0(y)=x$ and $[1]=1$. A final product rule givesFor $s=0$ the answer is zero. By contrast the literal printed input has $\boxed{p(E)(x^r{x}^s)=0}$; it cannot satisfy the printed right-hand side for $s=1$.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

An $R$-point of a unital algebra over a field $A$ is a unital algebra homomorphism over a field $A\to R$. For the quantum plane over the complex numbers, scalar points lie on the two coordinate axes when $q\ne 1$, while matrix-valued points are pairs $X,Y$ with $YX=qXY$.

For $yx=qxy$, the quantum enveloping algebra of sl2 acts by $K(x^r{y}^s)=q^{r-s}{x}^r{y}^s$, $E(x^r{y}^s)=[s{]}_q{x}^{r+1}{y}^{s-1}$ and $F(x^r{y}^s)=[r{]}_q{x}^{r-1}{y}^{s+1}$. Negative exponents here have zero coefficients and are omitted. The quantum integer identity $[r][s+1]-[s][r+1]=[r-s]$ verifies the commutator. These operations preserve total degree and make the quantum plane a module algebra.

For $q\ne 0$, define a vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. Bilinearity of the exponent verifies associativity. This constructs the algebra and proves the basis independently of its presentation. It localizes the quantum plane at powers of $u$ and gives a quotient witness for noncommutativity and noncocommutativity of the coordinate Hopf algebra of quantum SL2.

The two-dimensional quantum torus is the algebra with invertible generators $X,Y$ and relation $YX=qXY$, for $q\in k^{\times }$. Its basis consists of $X^i{Y}^j$ with $i,j\in \mathbb{Z}$, with the same multiplication rule as the quantum plane. It is a right Noetherian ring: successive adjunction of $X^{-1}$, $Y$, and $Y^{-1}$ satisfies the hypotheses of the noncommutative Hilbert basis theorem.