Heavily Separable Functors

TL;DR

This paper introduces the concept of heavily separable functors, a stronger version of separable functors, and explores their properties through examples related to tensor algebra.

Contribution

It defines and investigates heavily separable functors, extending the theory of separability in functor categories with new stronger conditions.

Findings

Heavily separable functors are a strict strengthening of separable functors.

The paper applies the concept to functors connected to tensor algebra.

Examples demonstrate the differences and potential advantages of the new notion.

Abstract

Prompted by an example related to the tensor algebra, we introduce and investigate a stronger version of the notion of separable functor that we call heavily separable. We test this notion on several functors traditionally connected to the study of separability.

Equations

$$F_{X,Y}:{\mathrm{Hom}}_{\mathcal{B}}\left(X,Y\right)\to {\mathrm{Hom}}_{\mathcal{A}}\left(FX,FY\right):f\mapsto Ff$$

$$P_{-,-}:=P_{-,-}^F:{\mathrm{Hom}}_{\mathcal{A}}\left(F-,F-\right)\to {\mathrm{Hom}}_{\mathcal{B}}\left(-,-\right)$$

$${\mathrm{Hom}}_{\mathcal{A}}\left(FX,FY\right)\times {\mathrm{Hom}}_{\mathcal{A}}\left(FY,FZ\right)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{ccc}\mathrm{Hom}_{\mathcal{A}}\left(FX,FY\right)\times\mathrm{Hom}_{\mathcal{A}}\left(FY,FZ\right)&\overset{P_{X,Y}\times P_{Y,Z}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{B}}\left(X,Y\right)\times\mathrm{Hom}_{\mathcal{B}}\left(Y,Z\right)\\ \circ\downarrow&&\downarrow\circ\\ \mathrm{Hom}_{\mathcal{A}}\left(FX,FZ\right)&\overset{P_{X,Z}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{B}}\left(X,Z\right)\end{array}$$

$${\mathrm{Hom}}_{\mathcal{A}}\left(GFX,GFY\right)\times {\mathrm{Hom}}_{\mathcal{A}}\left(GFY,GFZ\right)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{ccccc}\mathrm{Hom}_{\mathcal{A}}\left(GFX,GFY\right)\times\mathrm{Hom}_{\mathcal{A}}\left(GFY,GFZ\right)&\overset{P_{FX,FY}^{G}\times P_{FY,FZ}^{G}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{B}}\left(FX,FY\right)\times\mathrm{Hom}_{\mathcal{B}}\left(FY,FZ\right)&\overset{P_{X,Y}^{F}\times P_{Y,Z}^{F}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{C}}\left(X,Y\right)\times\mathrm{Hom}_{\mathcal{C}}\left(Y,Z\right)\\ \circ\downarrow&&\circ\downarrow&&\downarrow\circ\\ \mathrm{Hom}_{\mathcal{A}}\left(GFX,GFZ\right)&\overset{P_{FX,FZ}^{G}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{B}}\left(FX,FZ\right)&\overset{P_{X,Z}^{F}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{C}}\left(X,Z\right)\end{array}$$

$${\mathrm{Hom}}_{\mathcal{B}}\left(FX,FY\right)\times {\mathrm{Hom}}_{\mathcal{B}}\left(FY,FZ\right)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{ccccc}\mathrm{Hom}_{\mathcal{B}}\left(FX,FY\right)\times\mathrm{Hom}_{\mathcal{B}}\left(FY,FZ\right)&\overset{G_{FX,FY}\times G_{FY,FZ}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{A}}\left(GFX,GFY\right)\times\mathrm{Hom}_{\mathcal{A}}\left(GFY,GFZ\right)&\overset{P_{X,Y}^{GF}\times P_{Y,Z}^{GF}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{C}}\left(X,Y\right)\times\mathrm{Hom}_{\mathcal{C}}\left(Y,Z\right)\\ \circ\downarrow&&\circ\downarrow&&\downarrow\circ\\ \mathrm{Hom}_{\mathcal{B}}\left(FX,FZ\right)&\overset{G_{FX,FZ}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{A}}\left(GFX,GFZ\right)&\overset{P_{X,Z}^{GF}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{C}}\left(X,Z\right)\end{array}$$

$${\mathrm{Hom}}_{\mathcal{B}}\left(X,Y\right)\times {\mathrm{Hom}}_{\mathcal{B}}\left(Y,Z\right)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{ccc}\mathrm{Hom}_{\mathcal{B}}\left(X,Y\right)\times\mathrm{Hom}_{\mathcal{B}}\left(Y,Z\right)&\overset{F_{X,Y}\times F_{Y,Z}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{A}}\left(FX,FY\right)\times\mathrm{Hom}_{\mathcal{A}}\left(FY,FZ\right)\\ \downarrow\circ&&\circ\downarrow\\ \mathrm{Hom}_{\mathcal{B}}\left(X,Z\right)&\overset{F_{X,Z}}{\longrightarrow}&\mathrm{Hom}_{\mathcal{A}}\left(FX,FZ\right)\end{array}$$

$$\gamma \gamma =\gamma \circ RϵL.$$

$$\delta \delta =L\eta R\circ \delta .$$

$$\gamma X=P_{RLX,X}\left(ϵLX\right)$$

$$\gamma Y\circ Rg\circ \eta X=P_{RLY,Y}\left(ϵLY\right)\circ Rg\circ \eta X=P_{X,Y}\left(ϵLY\circ LRg\circ L\eta X\right)=P_{X,Y}\left(g\circ ϵLX\circ L\eta X\right)=P_{X,Y}(g).$$

$$P_{Y,Z}\left(g\right)\circ P_{X,Y}\left(f\right)$$

$$\gamma \gamma X=\gamma X\circ \gamma RLX\stackrel{(\text{ref}form:gammaP)}{=}{P}_{RLX,X}\left(ϵLX\right)\circ P_{RLRLX,RLX}\left(ϵLRLX\right)$$

$$RLRL\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\begin{array}[]{ccccc}RLRL&\overset{RL\gamma}{\underset{R\epsilon L}{\rightrightarrows}}&RL&\overset{\gamma}{\longrightarrow}&\mathrm{Id}\end{array}$$

$$\left(\nu M\right)\left(ma\right)=\left(f_m{\otimes }_A\mathcal{C}\right)\left(\left(\nu A\right)\left(a\right)\right).$$

$$\left(\nu M\right)\left(ma\right)=\left(f_m{\otimes }_A\mathcal{C}\right)\left(1_A{\otimes }_{A}ea\right)=m{\otimes }_{A}ea$$

$$m{\otimes }_{A}e{\otimes }_{A}e=m{\otimes }_A{\Delta }_{\mathcal{C}}\left(e\right)$$

$$\left(\gamma M\right)\left(s{\otimes }_{R}m\right)=\left(\gamma M\right)\left(S{\otimes }_R{f}_m\right)\left(s{\otimes }_R{1}_R\right)=f_m\left(\gamma R\right)\left(s{\otimes }_R{1}_R\right)=\left(\gamma R\right)\left(s{\otimes }_R{1}_R\right)\cdot m.$$

$$\sum _i{a}_i{b}_i=1,\sum _{i}s{a}_i{\otimes }_R{b}_i=\sum _i{a}_i{\otimes }_R{b}_{i}s\text{ for every }s\in S,$$

$$\sum _{i,j}{a}_i{\otimes }_R{b}_i{a}_j{\otimes }_R{b}_j=\sum _i{a}_i{\otimes }_R{1}_S{\otimes }_R{b}_i.$$

$$\left(S{\otimes }_R\delta M\right)\circ \delta M=\left(S{\otimes }_R\eta M\right)\circ \delta M.$$

$$\left(S{\otimes }_R\delta M\right)\left(\delta M\right)\left(m\right)=\sum _i{a}_i{\otimes }_R\left(\delta M\right)\left(b_{i}m\right)=\sum _{i,j}{a}_i{\otimes }_R{a}_j{\otimes }_R{b}_j{b}_{i}m\stackrel{\text{eqref}hvspdmp1}{=}\sum _{i,j}{a}_i{\otimes }_R{b}_i{a}_j{\otimes }_R{b}_{j}m,$$

$$\left(S{\otimes }_R\eta M\right)\left(\delta M\right)\left(m\right)=\sum _i{a}_i{\otimes }_R\left(\eta M\right)\left(b_{i}m\right)=\sum _i{a}_i{\otimes }_R{1}_S{\otimes }_R{b}_{i}m.$$

$$\left(h\circ m\right)\left(s^{\prime }{\otimes }_{R}s\right)=s^{\prime }s{\otimes }_R{1}_S=\left(m{\otimes }_{R}S\right)\left(s^{\prime }{\otimes }_{R}s{\otimes }_R{1}_S\right)=\left(m{\otimes }_{R}S\right)\left(s^{\prime }{\otimes }_R{1}_S{\otimes }_{R}s\right)=s^{\prime }{\otimes }_{R}s$$

$$\sum _i{a}_i{\otimes }_R{b}_i=\sum _i{a}_i{1}_S{\otimes }_R{1}_S{b}_i=1_S{\otimes }_R{1}_S\sum _i{a}_i{b}_i\stackrel{\text{eqref}hvspdmp1}{=}{1}_S{\otimes }_R{1}_S.$$

$$\frac{r}{s}{\otimes }_{R}1=\frac{r}{s}{\otimes }_R\frac{s}{s}=\frac{s}{s}{\otimes }_R\frac{r}{s}=1{\otimes }_R\frac{e}{s}.$$

$$\alpha \left(E_{ij}\right)$$

Full text

This paper was written while both authors were members of the "National Group for Algebraic and Geometric Structures and their Applications" (GNSAGA-INdAM)

Introduction

Given a field $\Bbbk$, the functor $\mathbf{P}:\mathrm{Bialg}_{\Bbbk}\to\mathrm{Vec}_{\Bbbk}$, assigning to a $\Bbbk$-bialgebra $B$ the $\Bbbk$-vector space of its primitive elements, admits a left adjoint $\mathbf{T}$, assigning to a vector space $V$ the tensor algebra $\mathbf{T}V$ endowed with its canonical bialgebra structure such that the elements in $V$ becomes primitive. By investigating the properties of the adjunction $(\mathbf{T},\mathbf{P})$, together with its unit $\boldsymbol{\eta}$ and counit $\boldsymbol{\epsilon}$, we discovered that there is a natural retraction $\boldsymbol{\gamma}:\mathbf{T}\mathbf{P}\to\mathrm{Id}$ of $\boldsymbol{\eta}$, i.e. $\boldsymbol{\gamma}\circ\boldsymbol{\eta}=\mathrm{Id}$, fulfilling the condition $\boldsymbol{\gamma}\boldsymbol{\gamma}=\boldsymbol{\gamma}\circ\mathbf{P}\boldsymbol{\epsilon}\mathbf{T}$. The existence of a natural retraction of the unit of an adjunction is, by Rafael Theorem, equivalent to the fact that the left adjoint is a separable functor. It is then natural to wonder if the above extra condition on the retraction $\boldsymbol{\gamma}$ corresponds to a stronger notion of separability. In the present paper, we show that an affirmative answer to this question is given by what we call a heavily separable (h-separable for short) functor and we investigate this notion in case of functors usually connected to the study of separability.

Explicitly, in Section 1 we introduce the concept of h-separable functor and we recover classical results in the h-separable case such as their behaviour with respect to composition (Lemma 1.4). In Section 2, we obtain a Rafael type Theorem 2.1. As a consequence we characterize the h-separability of a left (respectively right) adjoint functor either with respect to the forgetful functor from the Eilenberg-Moore category of the associated monad (resp. comonad) in Proposition 2.3 or by the existence of an augmentation (resp. grouplike morphism) of the associated monad (resp. comonad) in Corollary 2.7. In Theorem 2.8, we prove that the induced functor attached to an $A$-coring is h-separable if and only if this coring has an invariant grouplike element.

Section 3 is devoted to the investigation of the h-separability of the induction functor $\varphi^{*}$ and of the restriction of scalars functors $\varphi_{*}$ attached to a ring homomorphism $\varphi:R\to S.$ In Proposition 3.1, we prove that $\varphi^{*}$ is h-separable if and only if there is a ring homomorphism $E:S\to R$ such that $E\circ\varphi=\mathrm{Id}$. Characterizing whether $\varphi_{*}$ is h-separable (in this case we say that $S/R$ is h-separable) is more laborious. In Proposition 3.4, we prove that $S/R$ is h-separable if and only if it is endowed with what we call a $h$-separability idempotent, a stronger version of a separability idempotent. In Lemma 3.7 we show that the ring epimorphisms (by this we mean epimorphisms in the category of rings) provide particular examples of h-separability. In Lemma 3.11 we show that the ring of matrices is never h-separable over the base ring except in trivial cases. In the rest of the present section we investigate the particular case when $S$ is an $R$-algebra i.e. $\mathrm{Im}(\varphi)\subseteq Z(S).$ In Theorem 3.13 we discover that, in this case, $S/R$ is h-separable if and only if $\varphi$ is a ring epimorphism. Moreover $S$ becomes commutative. As a consequence, in Proposition 3.14 we show that a h-separable algebra over a field $\Bbbk$ is necessarily trivial.

Finally in Section 4 we provide a more general version of our starting example $(\mathbf{T},\mathbf{P})$ involving monoidal categories and bialgebras therein.

1. Heavily separable functors

In this section we collect general facts about heavily separable functors.

Definition 1.1**.**

For every functor $F:\mathcal{B}\rightarrow\mathcal{A}$ we set

[TABLE]

Recall that $F$ is called separable if there is a natural transformation

[TABLE]

such that $P_{X,Y}\circ F_{X,Y}=\mathrm{Id}$ for every $X,Y$ objects in $\mathcal{B}$.

We say that $F$ is **heavily separable ** (h-separable for short) if it is separable and the $P_{X,Y}$’s make commutative the following diagram for every $X,Y,Z\in\mathcal{B}.$

[TABLE]

Invisible (To be hidden):

[TABLE]

where the vertical arrows are the obvious compositions. On elements the above diagram means that $P_{X,Z}(f\circ g)=P_{Y,Z}(f)\circ P_{X,Y}(g).$

Remark 1.2*.*

We were tempted to use the word "strongly" at first, instead of "heavily", but a notion of "strongly separable functor" already appeared in the literature in connection with graded rings in [CGN, Definition 3.1].

Remark 1.3*.*

The Maschke’s Theorem for separable functors asserts that for a separable functor $F:\mathcal{B}\rightarrow\mathcal{A}$ a morphism $f:X\rightarrow Y$ splits (resp. cosplits) if and only if $F\left(f\right)$ does. Explicitly, if $F\left(f\right)\circ g=\mathrm{Id}$ (resp. $g\circ F\left(f\right)=\mathrm{Id}$) for some morphism $g$ then $f\circ P_{Y,X}\left(g\right)=\mathrm{Id}$ (resp. $P_{Y,X}\left(g\right)\circ f=\mathrm{Id}$). If $F\left(f\right)\circ g=\mathrm{Id}$ and $F\left(f^{\prime}\right)\circ g^{\prime}=\mathrm{Id}$ for $f:X\rightarrow Y,f^{\prime}:Y\rightarrow Z,$ then $f\circ P_{Y,X}\left(g\right)=\mathrm{Id}$ and $f^{\prime}\circ P_{Z,Y}\left(g^{\prime}\right)=\mathrm{Id}$ so that $f^{\prime}\circ f\circ P_{Y,X}\left(g\right)\circ P_{Z,Y}\left(g^{\prime}\right)=\mathrm{Id}$ so that $P_{Y,X}\left(g\right)\circ P_{Z,Y}\left(g^{\prime}\right)$ is a section of $f^{\prime}\circ f.$ Since $F\left(f^{\prime}\circ f\right)\circ g\circ g^{\prime}=\mathrm{Id},$ we also have $f^{\prime}\circ f\circ P_{Z,X}\left(g\circ g^{\prime}\right)=\mathrm{Id}$ so that $P_{Z,X}\left(g\circ g^{\prime}\right)$ is another section of $f^{\prime}\circ f.$ In general these two section may differ but not in case $F$ is h-separable. Thus in some sense we get a sort of functoriality of the splitting. A similar remark holds for cosplittings. We thank J. Vercruysse for this observation.

Lemma 1.4**.**

Let $F:\mathcal{C}\rightarrow\mathcal{B}$ and $G:\mathcal{B}\rightarrow\mathcal{A}$ be functors.

1. i)

If $F$ and $G$ are h-separable so is $GF.$ 2. ii)

If $GF$ is h-separable so is $F.$ 3. iii)

If $G$ is h-separable, then $F$ is h-sparable if and only if so is $GF.$

Proof.

i). By [NVV, Lemma 1] we know that $GF$ is separable with respect to $P_{X,Y}^{GF}:=P_{X,Y}^{F}\circ P_{FX,FY}^{G}.$ As a consequence, since $F$ and $G$ are h-separable, the following diagram

[TABLE]

Invisible (To be hidden):

[TABLE]

commutes so that $GF$ is h-separable.

ii). By [NVV, Lemma 1] we know that $F_{X,Y}$ cosplits naturally through $P_{X,Y}^{F}:=P_{X,Y}^{GF}\circ G_{FX,FY}.$ On the other hand, since $G$ is a functor and $GF$ is h-separable the following diagram commutes

[TABLE]

Invisible (To be hidden):

[TABLE]

so that $F$ is h-separable.

iii) It follows trivially from i) and ii).

Remark 1.5*.*

The present remark was pointed out by J. Vercruysse. If the functor $F:\mathcal{B}\rightarrow\mathcal{A}$ is a split monomorphism, meaning that there is a functor $G:\mathcal{A}\rightarrow\mathcal{B}$ such that $G\circ F=\mathrm{Id},$ then $F$ is h-separable. This follows by setting $P_{X,Y}:=G_{X,Y}$ as in Definition 1.1. It can also be proved by means of Lemma 1.4,ii).

Lemma 1.6**.**

A full and faithful functor is h-separable.

Proof.

If $F:\mathcal{B}\rightarrow\mathcal{A}$ is full and faithful we have that the canonical map $F_{X,Y}:\mathrm{Hom}_{\mathcal{B}}\left(X,Y\right)\rightarrow\mathrm{Hom}_{\mathcal{A}}\left(FX,FY\right)$ is invertible so that we can take $P_{X,Y}:=F_{X,Y}^{-1}.$ Since $F$ is a functor, the following diagram commutes

[TABLE]

Invisible (To be hidden):

[TABLE]

Reversing the horizontal arrows we obtain that $F$ h-separable.

2. Heavily separable adjoint functors

In this section we investigate h-separable functors which are adjoint functors.

Theorem 2.1** (Rafael type Theorem).**

Let $\left(L,R,\eta,\epsilon\right)$ be an adjunction where $L:\mathcal{B}\rightarrow\mathcal{A}.$

1. (i)

$L$* is h-separable if and only if there is a natural transformation $\gamma:RL\rightarrow\mathrm{Id}_{\mathcal{B}}$ such that $\gamma\circ\eta=\mathrm{Id}$ and*

[TABLE] 2. (ii)

$R$* is h-separable if and only if there is a natural transformation $\delta:\mathrm{Id}_{\mathcal{A}}\rightarrow LR$ such that $\epsilon\circ\delta=\mathrm{Id}$ and*

[TABLE]

Proof.

The second part of the statement follows from the first by duality so we only have to establish (i).

First recall that, by Rafael Theorem [Ra, Theorem 1.2], the functor $L$ is separable if and only if there is a natural transformation $\gamma:RL\rightarrow\mathrm{Id}_{\mathcal{B}}$ such that $\gamma\circ\eta=\mathrm{Id.}$ Moreover, by construction

[TABLE]

so that, by naturality of $P_{-,-}$ one has

[TABLE]

Assume that (1) holds. Then, for all $f\in\mathrm{Hom}_{\mathcal{A}}\left(LX,LY\right)$ and $g\in\mathrm{Hom}_{\mathcal{A}}\left(LY,LZ\right),$ we have

[TABLE]

so that $P_{Y,Z}\left(g\right)\circ P_{X,Y}\left(f\right)=P_{X,Z}\left(g\circ f\right)$ and hence $L$ is h-separable. Conversely, if the latter condition holds for every $X,Y,Z$ and $f,g$ as above, we have

[TABLE]

so that (1) holds.

Remark 2.2*.*

Let $\gamma$ be as in Theorem 2.1. Then

[TABLE]

Invisible (To be hidden):

[TABLE]

is a split coequalizer (in the sense of [Bo, Definition 4.4.2 ]) taking $u=R\epsilon L,v=RL\gamma,r=\eta RL,q=\gamma$ and $s=\eta.$ In particular the coequalizer above is absolute, i.e. preserved by every functor defined on $\mathcal{B}$. One can see this as a consequence of the fact that, for every $B\in\mathcal{B},$ the pair $\left(B,\gamma B\right)$ belongs to ${}_{RL}\mathcal{B}$, i.e. the Eilenberg-Moore category of the monad $\left(RL,R\epsilon L,\eta\right)$, and hence [Bo, Lemma 4.4.3 ] applies to this pair. A similar remark holds for $\delta$ as in Theorem 2.1 in connection with the Eilenberg-Moore category $\mathcal{B}^{LR}$ of the comonad $\left(LR,L\eta R,\epsilon\right)$.

Proposition 2.3**.**

Let $\left(L,R\right)$ be an adjunction.

• $\left(1\right)$

The functor $L$ is h-separable if and only if the forgetful functor $U:{{}_{RL}\mathcal{B}}\rightarrow\mathcal{B}$ is a split epimorphism i.e. there is a functor $\Gamma:\mathcal{B}\rightarrow{{}_{RL}\mathcal{B}}$ such that $U\circ\Gamma=\mathrm{Id}_{\mathcal{B}}.$

• $\left(2\right)$

The functor $R$ is h-separable if and only if the functor forgetful $U:\mathcal{B}^{LR}\rightarrow\mathcal{B}$ is a split epimorphism i.e. there is a functor $\Gamma:\mathcal{B}\rightarrow\mathcal{B}^{LR}$ such that $U\circ\Gamma=\mathrm{Id}_{\mathcal{B}}.$

Proof.

We just prove $\left(1\right),$ the proof of $\left(2\right)$ being similar. By Theorem 2.1, $L$ is h-separable if and only if there is a natural transformation $\gamma:RL\rightarrow\mathrm{Id}_{\mathcal{B}}$ such that $\gamma\circ\eta=\mathrm{Id}$ and (1) holds. For every $B\in\mathcal{B},$ define $\Gamma B:=\left(B,\gamma B\right).$ Then $\Gamma B\in{{}_{RL}\mathcal{B}}$ by the properties of $\gamma.$ Moreover any morphism $f:B\rightarrow C$ fulfills $f\circ\gamma B=\gamma C\circ RLf$ by naturality of $\gamma.$ This means that $f$ induces a morphism $\Gamma f:\Gamma B\rightarrow\Gamma C$ such that $U\Gamma f=f.$ We have so defined a functor $\Gamma:\mathcal{B}\rightarrow\mathcal{B}_{RL}$ such that $U\circ\Gamma=\mathrm{Id}_{\mathcal{B}}.$

Conversely, let $\Gamma:\mathcal{B}\rightarrow{{}_{RL}\mathcal{B}}$ be a functor such that $U\circ\Gamma=\mathrm{Id}_{\mathcal{B}}.$ Then, for every $B\in\mathcal{B},$ we have that $\Gamma B=\left(B,\gamma B\right)$ for some morphism $\gamma B:RLB\rightarrow B.$ Since $\Gamma B\in\mathcal{B}_{RL}$ we must have that $\gamma B\circ\eta B=B$ and $\gamma B\circ RL\gamma B=\gamma B\circ R\epsilon LB.$ Given a morphism $f:B\rightarrow C,$ we have that $\Gamma f:\Gamma B\rightarrow\Gamma C$ is a morphism in ${{}_{RL}\mathcal{B}},$ which means that $f\circ\gamma B=\gamma C\circ RLf$ i.e. $\gamma:=\left(\gamma B\right)_{B\in\mathcal{B}}$ is a natural transformation. By the foregoing $\gamma\circ\eta=\mathrm{Id}$ and (1) holds.

Corollary 2.4**.**

Let $\left(L,R\right)$ be an adjunction.

• $\left(1\right)$

Assume that $R$ is strictly monadic (i.e. the comparison functor is an isomorphism of categories). Then the functor $L$ is h-separable if and only if $R$ is a split epimorphism.

• $\left(2\right)$

Assume that $L$ is strictly comonadic (i.e. the cocomparison functor is an isomorphism of categories). Then the functor $R$ is h-separable if and only if $L$ is a split epimorphism.

Proof.

We just prove $\left(1\right),$ the proof of $\left(2\right)$ being similar. Since the comparison functor $K:\mathcal{A}\rightarrow\mathcal{B}_{RL}$ is an isomorphism of categories and $U\circ K=R$, we get that $R$ is a split epimorphism if and only if $U$ is a split epimorphism. By Proposition 2.3, this is equivalent to the h-separability of $L.$

Example 2.5**.**

In Remark 4.3 we will obtain that the tensor algebra functor $T:\mathrm{Vec}_{\Bbbk}\rightarrow\mathrm{Alg}_{\Bbbk}$ is separable but not h-separable.

Definition 2.6**.**

Following [LMW, Section 4] we say that a grouplike morphism for a comonad $\left(C,\Delta:C\rightarrow CC,\epsilon:C\rightarrow\mathrm{Id}\right)$ is a natural transformation $\delta:\mathrm{Id}\rightarrow C$ such that $\epsilon\circ\delta=\mathrm{Id}$ and $\delta\delta=\Delta\circ\delta.$ Dually an augmentation for a monad $\left(M,m:MM\rightarrow M,\eta:\mathrm{Id}\rightarrow M\right)$ is a natural transformation $\gamma:M\rightarrow\mathrm{Id}$ such that $\gamma\circ\eta=\mathrm{Id}$ and $\gamma\gamma=\gamma\circ m.$

An immediate consequence of the previous definition and Theorem 2.1 is the following result.

Corollary 2.7**.**

Let $\left(L,R,\eta,\epsilon\right)$ be an adjunction.

1. (1)

$L$* is h-separable if and only if the monad $\left(RL,R\epsilon L,\eta\right)$ has an augmentation.* 2. (2)

$R$* is h-separable if and only if the comonad $\left(LR,L\eta R,\epsilon\right)$ has a grouplike morphism.*

Consider an $A$-coring $\mathcal{C}$ and its set of invariant elements $\mathcal{C}^{A}=\left\{c\in\mathcal{C}\mid ac=ca,\text{ for every }a\in A\right\}$. In [Br, Theorem 3.3], Brzeziński proved that the induction functor $R:=\left(-\right)\otimes_{A}\mathcal{C}:$ Mod-$A\rightarrow\mathcal{M}^{\mathcal{C}}$ is separable if and only if there is an invariant element $e\in\mathcal{C}^{A}$ such that $\varepsilon_{\mathcal{C}}\left(e\right)=1.$ Next result provides a similar characterization for the h-separable case.

Theorem 2.8**.**

Given an $A$-coring $\mathcal{C}$, the induction functor $R:=\left(-\right)\otimes_{A}\mathcal{C}:\mathrm{Mod}$-$A\rightarrow\mathcal{M}^{\mathcal{C}}$ is h-separable if and only if $\mathcal{C}$ has an invariant grouplike element.

Proof.

By [Br, Lemma 3.1], the functor $R$ is the right adjoint of the forgetful functor $L:\mathcal{M}^{\mathcal{C}}\rightarrow\mathcal{M}_{A}$. Thus by Corollary 2.7, $R$ is h-separable if and only if the comonad $\left(LR,L\eta R,\epsilon\right)$ has a grouplike morphism. A grouplike morphism for this particular comonad is equivalent to an invariant grouplike element for the coring $\mathcal{C}$ i.e. an element $e\in\mathcal{C}^{A}$ such that $\varepsilon_{\mathcal{C}}\left(e\right)=1$ and $\Delta_{\mathcal{C}}\left(e\right)=e\otimes_{A}e.$

Invisible (To be hidden):  The comonad $\left(LR,L\eta R,\epsilon\right)$ has a grouplike morphism if and only if there is a natural transformation $\nu:\mathrm{Id}\rightarrow LR$ such $\epsilon\circ\nu=\mathrm{Id}$ and $\nu\nu=L\eta R\circ\nu.$ Let $M\in\mathcal{M}_{A}$ and $m\in M.$ Consider $f_{m}:A\rightarrow M:a\mapsto ma.$ By naturality of $\nu$ we have that $\nu M\circ f_{m}=LRf_{m}\circ\nu A$ i.e.

[TABLE]

Since $\nu A:A\rightarrow A\otimes_{A}\mathcal{C}$ we can write $\left(\nu A\right)\left(a\right)=\left(\nu A\right)\left(1\right)a=\left(1_{A}\otimes_{A}e\right)a=1_{A}\otimes_{A}ea$ for some $e\in\mathcal{C}.$ Thus

[TABLE]

and hence $\nu M:M\rightarrow M\otimes_{A}\mathcal{C}:m\mapsto m\otimes_{A}e$ for every $M\in\mathcal{M}_{A}$. The proof of [Br, Theorem 3.3] shows that $\epsilon\circ\nu=\mathrm{Id}$ is equivalent to require that e\in\mathcal{C}^{A}\and $\varepsilon_{\mathcal{C}}\left(e\right)=1.$ On elements we can rewrite the equality $\nu\nu=L\eta R\circ\nu$ for every $m\in M$ as

[TABLE]

which is equivalent to $\Delta_{\mathcal{C}}\left(e\right)=e\otimes_{A}e.$

Remark 2.9*.*

Let $\mathcal{C}$ be an $A$-coring. We recall that, by [Br, Lemma 5.1], if $A$ is a right $\mathcal{C}$-comodule via $\rho_{A}:A\rightarrow A\otimes_{A}\mathcal{C}$, then $g=\rho_{A}\left(1_{A}\right)$ is a grouplike element of  $\mathcal{C}$. Conversely if $g$ is a grouplike element of  $\mathcal{C}$, then $A$ is a right $\mathcal{C}$-comodule via $\rho_{A}:A\rightarrow A\otimes_{A}\mathcal{C}$ defined by $\rho_{A}\left(a\right)=1_{A}\otimes_{A}\left(g\cdot a\right)$. Moreover, if $g$ is a grouplike element of  $\mathcal{C}$, then, by [Br, page 404], $g$ is an invariant element if and only if $A=A^{\text{co}\mathcal{C}}:=\left\{a\in A\mid ag=ga\right\}$

3. Heavily separable ring homomorphisms

Let $\varphi:R\rightarrow S$ be a ring homomorphism. Consider the induction functor $\varphi^{\ast}:=S\otimes_{R}(-):{}R$-$\mathrm{Mod}\to S$-$\mathrm{Mod}$.

Proposition 3.1**.**

Let $\varphi:R\rightarrow S$ be a ring homomorphism. Then the induction functor $\varphi^{\ast}$ is h-separable if and only if there is a ring homomorphism $E:S\rightarrow R$ such that $E\circ\varphi=\mathrm{Id}.$

Proof.

By [CMZ, Theorem 27], we know that $\varphi^{\ast}$ is separable if and only if there is a morphism of $R$-bimodules $E:S\rightarrow R$ such that $E\left(1_{S}\right)=1_{R}.$ Using $E$ one defines a natural transformation $\gamma M:S\otimes_{R}M\rightarrow M:s\otimes_{R}m\mapsto E\left(s\right)m,$ for every $M\in R$-$\mathrm{Mod,}$ such that $\gamma\circ\eta=\mathrm{Id}$ where the unit $\eta$ is defined by $\eta M:M\rightarrow S\otimes_{R}M:m\mapsto 1_{S}\otimes_{R}m.$ All natural transformations $\gamma$ such that $\gamma\circ\eta=\mathrm{Id}$ are of this form because $R$ is a generator in $R$-$\mathrm{Mod.}$

Invisible (To be hidden):  For every $m\in M$ set $f_{m}:R\rightarrow M:r\mapsto rm.$ Then by naturality of $\gamma$ we have

[TABLE]

Set $E\left(s\right):=\left(\gamma R\right)\left(s\otimes_{R}1_{R}\right)\in R.$ Then $\left(\gamma M\right)\left(s\otimes_{R}m\right)=E\left(s\right)m.$ Now $m=\left(\gamma M\right)\left(\eta M\right)\left(m\right)=\left(\gamma M\right)\left(1_{S}\otimes_{R}m\right)=E\left(1_{S}\right)m$ so that $E\left(1_{S}\right)=1_{R}.$ Moreover $E\left(rs\right)=\left(\gamma R\right)\left(rs\otimes_{R}1_{R}\right)=\left(\gamma R\right)\left(r\left(s\otimes_{R}1_{R}\right)\right)=r\left(\gamma R\right)\left(s\otimes_{R}1_{R}\right)=rE\left(s\right)$ since $\gamma R\in R$-$\mathrm{Mod.}$ Furthermore $E\left(sr\right)=\left(\gamma R\right)\left(sr\otimes_{R}1_{R}\right)=\left(\gamma R\right)\left(s\otimes_{R}r\right)=\left(\gamma R\right)\left(s\otimes_{R}r\right)=E\left(s\right)r.$ Hence $E$ is $R$-bilinear.

Finally (1) rewrites as $E\left(x\right)E\left(y\right)m=E\left(xy\right)m$ for every $x,y\in S$ and $m\in M,$ for every $M\in R$-$\mathrm{Mod.}$ Thus it is equivalent to ask that $E$ is multiplicative. Summing up $\varphi^{\ast}$ is h-separable if and only if there is a morphism of $R$-bimodules $E:S\rightarrow R$ which is a ring homomorphism. This is equivalent to ask that $E:S\rightarrow R$ is a ring homomorphism such that $E\circ\varphi=\mathrm{Id}.$

Invisible (To be hidden):  If $E$ is $R$-bilinear, then $E\left(\varphi\left(r\right)\right)=E\left(r1_{S}\right)=rE\left(1_{S}\right)=r1_{R}=r$ so that $E\circ\varphi=\mathrm{Id}.$ Conversely, if $E\circ\varphi=\mathrm{Id}$ then $E\left(rsr^{\prime}\right)=E\left(\varphi\left(r\right)s\varphi\left(r^{\prime}\right)\right)=E\varphi\left(r\right)E\left(s\right)E\varphi\left(r^{\prime}\right)=rE\left(s\right)r^{\prime}.$

It is well-known that the restriction of scalars functor $\varphi_{\ast}:S$-$\mathrm{Mod}\rightarrow R$-$\mathrm{Mod}$ is the right adjoint of the induction functor $\varphi^{\ast}.$ Moreover $\varphi_{\ast}$ is separable if and only if $S/R$ is separable (see [NVV, Proposition 1.3]) if and only if it admits a separability idempotent.

We are so lead to the following definition.

Definitions 3.2**.**

1. $S/R$ is h-separable if the functor $\varphi_{\ast}:S$-$\mathrm{Mod}\rightarrow R$-$\mathrm{Mod}$ is h-separable.

2. A heavy separability idempotent (h-separability idempotent for short) of $S/R$ is an element $\sum_{i}a_{i}\otimes_{R}b_{i}\in S\otimes_{R}S$ such that $\sum_{i}a_{i}\otimes_{R}b_{i}$ is a separability idempotent, i.e.

[TABLE]

which moreover fulfills

[TABLE]

Remark 3.3*.*

Note that a h-separability idempotent $e:=\sum_{i}a_{i}\otimes_{R}b_{i}$ is exactly a grouplike element in the Sweedler’s coring $\mathcal{C}:=S\otimes_{R}S$ such that $se=es$ for every $s\in S$ i.e. which is invariant. Note that $1_{S}\otimes_{R}1_{S}$ is always a grouplike element in $\mathcal{C}$ but it is not invariant in general.

Proposition 3.4**.**

$S/R$* is h-separable if and only if it has a h-separability idempotent.*

Proof.

We observed that $\varphi_{\ast}$ is the right adjoint of the induction functor $\varphi^{\ast}:=S\otimes_{R}\left(-\right)$. Recall that $S/R$ is separable if and only if the map $S\otimes_{R}S\rightarrow S$ splits as an $S$-bimodule map. The splitting is uniquely determined by a so-called separability idempotent i.e. an element $\sum_{i}a_{i}\otimes_{R}b_{i}\in S\otimes_{R}S$ such that (4) hold. Using this element we can define $\delta:\mathrm{Id}\rightarrow\varphi^{\ast}\varphi_{\ast}$ such that $\epsilon\circ\delta=\mathrm{Id}$ by $\delta M:M\rightarrow S\otimes_{R}M:m\mapsto\sum_{i}a_{i}\otimes_{R}b_{i}m.$ This natural transformation satisfies (2) if and only if

[TABLE]

Let us compute separately the two terms of this equality on any $m\in M,$

[TABLE]

Thus $\delta$ satisfies (2) if and only if (5) holds true.

Remark 3.5*.*

Let $\varphi:R\rightarrow S$ be a ring homomorphism and let $\mathcal{C}:=S\otimes_{R}S$ be the Sweedler coring. In view of Theorem 2.8 and Remark 3.3 we obtain that $S/R$ is h-separable (i.e. the functor $\varphi_{\ast}:S$-$\mathrm{Mod}\rightarrow R$-$\mathrm{Mod}$ is h-separable) if and only if the induction functor $R:=\left(-\right)\otimes_{S}\mathcal{C}:\mathrm{Mod}$-$S\rightarrow\mathcal{M}^{\mathcal{C}}$ is h-separable. Note that here $\mathcal{M}^{\mathcal{C}}$ is isomorphic (see e.g. [BW, page 252-253.]) to the category $\mathrm{Desc}\left(S/R\right)$ of descent data associated to the ring homomorphism $\varphi.$

Corollary 3.6**.**

Let $\varphi:R\rightarrow S$ and $\psi:S\rightarrow T$ be ring homomorphisms.

If $T/S$ and $S/R$ are h-separable so is $T/R$.

• 2)

If $T/R$ is h-separable so is $T/S$.

• 3)

If $S/R$ is h-separable then $T/S$ is h-separable if and only if so is $T/R.$

Proof.

It follows by Definition 3.2 and Lemma 1.4. Invisible (To be hidden): 

1. $T/S$ and $S/R$ are h-separable $\Leftrightarrow\psi_{\ast}$ and $\varphi_{\ast}$ are h-separable. By Lemma 1.4, this implies that $\varphi_{\ast}\circ\psi_{\ast}$ h-separable. Thus $\left(\psi\varphi\right)_{\ast}$ is h-separable i.e. so is $T/R.$

2. $T/R$ is h-separable means that the functor $\left(\psi\varphi\right)_{\ast}$ is h-separable i.e. $\varphi_{\ast}\circ\psi_{\ast}$ h-separable. By Lemma 1.4, this implies that $\psi_{\ast}$ is h-separable i.e. $T/S$ is h-separable.

Lemma 3.7**.**

Let $\varphi:R\rightarrow S$ be a ring homomorphism. The following are equivalent.

1. (1)

The map $\varphi$ is a ring epimorphism (i.e. an epimorphism in the category of rings); 2. (2)

the multiplication $m:S\otimes_{R}S\rightarrow S$ is invertible; 3. (3)

$1_{S}\otimes_{R}1_{S}$* is a separability idempotent for $S/R;$* 4. (4)

$1_{S}\otimes_{R}1_{S}$* is a h-separability idempotent for $S/R.$*

If these equivalent conditions hold true then $S/R$ is h-separable.

Moreover $1_{S}\otimes_{R}1_{S}$ is the unique separability idempotent for $S/R.$

Proof.

$\left(1\right)\Leftrightarrow\left(2\right)$ follows by [St, Proposition XI.1.2 page 225].

$\left(1\right)\Leftrightarrow\left(3\right)$ follows by [St, Proposition XI.1.1 page 226].

Invisible (To be hidden):  If $m$ is invertible, from $m\left(s\otimes_{R}1_{S}\right)=s=m\left(1_{S}\otimes_{R}s\right)$ and the injectivity of $m$ we deduce the claimed equality. Conversely if the equality holds let $h:S\rightarrow S\otimes_{R}S:s\mapsto s\otimes_{R}1_{S}.$ Then $m\circ h=\mathrm{Id}$ and

[TABLE]

and hence $h\circ m=\mathrm{Id}.$ Hence $m$ is invertible.

$\left(3\right)\Leftrightarrow\left(4\right)$ depends on the fact that $1_{S}\otimes_{R}1_{S}$ always fulfills (5).

By Proposition 3.4, $\left(4\right)$ implies that $S/R$ is h-separable.

Let us check the last part of the statement. If $\sum_{i}a_{i}\otimes_{R}b_{i}$ is another separability idempotent, we get

[TABLE]

Example 3.8**.**

We now give examples of ring epimorphisms $\varphi:R\rightarrow S$.

1. Let $S$ be a multiplicative closed subset of a commutative ring $R.$ Then the canonical map $\varphi:R\rightarrow S^{-1}R$ is a ring epimorphism, c.f. [AMa, Proposition 3.1]. More generally we can consider a perfect right localization of $R$ as in [St, page 229].

Invisible (To be hidden):  The case $\left(S^{-1}R\right)/R$ where $S$ is a multiplicative closed subset of a commutative ring $R.$ In fact

[TABLE]

In particular $\mathbb{Q}/\mathbb{Z}$ is of this form.

2. Consider the ring of matrices $\mathrm{M}_{n}\left(R\right)$ and the ring $\mathrm{T}_{n}\left(R\right)$ of $n\times n$ upper triangular matrices over a ring $R.$ Then the inclusion $\varphi:\mathrm{T}_{n}\left(R\right)\rightarrow\mathrm{M}_{n}\left(R\right)$ is a ring epimorphism. In fact, given ring homomorphisms $\alpha,\beta:\mathrm{M}_{n}\left(R\right)\rightarrow B$ that coincide on $\mathrm{T}_{n}\left(R\right)$ then they coincide on all matrices. To see this we first check that $\alpha\left(E_{ij}\right)=\beta\left(E_{ij}\right)$ for all $i>j$,

[TABLE]

Thus $\alpha\left(E_{ij}\right)=\beta\left(E_{ij}\right)$ for every $i,j.$ Now, given $r\in R$ we have

[TABLE]

As a consequence $\alpha\left(M\right)=\beta\left(M\right)$ for all $M\in\mathrm{M}_{n}\left(R\right)$ as desired.

3. Any surjective ring homomorphism $\varphi:R\rightarrow S$ is trivially a ring epimorphism.

Remark 3.9*.*

A kind of dual to Lemma 3.7, establishes that a $\Bbbk$-coalgebra homomorphism $\varphi:C\rightarrow D$ is a coalgebra epimorphism if and only if the induced functor $\mathcal{M}^{C}\rightarrow\mathcal{M}^{D}$ is full, see [NT, Theorem 3.5]. Since this functor is always faithful, by Lemma 1.6 it is in this case h-separable.

Proposition 3.10**.**

Let $\varphi:R\rightarrow S$ be a ring homomorphism. Then $S/R$ is h-separable if and only if $S/\varphi\left(R\right)$ is h-separable.

Proof.

Write $\varphi=i\circ\overline{\varphi}$ where $i:\varphi\left(R\right)\rightarrow S$ is the canonical inclusion and $\overline{\varphi}:R\rightarrow\varphi\left(R\right)$ is the corestriction of $\varphi$ to its image $\varphi\left(R\right).$ By Lemma 3.7, we have that $\varphi\left(R\right)/R$ is h-separable.

By Corollary 3.6, $S/R$ is h-separable if and only if $S/\varphi\left(R\right)$ is h-separable.

It is well-known that the ring of matrices is separable, see e.g. [DI, Example II, page 41]. In the following result we show that however it is never h-separable.

Lemma 3.11**.**

Let $R$ be a ring and $S:=\mathrm{M}_{n}\left(R\right)$. If $S/R$ is h-separable, then either $n=1$ or $R=0.$

Proof.

By Proposition 3.4, $S/R$ admits a h-separability idempotent $e=\sum_{i,j,s,t}r_{i,j}^{s,t}E_{i,j}\otimes_{R}E_{s,t}$ where $r_{i,j}^{s,t}\in R$ and $E_{i,j}$ is the canonical matrix having $1$ in the entry $\left(i,j\right)$ and zero elsewhere ($e$ can be written in the given form since the tensor product is over $R$ and the $E_{i,j}$’s are $R$-invariant). Then

[TABLE]

Note that the elements $E_{i,j}\otimes_{R}E_{i^{\prime},j^{\prime}}$’s form a basis of $S\otimes_{R}S$ as a free left $R$-module. Thus, the equalities $E_{a,b}e=eE_{a,b}$ implies $r_{b,j}^{s,t}=0$ for $t\neq b.$ Moreover if $t=b$ then $r_{b,j}^{s,t}=r_{a,j}^{s,a}.$ Thus $r_{b,j}^{s,t}=\delta_{t,b}r_{a,j}^{s,a}$ for every $a,b,j,s.$ Thus, if we set $r_{j}^{s}:=r_{1,j}^{s,1},$ we obtain $r_{b,j}^{s,t}=\delta_{t,b}r_{1,j}^{s,1}=\delta_{t,b}r_{j}^{s}.$ We can now rewrite

[TABLE]

Now

[TABLE]

so that

[TABLE]

From (5) and the fact that the $E_{i,j}$’s are $R$-invariant we deduce that

[TABLE]

i.e.

[TABLE]

Equivalently for all $i,j$ we have

[TABLE]

i.e., replacing $s$ with $s^{\prime}$ in the second term,

[TABLE]

Thus for every $s^{\prime},i,j$

[TABLE]

From this equality we deduce $r_{j}^{s}r_{j^{\prime}}^{s^{\prime}}=\delta_{s,j^{\prime}}r_{j}^{s^{\prime}}$ for all $s,j,s^{\prime},j^{\prime}.$ Now

[TABLE]

so that we can set $r_{j}:=r_{j}^{1}$ and we get $r_{j}^{s}=r_{j}$ for each $s,j.$ Hence the equality $r_{j}^{s}r_{j^{\prime}}^{s^{\prime}}=\delta_{s,j^{\prime}}r_{j}^{s^{\prime}}$ rewrites as $r_{j}r_{j^{\prime}}=\delta_{s,j^{\prime}}r_{j}$ for all $s,j,j^{\prime}.$

If $n\geq 2,$ then for every $j^{\prime}$ there is always $s\neq j^{\prime}\ so$ that we obtain $r_{j}r_{j^{\prime}}=0$ for all $j,j^{\prime}.$ Now $0=\sum_{j}\sum_{j^{\prime}}r_{j}r_{j^{\prime}}=\sum_{j}r_{j}\sum_{j^{\prime}}r_{j^{\prime}}=1_{R}\cdot 1_{R}=1_{R}$, a contradiction.

3.1. Heavily separable algebras

3.12*.*

Let $R$ be a commutative ring, let $S$ be a ring and let $Z\left(S\right)$ be its center. We recall that a $S$ is said to be an $R$-algebra, or that $S$ is an algebra over $R$, if there is a unital ring homomorphism $\varphi:R\rightarrow S$ such that $\varphi\left(R\right)\subseteq Z\left(S\right)$. In this case we set

[TABLE]

Since ${Im}\left(\varphi\right)\subseteq Z\left(S\right),$ we have $r\cdot s=s\cdot r$ for every $r\in R$ and $s\in S$ and

[TABLE]

Theorem 3.13**.**

Let $S$ be and $R$-algebra. Then $S/R$ is h-separable if and only if the canonical map $\varphi:R\rightarrow S$ is a ring epimorphism. Moreover if one of these conditions holds, then $S$ is commutative.

Proof.

$\left(\Rightarrow\right).$ Let $\sum_{i}a_{i}\otimes_{R}b_{i}$ be an h-separability idempotent. Since $\varphi\left(R\right)\subseteq Z\left(S\right),$ we get that the map $\tau:A\otimes_{R}A\rightarrow A\otimes_{R}A,\tau\left(a\otimes_{R}b\right)=b\otimes_{R}a,$ is well-defined and left $R$-linear. Hence we can apply $A\otimes_{R}\tau$ on both sides of (5) to get $\sum_{i,j}a_{i}\otimes_{R}b_{j}\otimes_{R}b_{i}a_{j}=\sum_{i}a_{i}\otimes_{R}b_{i}\otimes_{R}1_{S}.$ By multiplying, we obtain $\sum_{i,j}a_{i}b_{j}\otimes_{R}b_{i}a_{j}=\sum_{i}a_{i}b_{i}\otimes_{R}1_{S}.$ By (4), we get

[TABLE]

By (4), we get that $\sum_{t}a_{t}sb_{t}\in Z\left(S\right)$, for all $s\in S$. Using this fact we have

[TABLE]

We have so proved that $S\subseteq Z\left(S\right)$ and hence $S$ is commutative. Now, we compute

[TABLE]

so that $\sum_{i}a_{i}\otimes_{R}b_{i}=1_{S}\otimes_{R}1_{S}$ by (7). We conclude by Lemma 3.7.

$\left(\Leftarrow\right)$ It follows by Lemma 3.7.

The following result establishes that there is no non-trivial h-separable algebra over a field $\Bbbk$.

Proposition 3.14**.**

Let $A$ be a h-separable algebra over a field $\Bbbk$. Then either $A=\Bbbk$ or $A=0.$

Proof.

By Theorem 3.13, the unit $u:\Bbbk\rightarrow A$ is a ring epimorphism. By Lemma 3.7, we have that $A\otimes_{\Bbbk}A\cong A$ via multiplication. Since $A$ is h-separable over $\Bbbk$ it is in particular separable over $\Bbbk.$ By [Pi, page 184], the separable $\Bbbk$-algebra $A$ is finite-dimensional. Thus, from $A\otimes_{\Bbbk}A\cong A$ we deduce that $A$ has either dimensional one or zero over $\Bbbk$.

Example 3.15**.**

$\mathbb{C}/\mathbb{R}$ is separable but not h-separable. In fact, by Proposition 3.14, $\mathbb{C}/\mathbb{R}$ is not h-separable. On the other hand $e=\frac{1}{2}\left(1\otimes 1-i\otimes i\right)$ is a separability idempotent (it is the only possible one). It is clear that $e$ is not a h-separability idempotent.

Invisible (To be hidden):  We now that $\mathbb{C}/\mathbb{R}$ is classically separable because it is a finite extension. Let us see if it is h-separable as well. Let $e=a_{11}1\otimes 1+a_{12}1\otimes i+a_{21}i\otimes 1+a_{22}i\otimes i$ be a h-separability idempotent. Then $1=m\left(e\right)=a_{11}-a_{22}+a_{12}i+a_{21}i$ so that $a_{11}=a_{22}+1$ and $a_{12}+a_{21}=0.$ Moreover

[TABLE]

Since $ie=ei$ we get $a_{11}=-a_{22},a_{12}=a_{21}.$ Thus $a_{12}=a_{21}=0,$ $a_{22}=-\frac{1}{2}$ and $a_{11}=\frac{1}{2}$ so that

[TABLE]

Moreover the condition $\left(\ref{hvspdmp2}\right)$ rewrites as

[TABLE]

a contradiction. Thus $\mathbb{C}/\mathbb{R}$ is not h-separable.

Remark 3.16*.*

Let $\Bbbk$ be a field. Set $A=B=\Bbbk,R=\Bbbk\times\Bbbk=S.$ Then $A$ and $B$ are $R$-algebras and $S=A\times B$ is their product in the category of $R$-algebras. Moreover $S/R$ is h-separable as $S=R.$ Hence the product of $R$-algebras may be h-separable.

Lemma 3.17**.**

Let $A$ and $B$ be $R$-algebras and let $S=A\times B$ be their product in the category of $R$-algebras. Set $e_{1}:=\left(1_{A},0_{B}\right)\in S$ and $e_{2}:=\left(0_{A},1_{B}\right)\in S.$ The following are equivalent.

• $\left(i\right)$

$S/R$* is h-separable.*

• $\left(ii\right)$

$A/R$* and $B/R$ are h-separable and $e_{1}\otimes_{R}e_{2}=0=e_{2}\otimes_{R}e_{1}.$*

Proof.

First, by Theorem 3.13 and Lemma 3.7, the conditions $\left(i\right)$ and $\left(ii\right)$ can be replaced respectively by

• •

$1_{S}\otimes_{R}1_{S}$ is a separability idempotent of $S/R$

• •

$1_{A}\otimes_{R}1_{A}$ and $1_{B}\otimes_{R}1_{B}$ are separability idempotents of $A/R$ and $B/R$ respectively and $e_{1}\otimes_{R}e_{2}=0=e_{2}\otimes_{R}e_{1}.$

Note also that, if the first condition holds, then, for $i\neq j$ we get

[TABLE]

so that $e_{1}\otimes_{R}e_{2}=0=e_{2}\otimes_{R}e_{1}.$ Thus the latter condition can be assumed to hold.

Denote by $p_{A}:S\rightarrow A$ and $p_{B}:S\rightarrow B$ the canonical projections.

Since $1_{S}=e_{1}+e_{2}$ and $e_{1}\otimes_{R}e_{2}=0=e_{2}\otimes_{R}e_{1},$ we get that $1_{S}\otimes_{R}1_{S}=e_{1}\otimes_{R}e_{1}+e_{2}\otimes_{R}e_{2}$. For every $s\in S$ we have

[TABLE]

Similarly $1_{S}\otimes_{R}1_{S}s=\left(i_{A}\otimes_{R}i_{A}\right)\left(1_{A}\otimes_{R}1_{A}p_{A}\left(s\right)\right)+\left(i_{B}\otimes_{R}i_{B}\right)\left(1_{B}\otimes_{R}1_{B}p_{B}\left(s\right)\right).$

Invisible (To be hidden):  Here is the computation

[TABLE]

As a consequence, for every $s\in S$

[TABLE]

[TABLE]

[TABLE]

Since $p_{A}$ and $p_{B}$ are surjective, we get that to require that $s1_{S}\otimes_{R}1_{S}=1_{S}\otimes_{R}1_{S}s$ for every $s\in S$ is equivalent to require that

[TABLE]

for every $a\in A,b\in B.$ We have so proved that $1_{S}\otimes_{R}1_{S}$ is a separability idempotent of $S/R$ if and only if $1_{A}\otimes_{R}1_{A}$ and $1_{B}\otimes_{R}1_{B}$ are separability idempotents of $A/R$ and $B/R$ under the assumption $e_{1}\otimes_{R}e_{2}=0=e_{2}\otimes_{R}e_{1}.$

The following result is similar to [DI, Corollary 1.7 page 44].

Lemma 3.18**.**

Let $R$ be a commutative ring. Let $A$ and $B$ be $R$-algebras. Then, if $B/R$ is h-separable, so is $\left(A\otimes_{R}B\right)/A$. As a consequence if both $A/R$ and $B/R$ are h-separable, so is $\left(A\otimes_{R}B\right)/R$ .

Proof.

Since $B/R$ is h-separable, by Theorem 3.13, we have that the unit $u_{B}:R\rightarrow B$ is a ring epimorphism. By Lemma 3.7 this means that $1_{B}\otimes_{R}1_{B}$ is a separability idempotent. Thus also $\left(1_{A}\otimes_{R}1_{B}\right)\otimes_{A}\left(1_{A}\otimes_{R}1_{B}\right)$ is a separability idempotent.

Invisible (To be hidden):  Let us check it:

$\left(a\otimes_{R}b\right)\left(1_{A}\otimes_{R}1_{B}\right)\otimes_{A}\left(1_{A}\otimes_{R}1_{B}\right)=\left(1_{A}\otimes_{R}b1_{B}\right)\left(a\otimes_{R}1_{B}\right)\otimes_{A}\left(1_{A}\otimes_{R}1_{B}\right)$

$=\left(1_{A}\otimes_{R}1_{B}\right)\otimes_{A}\left(a\otimes_{R}1_{B}\right)\left(1_{A}\otimes_{R}1_{B}b\right)=\left(1_{A}\otimes_{R}1_{B}\right)\otimes_{A}\left(1_{A}\otimes_{R}1_{B}\right)\left(a\otimes_{R}b\right).$

As a consequence also $A\otimes_{R}u_{B}:A\otimes_{R}R\rightarrow A\otimes_{R}B$ is a ring epimorphism by the same lemma. If $A/R$ is h-separable, then the unit $u_{A}:R\rightarrow A$ is a ring epimorphism too. Thus the composition

[TABLE]

Invisible (To be hidden):

[TABLE]

i.e. the unit of $A\otimes_{R}B$, is an epimorphism. By Theorem 3.13, $\left(A\otimes_{R}B\right)/R$ is h-separable. Invisible (To be hidden):  Another proof: $u_{B}$ epi$\Rightarrow A\otimes_{R}u_{B}$ epi.

$f,g:A\otimes_{R}B\rightarrow E$ ring map such that $f\circ\left(A\otimes_{R}u_{B}\right)=g\circ\left(A\otimes_{R}u_{B}\right).$

$f\left(a\otimes_{R}b\right)=f\left(\left(a\otimes_{R}1_{B}\right)\left(1_{A}\otimes_{R}b\right)\right)=f\left(a\otimes_{R}1_{B}\right)f\left(1_{A}\otimes_{R}b\right)=f\left(a\otimes_{R}u_{B}\left(1_{R}\right)\right)f\left(1_{A}\otimes_{R}b\right)$

$g\left(a\otimes_{R}b\right)=g\left(a\otimes_{R}u_{B}\left(1_{R}\right)\right)g\left(1_{A}\otimes_{R}b\right)$

$f\left(1_{A}\otimes_{R}b\right)=?=g\left(1_{A}\otimes_{R}b\right)$

$f\circ\left(A\otimes_{R}u_{B}\right)=g\circ\left(A\otimes_{R}u_{B}\right)\Rightarrow f\left(1_{A}\otimes_{R}u_{B}\left(r\right)\right)=g\left(1_{A}\otimes_{R}u_{B}\left(r\right)\right)$

$\Rightarrow f\left(1_{A}\otimes_{R}-\right)u_{B}\left(r\right)=g\left(1_{A}\otimes_{R}-\right)u_{B}\left(r\right)\Rightarrow f\left(1_{A}\otimes_{R}-\right)\circ u_{B}=g\left(1_{A}\otimes_{R}-\right)\circ u_{B}$

$\Rightarrow f\left(1_{A}\otimes_{R}-\right)=g\left(1_{A}\otimes_{R}-\right).$

Remark 3.19*.*

Let $R$ be a ring, $G$ be a group and consider $RG$, the group ring. S. Caenepeel posed the following problem: to characterize whether RG/R\is h-separable. In general we do not have an answer to this question. However, if $R$ is commutative, we can consider a maximal ideal $M$ of $R$ and take $\Bbbk:=R/M.$ By Lemma 3.18, we deduce that $\left(\Bbbk\otimes_{R}RG\right)/\Bbbk$ is h-separable i.e. $\Bbbk G/\Bbbk$ is h-separable. By Proposition 3.14, we conclude that $\Bbbk G=\Bbbk$ and hence $\left|G\right|=1.$

4. Example on monoidal categories

In the present section $\mathcal{M}$ denotes a preadditive braided monoidal category such that

• •

$\mathcal{M}$ has equalizers and denumerable coproducts;

• •

the tensor products are additive and preserve equalizers and denumerable coproducts.

In view of the assumptions above, we can apply [AM1, Theorem 4.6] to obtain an adjunction $\left(\mathbf{T},\mathbf{P}\right)$ as in the following diagram Invisible (To be hidden):

[TABLE]

[TABLE]

Here $\mathrm{Alg}(\mathcal{M})$ denotes the category of algebras (or monoids) in $\mathcal{M}$, $\mathrm{Bialg}(\mathcal{M})$ is the category of bialgebras (or bimonoids) in $\mathcal{M}$, the functors $\mho$ and $\Omega$ are the obvious forgetful functors and, by construction of $\mathbf{T}$, we have $\mho\circ\mathbf{T}=T.$

It is noteworthy that, since $\Omega$ has a left adjoint $T$, then $\Omega$ is strictly monadic (the comparison functor is a category isomorphism), see [AM2, Theorem A.6].

Let $V\in\mathcal{M}.$ By construction $\Omega TV=\oplus_{n\in\mathbb{N}}V^{\otimes n},$ see [AM1, Remark 1.2]. Denote by $\alpha_{n}V:V^{\otimes n}\rightarrow\Omega TV$ the canonical inclusion. The unit $\eta:\mathrm{Id}_{\mathcal{M}}\rightarrow\Omega T$ of the adjunction $\left(T,\Omega\right)$ is defined by $\eta V:=\alpha_{1}V$ while the counit $\epsilon:T\Omega\rightarrow\mathrm{Id}$ is uniquely defined by the equality

[TABLE]

where $m^{n-1}:A^{\otimes n}\rightarrow A$ denotes the iterated multiplication of an algebra $\left(A,m,u\right)$ defined by $m^{-1}=u,m^{0}=\mathrm{Id}_{A}$ and, for $n\geq 2,m^{n-1}=m\circ\left(m^{n-2}\otimes A\right).$

Denote by $\boldsymbol{\eta},\boldsymbol{\epsilon}$ the unit and counit of the adjunction $\left(\mathbf{T},\mathbf{P}\right)$.

Consider the natural transformation $\xi:\mathbf{P}\rightarrow\Omega\mho$ defined by

[TABLE]

Invisible (To be hidden):

[TABLE]

We have $\epsilon\mho\circ T\xi=\epsilon\mho\circ T\Omega\mho\boldsymbol{\epsilon}\circ T\eta P=\mho\boldsymbol{\epsilon}\circ\epsilon T\mathbf{P}\circ T\eta\mathbf{P}=\mho\boldsymbol{\epsilon}$ i.e.

[TABLE]

so that $\xi$ is exactly the natural transformation of [AM1, Theorem 4.6], whose components are the canonical inclusions of the subobject of primitives of a bialgebra $B$ in $\mathcal{M}$ into $\Omega\mho B$ and hence they are regular monomorphisms.

Define the functor

[TABLE]

that assigns to every bialgebra $A$ the kernel $\left(A^{+},\zeta A:A^{+}\rightarrow\Omega\mho A\right)$ of $\varepsilon_{\Omega\mho A}$ (i.e. the equalizer of $\varepsilon_{\Omega\mho A}:\Omega\mho A\rightarrow\mathbf{1}$ and the zero morphism) and to every morphism $f$ the induced morphism $f^{+}.$

Since $\zeta A$ is natural in $A$ we get a natural transformation $\zeta:\left(-\right)^{+}\rightarrow\Omega\mho$ which is by construction a monomorphism on components.

Lemma 4.1**.**

The natural transformation $\xi:\mathbf{P}\rightarrow\Omega\mho$ factors through the natural transformation $\zeta:\left(-\right)^{+}\rightarrow\Omega\mho$ (i.e. there is $\widehat{\xi}:\mathbf{P}\rightarrow\left(-\right)^{+}$ such that $\xi=\zeta\circ\widehat{\xi}$) which is a monomorphism on components.

Proof.

Given $A\in\mathrm{Bialg}\left(\mathcal{M}\right)$ we have that $\xi$ and $\zeta$ are defined by the following kernels.

[TABLE]

Invisible (To be hidden):

[TABLE]

Since the right square above commutes, there is a unique morphism $\widehat{\xi}A:\mathbf{P}A\rightarrow A^{+}$ such that $\zeta A\circ\widehat{\xi}A=\xi A.$ The naturality of $\zeta A$ and $\xi A$ in $A$ implies the one of \widehat{\xi}A\so that $\zeta\circ\widehat{\xi}=\xi.$

There is a unique morphism $\omega V:\Omega TV\rightarrow V$ such that

[TABLE]

Given $f:V\rightarrow W$ a morphism in $\mathcal{M},$ we get for every $n\in\mathbb{N},$

[TABLE]

so that $\omega W\circ\Omega Tf=f\circ\omega V$ which means that $\omega:=\left(\omega V\right)_{V\in\mathcal{M}}$ is a natural transformation $\omega:\Omega T\rightarrow\mathrm{Id}_{\mathcal{M}}.$

Lemma 4.2**.**

The natural transformation $\omega$ fulfills $\omega\circ\eta=\mathrm{Id}$ and

[TABLE]

Proof.

We have

[TABLE]

and hence $\omega\circ\eta=\mathrm{Id}.$ Let us check (11). For every $V\in\mathcal{M}$ we compute

[TABLE]

On the other hand

[TABLE]

Hence we have to check that

[TABLE]

For $n=0$ we have

[TABLE]

For $n=1$ we have

[TABLE]

For $n\geq 2$ we have $\delta_{n,1}\omega V\circ\zeta\mathbf{T}V=0$. In order to prove that also $\omega\circ m_{\Omega TV}^{n-1}\circ\left(\zeta\mathbf{T}V\right)^{\otimes n}=0$ we need first to give a different expression for $\omega V\circ m_{\Omega TV}.$ To this aim we compute (we use the identifications $V\otimes\mathbf{1}\cong V\cong\mathbf{1}\otimes V$)

[TABLE]

Since the tensor products preserve denumerable coproducts, the equalities above yield the identity

[TABLE]

Using it, we obtain

[TABLE]

The last two summands are zero as $\varepsilon_{\Omega TV}\circ\zeta\mathbf{T}V=\varepsilon_{\Omega\mho\mathbf{T}V}\circ\zeta\mathbf{T}V=0$ by definition of $\zeta$.

Remark 4.3*.*

As observed the comparison functor $K:\mathrm{Alg}\left(\mathcal{M}\right)\rightarrow\mathcal{M}_{\Omega T}$ is an isomorphism of categories. By Corollary 2.4, $T$ is h-separable if and only if $\Omega:\mathrm{Alg}\left(\mathcal{M}\right)\rightarrow\mathcal{M}$ is a split epimorphism. Let us prove, by contradiction, that this is not the case. Assume that there is a functor $\Gamma:\mathcal{M}\rightarrow\mathrm{Alg}\left(\mathcal{M}\right)$ such that $\Omega\Gamma=\mathrm{Id}.$ Let $V\in\mathcal{M}$. Then $\Gamma V=\left(V,mV,uV\right)$ for some morphisms $mV:V\otimes V\rightarrow V$ and $uV:\mathbf{1}\rightarrow V.$ Let $f:V\rightarrow V$ be the zero morphism. Then $\Gamma f:\Gamma V\rightarrow\Gamma V$ is an algebra morphism and $\Omega\Gamma f=f$. Thus $f$ is unitary i.e. $uV=f\circ uV=0\circ uV=0$. Hence $\mathrm{Id}_{V}=mV\circ\left(V\otimes uV\right)\circ r_{V}^{-1}=0.$ As a consequence any morphism $h:V\rightarrow W$ would be zero as $h=h\circ\mathrm{Id}_{V}$ for every $V,W\in\mathcal{M}$. Hence $\mathrm{Hom}_{\mathcal{M}}\left(V,W\right)=\left\{0\right\}.$ This happens only if all objects are isomorphic to the unit object $\mathbf{1}$, i.e. if the skeleton of $\mathcal{M}$ is the trivial monoidal category $\left(\mathcal{T},\otimes,\mathbf{1}\right)$, where $\mathrm{Ob}\left(\mathcal{T}\right)=\left\{\mathbf{1}\right\},$ $\mathrm{Hom}_{\mathcal{T}}\left(\mathbf{1,1}\right)=\left\{\mathrm{Id}_{\mathbf{1}}\right\}$ and the tensor product is given by $\mathbf{1\otimes 1=1}$ and $\mathrm{Id}_{\mathbf{1}}\otimes\mathrm{Id}_{\mathbf{1}}=\mathrm{Id}_{\mathbf{1}}.$ This is evidently a restrictive condition on $\mathcal{M}$. Thus, in general $T:\mathcal{M}\rightarrow\mathrm{Alg}\left(\mathcal{M}\right)$ is not heavily separable. On the other hand the equality $\omega\circ\eta=\mathrm{Id}$ obtained in Lemma 4.2 means that the functor $T:\mathcal{M}\rightarrow\mathrm{Alg}\left(\mathcal{M}\right)$ is separable.

As a particular case, we get that the functor $T:\mathrm{Vec}_{\Bbbk}\rightarrow\mathrm{Alg}_{\Bbbk}$ is separable but not h-separable.

Theorem 4.4**.**

Set $\boldsymbol{\gamma}:=\omega\circ\xi\mathbf{T:PT}\rightarrow\mathrm{Id}_{\mathcal{M}}.$ Then $\boldsymbol{\gamma}\circ\boldsymbol{\eta}=\mathrm{Id}$ and $\boldsymbol{\gamma}\boldsymbol{\gamma}=\boldsymbol{\gamma}\circ\mathbf{\mathbf{P\boldsymbol{\epsilon}T.}}$ Hence the functor $\mathbf{T}:\mathcal{M}\rightarrow\mathrm{Bialg}\left(\mathcal{M}\right)$ is h-separable.

Proof.

We compute

[TABLE]

Moreover

[TABLE]

so that

[TABLE]