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An associative algebra is an algebra over a field whose multiplication satisfies $(ab)c=a(bc)$. Unless stated otherwise, the algebras considered here have a multiplicative identity.

An algebraic derivation of an algebra over a field is a linear map satisfying the Leibniz rule for its multiplication: $D(xy)=D(x)y+xD(y)$. This definition makes sense even when the multiplication is not associative. A derivation of an algebra uses an associative product; a derivation of a Lie algebra uses the Lie bracket as its product.

Over an algebra over a field whose scalar field is $\mathbb{Q}$, the series $\exp (X)-1$ defines an isomorphism from the formal additive group to the formal multiplicative group, with inverse $\log (1+X)$.

A presentation of an algebra over a field with finitely many generators and relations. The Hilbert basis theorem ensures that every finitely generated algebra over a field admits such a presentation. Its module of Kähler differentials is presented by the Jacobian matrix of the relations.

An algebra over a field acted on by a bialgebra is a module algebra when multiplication and unit are module maps: the displayed product identity holds and $h1=\epsilon (h)1$. This expresses the product rule determined by comultiplication.

A normed algebra is an algebra over a field with a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra, since multiplication extends continuously. Completeness is additional structure, rather than part of the definition of a normed algebra.

For an invertible braiding $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative algebra over a field on matrix coefficients $t_{ij}$, with $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$ and $\epsilon (t_{ij})={\delta }_{ij}$. Impose precisely the coefficients ofThis quotient is the FRT bialgebra, or Faddeev-Reshetikhin-Takhtajan construction. The relations form a coideal because composing two coactions that preserve $C$ again preserves $C$; the trivial coaction also preserves it, so the counit descends. Universality follows because any such coaction must satisfy exactly these coefficient relations.

For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta (e_1)=e_1\otimes a+e_2\otimes c$ and $\delta (e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, areThese are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the quantum determinant equal to one. They therefore define mutually inverse algebra homomorphisms: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix comultiplication and counit are the same, so the isomorphism is also one of bialgebras.

Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}{e}_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta (e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}{e}_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an FRT bialgebra.

With the right-comodule braiding convention in part (a), the defining formula isConsequently the scalar form on generators is the following table, with row the first argument and column the second:For example, the extra $e_2\otimes e_1$ component of $C(e_2\otimes e_1)$ is $r(b,c)=q^{-1/2}(q-q^{-1})$. The other off-diagonal component gives $r(d,a)=q^{-1/2}$.

For arbitrary elements of the FRT bialgebra, extend this table by bilinearity, the unit rule $r(1,z)=r(z,1)=\epsilon (z)$, andThese are an explicit recursive algorithm for any two words in the generators, so the table determines the entire cobraiding. The braid equation for $C$ ensures the algorithm annihilates the defining coefficient relations in either argument: moving one crossing past two successive crossings gives precisely the two reductions of a quadratic colinearity relation. Thus it descends from the free algebra over a field to the FRT bialgebra. For the convolution inverse, use the inverse of $R=\tau C$, where $\tau$ flips the tensor factors: its coefficients are $\bar{r}(t_{ki},t_{lj})=(R^{-1}{)}_{kl,ij}$. Equivalently, in the flipped coefficient formula use $\tau C^{-1}\tau$, not $C^{-1}$ alone. This gives $\bar{r}(a,a)=\bar{r}(d,d)=q^{-1/2}$, $\bar{r}(a,d)=\bar{r}(d,a)=q^{1/2}$, $\bar{r}(b,c)=-q^{1/2}(q-q^{-1})$, and all other generator values zero. This specifies the coquasitriangular structure completely, not just its values on degree-one elements.

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

The real Stone-Weierstrass theorem states that if $X$ is compact Hausdorff and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$. Let $B$ be its uniform closure, still an algebra over a field. If $h\in B$ and $M\ge \Vert h{\Vert }_{\infty }$, scale the permitted polynomial approximation of absolute value on $[-1,1]$ to $[-M,M]$. Applying these polynomials to $h$ proves $|h|\in B$. Thereforebelong to $B$: it is a lattice.

Fix $f\in C(X,\mathbb{R})$ and $\epsilon >0$. For every $x,y\in X$, point separation and an affine adjustment give $g_{x,y}\in A$ with $g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$; if $x=y$, a constant suffices. For fixed $x$, the open sets where $g_{x,y}>f-\epsilon$ cover $X$. Select finitely many and take their maximum $h_x\in B$. Then $h_x>f-\epsilon$ everywhere and $h_x(x)=f(x)$. By continuity, $h_x<f+\epsilon$ on some neighborhood $V_x$ of $x$. Choose finitely many $V_x$ covering $X$ and take the minimum of the corresponding $h_x$. The result $h\in B$ satisfies $f-\epsilon <h<f+\epsilon$ everywhere. Since $B$ is closed and $\epsilon$ arbitrary, $f\in B$. Thus $\overline{A}=C(X,\mathbb{R})$, proving the theorem.

The formal additive group and formal multiplicative group have lawsIf $R$ is an algebra over a field whose scalar field is $\mathbb{Q}$, the serieshas linear coefficient one and satisfiesIt is therefore the exponential isomorphism between the additive and multiplicative formal groups, with inverse $\log (1+X)$.

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 $YX=qXY$.

A linear subspace of an algebra over a field closed under multiplication is a subalgebra. If a unital subalgebra is intended, it must also contain the ambient identity. A closed subalgebra of a normed algebra is closed in the norm topology.