2 Algebra
2.1 Tensor Product
Let \(R\) be a domain and \(M, N\) two \(R\)-semimodules. If \(f\) and \(g\) are linearly independent families of points in \(M\) and \(N\), then \((i, j) \mapsto f i \otimes g j\) is a linearly independent family of points in \(M \otimes N\).
We will prove the equivalent statement:
Let \(P, Q\) be two free \(R\) modules, \(f : P \to M\) and \(g : Q \to N\) be two \(R\)-linear injective maps. Then \(f \otimes g : P \otimes _R Q \to M \otimes _R N\) is injective.
Let \(K\) be the field of fractions of \(R\).
The map
is injective because \(R \to K\) is injective and all the modules involved are flat. The map
is injective because all the modules involved are \(K\)-flat (as \(K\) is a field).
\(P \otimes _R Q \to M \otimes _R N\) is now a factor of the composition of the two injections above, and is thus is injective.
2.2 Affine Monoids
Multivariate Laurent polynomials over an integral domain are an integral domain.
Come on.
An affine monoid is a finitely generated commutative monoid which is:
cancellative: if \(a + c = b + c\) then \(a = b\), and
torsion-free: if \(n a = n b\) then \(a = b\) (for \(n \geq 1\)).
If \(M\) is an affine monoid, then \(M\) can be embedded inside \({\mathbb Z}^n\) for some \(n\).
Embed \(M\) inside its Grothendieck group \(G\). Prove that \(G\) is finitely generated free.
If \(R\) is an integral domain \(M\) is an affine monoid, then \(R[M]\) is an integral domain and is a finitely generated \(R\)-algebra.
\(i : R[M] \hookrightarrow R[{\mathbb Z}M]\) injects into an integral domain so is an integral domain. It’s finitely generated by \(\chi ^{a_i}\) where \(\mathcal A = \{ a_1, \dotsc , a_s\} \) is a finite generating set for \(M\).
An element \(x\) of a monoid \(M\) is irreducible if \(x = y + z\) implies \(y = 0\) or \(z = 0\).
If \(M\) is a monoid with a single unit, and \(S\) is a set generating \(M\), then \(S\) contains all irreducible elements of \(M\).
Assume \(p\) is an irreducible element. Since \(S\) generates \(M\), write
where the \(a_i\) are finitely many elements (not necessarily distinct) elements of \(S\). Since \(p\) is irreducible, we must have
for some \(i\).
If \(M\) is a finitely generated monoid with a single unit, then only finitely many elements of \(M\) are irreducible.
Let \(S\) be a finite set generating \(M\). Write \(I\) the set of irreducible elements. By Proposition 2.2.6, \(I \subseteq S\). Hence \(I\) is finite.
If \(M\) is a finitely generated cancellative monoid with a single unit, then \(M\) is generated by its irreducible elements.
We do not follow the proof from [ 1 ] .
Let \(S\) be a finite minimal generating set and assume for contradiction that \(r \in S\) is reducible, say \(r = a + b\) where \(a, b\) are non-units. Write
for some \(m_s, n_s \in {\mathbb N}\), so that
We distinguish three cases
\(m_r + n_r = 0\). Then
\[ r = \sum _{s \in S \setminus \{ r\} } (m_s + n_s) s \in \langle S \setminus \{ r\} \rangle \]contradicting the minimality of \(S\).
\(m_r + n_r = 1\). Then
\begin{align*} & 0 = \sum _{s \in S \setminus \{ r\} } (m_s + n_s) s & \implies \forall s \in S \setminus \{ r\} , m_s s = n_s s = 0 \end{align*}Furthermore, either \(m_r = 0\) or \(n_r = 0\), so \(a = 0\) or \(b = 0\), contradicting the fact that \(a\) and \(b\) are non-units.
\(m_r + n_r \ge 2\). Then
\[ 0 = r + \sum _{s \in S \setminus \{ r\} } (m_s + n_s) s \]and \(r = 0\), contradicting the minimality of \(S\) once again.
2.3 Hopf algebras
2.3.1 Ideals and quotients
Let \(R\) be a commutative ring and \((C,\Delta ,\varepsilon )\) be a coalgebra over \(R\). An \(R\)-submodule \(I\) of \(C\) is a coideal of \(C\) if \(\Delta (I) \subseteq \bar{I \otimes _R C} + \bar{C \otimes _R I}\) and \(\varepsilon (I)=0\), where \(\bar{\cdot }\) denotes the image in \(C \otimes _R C\).
If \(C\) is a coalgebra over \(R\) and \(I\) is a coideal, the quotient \(C /I\) is equipped with a canonical \(R\)-coalgebra structure.
Straightforward.
If \(C\) is a coalgebra over \(R\) and \(I\) is a coideal, the quotient map \(C \to C / I\) is a coalgebra homomorphism.
Straightforward.
Let \(B\) be a bialgebra over a commutative ring \(R\). A bialgebra ideal \(I\) is an ideal which is also a coideal.
If \(B\) is a bialgebra over \(R\) and \(I\) is a bialgebra ideal, the quotient \(B / I\) is equipped with a canonical \(R\)-bialgebra structure.
Straightforward.
If \(B\) is a bialgebra over \(R\) and \(I\) is a bialgebra ideal, the quotient map \(B \to B / I\) is a bialgebra homomorphism.
Straightforward.
Let \(A\) be a Hopf algebra over a commutative ring \(R\). A Hopf ideal \(I\) is a bialgebra ideal such that \(S(I)=I\).
If \(A\) is a Hopf algebra over \(R\) and \(I\) is a Hopf ideal, the quotient \(A / I\) is equipped with a canonical Hopf algebra structure over \(R\).
Straightforward.
If \(A\) is a Hopf algebra over \(R\) and \(I\) is a Hopf ideal, the quotient map \(A \to A / I\) is a Hopf algebra homomorphism.
Follows immediately from Proposition 2.3.6.
2.3.2 Group algebras
Let \(R\) be a commutative ring. Let \(G, H\) be abelian groups and \(f : G \to H\) an injective group hom. Then \(R[H]\) is a free \(R[G]\)-module.
Pick a section \(\sigma : H / f(G) \to H\) and the unique map \(\varphi : H \to G\) such that \(h = \sigma (h f(G)) f(\varphi (h))\). We claim that \(R[H]\) is isomorphic to \(R[G]^{\oplus H / f(G)}\), from which the result follows, as such:
Those two functions are clearly inverse to each other, and the forward map is clearly \(R[G]\)-linear.
Let \(G\) be an abelian group generated by a set \(S\). Let \(A, B\) be arbitrary indexing types and \(f : A \to B\) a function. Write \(f^\oplus : G^{\oplus A} \to G^{\oplus B}\) the pushforward. Then
Write \(I = \operatorname{span}\{ gX^a_1 - gX^a_2 | g \in G, a_1, a_2 \in A, f(a_1) = f(a_2)\} \) for brevity.
Note that we can assume WLOG that \(f\) is surjective. Write \(\sigma : B \to A\) a section of \(f\).
Let’s prove by induction on \(x \in G^\oplus A\) that \(\sigma ^\oplus (f^\oplus (x)) \equiv x \mod I\):
\(x = 0\): \(\sigma ^\oplus (f^\oplus (0)) = 0\)
\(x = gX^a\): \(\sigma ^\oplus (f^\oplus (gX^a)) = gX^{\sigma (f(a))} \equiv gX^a \mod I\) as \(S\) generates
\(x + y\): Assume the induction hypothesis for \(x\) and \(y\). Then
\[ \sigma ^\oplus (f^\oplus (x + y)) = \sigma ^\oplus (f^\oplus (x)) + \sigma ^\oplus (f^\oplus (y)) \equiv x + y \mod I \]
Now, for any \(x \in G^\oplus A\),
and we are done.
Let \(R\) be a commutative ring. Let \(M\) be a commutative monoid and \(M'\) be its localization at some \(S \subseteq M\). Then \(R[M']\) is the localization of \(R[M]\) at \(\operatorname{span}\{ X^s | s \in S\} \).
Straightforward.
2.3.3 Group-like elements
An element \(a\) of a coalgebra \(A\) is group-like if \(\eta (a) = 1\) and \(\Delta (a) = a \otimes a\), where \(\eta \) is the counit and \(\Delta \) is the comultiplication map.
We write \(\operatorname{GrpLike}A\) for the set of group-like elements of \(A\).
Group-like elements \(\operatorname{GrpLike}A\) of a bialgebra \(A\) form a monoid.
Group-like elements \(\operatorname{GrpLike}A\) of a Hopf algebra \(A\) form a group.
Check that group-like elements are closed under unit, multiplication and inverses.
Let \(f : A \to B\) be a bi-algebra hom. If \(a \in A\) is group-like, then \(f(a)\) is group-like too.
\(a\) is a unit, so \(f(a)\) is a unit too. Then
so \(f(a)\) is group-like.
If \(R\) is a commutative semiring, \(A\) is a Hopf algebra over \(R\) and \(G\) is a group, then every element of the image of \(G\) in \(A[G]\) is group-like.
This is an easy check.
If \(R\) is a commutative semiring, \(A\) is a Hopf algebra over \(R\) and \(G\) is a group, then the group-like elements in \(A[G]\) span \(A[G]\) as an \(A\)-module.
This follows immediately from 2.3.16.
The group-like elements in a bialgebra \(A\) over a domain are linearly independent.
Let’s prove that any finite set \(s\) of group-like elements is linearly independent, by induction on \(s\).
\(\emptyset \) is clearly linearly independent.
Assume now that the finite set \(s\) of group-like elements is linearly independent, that \(a \notin s\) is group-like, and let’s show that \(s \cup \{ a\} \) is linearly independent too.
Assume there is some \(c : A \to R\) such that \(\sum _{x \in s} c_x x = c_a a\). Since \(a\) and all elements of \(s\) are group-like, we compute
By Lemma 2.1.1, the \(x \otimes y\) are linearly independent and therefore \(c_x ^2 = c_a c_x\) and \(c_x c_y = 0\) if \(x \ne y\).
If \(c_x = 0\) for all \(x \in s\), then we are clearly done. Else find \(x \in s\) such that \(c_x \ne 0\). From the above two equations, we get that \(c_x = c_a\) and \(c_y = 0\) for all \(y \in s, y \ne x\). Therefore
and \(x = a\). Contradiction.
Let \(R\) be a domain. The group-like elements of \(R[M]\) are exactly the image of \(M\).
See Lemma 12.4 in [ 2 ] .
Let \(R\) be a domain, \(G\) a commutative group and \(A\) a \(R\)-bialgebra. Then bialgebra homs \(R[G] \to A\) are in bijection with group homs \(G \to \operatorname{GrpLike}A\).
If \(f : G \to \operatorname{GrpLike}A\) is a group hom, then we get
This is clearly an algebra hom, so for it to be a bialgebra hom we only need to check comultiplication is preserved. We only need to check this on \(g \in G\), in which case
since \(f(g) \in \operatorname{GrpLike}A\).
If \(f : R[G] \to A\) is a bialgebra hom, then it restricts to a group hom \(\operatorname{GrpLike}R[G] \to \operatorname{GrpLike}A\) by Proposition 2.3.15. Now use that \(\operatorname{GrpLike}R[G] \cong G\) from Proposition 2.3.19.
Let \(A\) be a Hopf algebra, \(H\) be a subgroup of \(\operatorname{GrpLike}A\) and
be an ideal. Then \(I\) is a Hopf ideal.
It suffices to check the conditions of a Hopf ideal on generators.
For the comultiplication condition:
For the counit condition:
Finally, for the antipode condition:
2.3.4 Diagonalizable bialgebras
A bialgebra is called diagonalizable if it is isomorphic to a group algebra.
A diagonalizable bialgebra is spanned by its group-like elements.
This is true for a group algebra by 2.3.17, and the property of being spanned by its group-like elements is preserves by isomorphisms of bialgebras.
Let \(A\) be a bialgebra over a domain \(R\), let \(G\) be a subgroup of \(\operatorname{GrpLike}(A)\) (which is a monoid by 2.3.14). If \(A\) is generated by \(G\), then the unique bialgebra morphism from \(R[G]\) to \(A\) sending each element of \(G\) to itself is bijective.
This morphism is injective by the linear independence of group-like elements (2.3.18), and surjective by assumption.
Let \(R\) be a domain, \(G\) a commutative group, \(A\) a \(R\)-bialgebra and \(f : R[G] \to A\) a surjective bialgebra hom. Then \(R[f(G)] \cong A\) as bialgebras.
Note that \(R[G] \xrightarrow f A\) factors as \(R[G] \xrightarrow f R[f(G)] \xrightarrow \phi A\), where \(f(G)\) is a group by Proposition 2.3.14.
Since \(R[G] \xrightarrow f A\) is surjective, so is \(R[f(G)] \xrightarrow \phi A\). Therefore Proposition 2.3.24 applies to \(f(G)\) inside \(A\), and we get \(R[f(G)] \cong A\).
A bialgebra over a domain is diagonalizable if and only if it is spanned by its group-like elements.
Proposition 2.3.24 and Corollary 4.2.4 are false over a general commutative ring. Indeed, let \(R\) be a commutative ring and let \(G\) be a group. Then the group-like elements of \(R[G]\) correspond to locally constant maps from \(Spec R\) to \(G\) (with the discrete topology), hence they are of the form \(e_1 g_1+\cdots +e_r g_r\), with the \(g_i\) in \(G\) and \(e_1,\ldots ,e_r\) a family of pairwise orthogonal idempotent elements of \(R\) that sum to \(1\). So \(R[G]\) is not isomorphic to the group algebra over its group-like elements unless \(Spec R\) is connected. As for the corollary, a bialgebra of the form \(R_1[G_1]\times \cdots \times R_n[G_n]\), seen as a bialgebra over \(R_1\times \cdots \times R_n\), is generated by its group-like elements but not diagonalizable.
2.3.5 The group algebra functor
If \(A\) is a \(R\)-Hopf algebra, then the antipode map \(s : A \to A\) is anti-commutative, ie \(s(a * b) = s(b) * s(a)\). If further \(A\) is commutative, then \(s(a * b) = s(a) * s(b)\).
Any standard reference will have a proof.
The category of \(R\)-bialgebras is equivalent to comonoid objects in the category of \(R\)-algebras.
Turn the arrows around.
The category of \(R\)-Hopf algebras is equivalent to cogroup objects in the category of \(R\)-algebras.
Turn the arrows around. Most of the diagrams have been turned around in Proposition 2.3.28 already.
For a commutative ring \(R\), we have a functor \(G \rightsquigarrow R[G] : \operatorname{Grp}\to \operatorname{Hopf}_R\).
Let \(R\) be a domain. The functor \(G \rightsquigarrow R[G]\) from the category of groups to the category of Hopf algebras over \(R\) is fully faithful.
The functor is clearly faithful. Now for the full part, if \(f : R[G] \to R[H]\) is a Hopf algebra hom, then we get a series of maps
and each map separately is clearly multiplicative.