Toric

2 Algebra

2.1 Tensor Product

Lemma 2.1.1 The tensor product of linearly independent families
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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 \(M \otimes N\).

Proof

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

\[ P \otimes _R Q \to (K \otimes _R P) \otimes _R (K \otimes _R Q) = (K \otimes _R P) \otimes _K (K \otimes _R Q) \]

is injective because \(R \to K\) is injective and all the modules involved are flat. The map

\[ (K \otimes _R P) \otimes _K (K \otimes _R Q) \to (K \otimes _R M) \otimes _K (K \otimes _R N) \]

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

Lemma 2.2.1 Multivariate Laurent polynomials are an integral domain
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Multivariate Laurent polynomials over an integral domain are an integral domain.

Proof

Come on.

Definition 2.2.2 Affine monoid

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 \(n \geq 1\)).

Proposition 2.2.3 Embedding an affine monoid inside a lattice

If \(M\) is an affine monoid, then \(M\) can be embedded inside \({\mathbb Z}^n\) for some \(n\).

Proof

Embed \(M\) inside its Grothendieck group \(G\). Prove that \(G\) is finitely generated free.

Proposition 2.2.4 Affine monoid algebras are domains

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.

Proof

\(i : R[M] \hookrightarrow R[{\mathbb Z}M]\) 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 \(M\).

Definition 2.2.5 Irreducible element
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An element \(x\) of a monoid \(M\) is irreducible if \(x = y + z\) implies \(y = 0\) or \(z = 0\).

Proposition 2.2.6 Irreducible elements lie in all sets generating a salient monoid

If \(M\) is a monoid with a single unit, and \(S\) is a set generating \(M\), then \(S\) contains all irreducible elements of \(M\).

Proof

Assume \(p\) is an irreducible element. Since \(S\) generates \(M\), write

\[ p = \sum _i a_i \]

where the \(a_i\) are finitely many elements (not necessarily distinct) elements of \(S\). Since \(p\) is irreducible, we must have

\[ p = a_i \in S \]

for some \(i\).

Proposition 2.2.7 A salient finitely generated monoid has finitely many irreducible elements

If \(M\) is a finitely generated monoid with a single unit, then only finitely many elements of \(M\) are irreducible.

Proof

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.

Proposition 2.2.8 A salient finitely generated cancellative monoid is generated by its irreducible elements

If \(M\) is a finitely generated cancellative monoid with a single unit, then \(M\) is generated by its irreducible elements.

Proof

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

\[ a = \sum _{s \in S} m_s s, b = \sum _{s \in S} n_s s \]

for some \(m_s, n_s \in {\mathbb N}\), so that

\[ r = \sum _{s \in S} (m_s + n_s) s. \]

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

Definition 2.3.1 Coideal
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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 \(C \otimes _R C\).

Proposition 2.3.2 Quotient coalgebra

If \(C\) is a coalgebra over \(R\) and \(I\) is a coideal, the quotient \(C /I\) is equipped with a canonical \(R\)-coalgebra structure.

Proof

Straightforward.

Proposition 2.3.3 Quotient coalgebra map

If \(C\) is a coalgebra over \(R\) and \(I\) is a coideal, the quotient map \(C \to C / I\) is a coalgebra homomorphism.

Proof

Straightforward.

Definition 2.3.4 Bialgebra ideal

Let \(B\) be a bialgebra over a commutative ring \(R\). A bialgebra ideal \(I\) is an ideal which is also a coideal.

Proposition 2.3.5 Quotient bialgebra

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.

Proof

Straightforward.

Proposition 2.3.6 Quotient bialgebra map

If \(B\) is a bialgebra over \(R\) and \(I\) is a bialgebra ideal, the quotient map \(B \to B / I\) is a bialgebra homomorphism.

Proof

Straightforward.

Definition 2.3.7 Hopf ideal

Let \(A\) be a Hopf algebra over a commutative ring \(R\). A Hopf ideal \(I\) is a bialgebra ideal such that \(S(I)=I\).

Proposition 2.3.8 Quotient Hopf algebra

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 \(R\).

Proof

Straightforward.

Proposition 2.3.9 Quotient Hopf algebra map

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.

Proof

Follows immediately from Proposition 2.3.6.

2.3.2 Group algebras

Proposition 2.3.10 Freeness of group algebras under an injective hom

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.

Proof

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:

\begin{align*} R[G]^{\oplus H / f(G)} & \simeq R[H] \\ \varphi (h) e_{h f(G)} & \mapsto h g e_x & \mapsfrom \sigma (x) f(g) \end{align*}

Those two functions are clearly inverse to each other, and the forward map is clearly \(R[G]\)-linear.

Proposition 2.3.11 The kernel of a map on direct sums

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

\[ \ker f^\oplus = \operatorname{span}\{ gX^a_1 - gX^a_2 | g \in S, a_1, a_2 \in A, f(a_1) = f(a_2)\} . \]
Proof

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 \(x \in G^\oplus A\),

\[ x \in \ker f^\oplus \iff f^\oplus (x) = 0 \iff \sigma ^\oplus (f^\oplus (x)) \equiv 0 \mod I \iff x \equiv 0 \mod I \]

and we are done.

Proposition 2.3.12 Localising a monoid algebra

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 \(\operatorname{span}\{ X^s | s \in S\} \).

Proof

Straightforward.

2.3.3 Group-like elements

Definition 2.3.13 Group-like elements
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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 \(A\).

Proposition 2.3.14 Group-like elements form a group

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.

Proof

Check that group-like elements are closed under unit, multiplication and inverses.

Lemma 2.3.15 Bialgebra homs preserve group-like elements
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Let \(f : A \to B\) be a bi-algebra hom. If \(a \in A\) is group-like, then \(f(a)\) is group-like too.

Proof

\(a\) is a unit, so \(f(a)\) is a unit too. Then

\[ f(a) \otimes f(a) = (f \otimes f)(\Delta _A(a)) = \Delta _B(f(a)) \]

so \(f(a)\) is group-like.

Lemma 2.3.16
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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.

Proof

This is an easy check.

Lemma 2.3.17

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.

Proof

This follows immediately from 2.3.16.

Lemma 2.3.18 Independence of group-like elements

The group-like elements in a bialgebra \(A\) over a domain are linearly independent.

Proof

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

\begin{align*} \sum _{x, y \in s} c_x c_y x \otimes y & = c_a ^2 a \otimes a \\ & = c_a ^2 \Delta (a) \\ & = c_a \Delta \left(\sum _{x \in s} c_x x\right) \\ & = \sum _{x \in s} c_a c_x \Delta (x) \\ & = \sum _{x \in s} c_a c_x x \otimes x \end{align*}

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

\[ c_x x = \sum _{y \in s} c_y y = c_a a = c_x a \]

and \(x = a\). Contradiction.

Lemma 2.3.19 Group-like elements in a group algebra

Let \(R\) be a domain. The group-like elements of \(R[M]\) are exactly the image of \(M\).

Proof

See Lemma 12.4 in [ 2 ] .

Proposition 2.3.20 Galois connection between group algebra and group-like elements

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 \(G \to \operatorname{GrpLike}A\).

Proof

If \(f : G \to \operatorname{GrpLike}A\) is a group hom, then we get

\begin{align*} R[G] \to A g \mapsto f(g) \end{align*}

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

\[ (f \otimes f)(\Delta (g)) = (f \otimes f)(g \otimes g) = f(g) \otimes f(g) = \Delta (f(g)) \]

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.

Proposition 2.3.21 Quotients by binomial ideals

Let \(A\) be a Hopf algebra, \(H\) be a subgroup of \(\operatorname{GrpLike}A\) and

\[ I = \langle h_1 - h_2 : h_1,h_2 \in H \rangle \]

be an ideal. Then \(I\) is a Hopf ideal.

Proof

It suffices to check the conditions of a Hopf ideal on generators.

For the comultiplication condition:

\begin{align*} \Delta (h_1-h_2) & = \Delta (h_1) - \Delta (h_2) \\ & = h_1 \otimes h_1 - h_2 \otimes h_2 \\ & = h_1 \otimes h_1 - h_1 \otimes h_2 + h_1 \otimes h_2 - h_2 \otimes h_2 \\ & = h_1 \otimes (h_1 - h_2) + (h_1 - h_2) \otimes h_2 \in \bar{A \otimes I} + \bar{I \otimes A}. \end{align*}

For the counit condition:

\[ \varepsilon (h_1 - h_2) = \varepsilon (h_1) - \varepsilon (h_2) = 1 - 1 = 0. \]

Finally, for the antipode condition:

\[ S(h_1 - h_2) = S(h_1) - S(h_2) = h_1^{-1} - h_2^{-1} \in I. \]

2.3.4 Diagonalizable bialgebras

Definition 2.3.22 Diagonalizable bialgebras
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A bialgebra is called diagonalizable if it is isomorphic to a group algebra.

A diagonalizable bialgebra is spanned by its group-like elements.

Proof

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.

Proof

This morphism is injective by the linear independence of group-like elements (2.3.18), and surjective by assumption.

Proposition 2.3.25 Quotient of a diagonalisable bialgebra is diagonalisable

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.

Proof

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.

Proof

We know that a diagonalizable bialgebra is spanned by its group-like elements by 2.3.23, and that a bialgebra over a domain that is spanned by its group-like elements is diagonalizable by 2.3.25 (and by the fact that a bijective morphism of bialgebras is an isomorphism).

Proposition 2.3.24 and Corollary ?? 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

Proposition 2.3.27 The antipode is a antihomomorphism

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 \(s(a * b) = s(a) * s(b)\).

Proof

Any standard reference will have a proof.

Proposition 2.3.28 Bialgebras are comonoid objects in the category of algebras
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The category of \(R\)-bialgebras is equivalent to comonoid objects in the category of \(R\)-algebras.

Proof

Turn the arrows around.

Proposition 2.3.29 Hopf algebras are cogroup objects in the category of algebras

The category of \(R\)-Hopf algebras is equivalent to cogroup objects in the category of \(R\)-algebras.

Proof

Turn the arrows around. Most of the diagrams have been turned around in Proposition 2.3.28 already.

Definition 2.3.30 The group algebra functor
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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.

Proof

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

\[ G \simeq \text{ group-like elements of } R[G] \to \text{ group-like elements of } R[H] \simeq H \]

and each map separately is clearly multiplicative.