Let $C$ be a coalgebra with coproduct $\Delta$ and counit $\epsilon$. Then a subset $I\subseteq C$ is a coideal if $\Delta(I)\subseteq I\otimes C+C\otimes I$ and $\epsilon(I)=0$.

My question is why the coideal should be defined by this way? Does it show the absorptivity ?

  • 14
    $\begingroup$ Please don't correct the typo in the title -- priceless! $\endgroup$
    – saulspatz
    Commented Mar 30, 2018 at 2:01
  • $\begingroup$ @saulspatz , thank you for your comment, do you mean my typo is not right? $\endgroup$
    – Daisy
    Commented Mar 30, 2018 at 2:19
  • 2
    $\begingroup$ You typed "coidea" when I imagine you meant "coideal" but it's very funny. You should leave it as is. If it was intentional, my compliments. $\endgroup$
    – saulspatz
    Commented Mar 30, 2018 at 2:24
  • 1
    $\begingroup$ @saulspatz, yes, you are right, I want to know who give this (coidea) to define coideal and why? $\endgroup$
    – Daisy
    Commented Mar 30, 2018 at 2:30
  • 2
    $\begingroup$ An ideal of an algebra $A$ is a submodule $I$ such that the quotient $A/I$ becomes an algebra by the usual rule ($\overline{x} \cdot \overline{y} = \overline{xy}$, etc.) Similarly, a coideal of a coalgebra $C$ is a submodule (not just subset!) $I$ such that the quotient $C/I$ becomes a koalagebra by the usual rule. It is not hard to show (at least when the ground field is a field) that this boils down to the conditions you gave. $\endgroup$ Commented Mar 30, 2018 at 2:39

1 Answer 1


Supossing that we are speaking about coalgebras over a field $k$, the definition of the notion of a coideal of a $k$-coalgebra has been much inspired by the relation between ideals of a $k$-algebra and kernels of $k$-algebra morphisms: we are actually trying to "mimic" this relation and define the notion of a coideal in a way which would allow for a similar relation between coideals of a $k$-coalgebra and kernels of $k$-coalgebra morphisms:

Recall that, given a field $k$ and the two $k$-coalgebras $C$, $D$, if the $k$-linear map $f:C\to D$ is a coalgebra map, then the following diagram is commutative: $$ \require{AMScd} \begin{CD} C@>f>> D \\ @V\Delta_C VV @VV \Delta_D V\\ C\otimes C @>{f\ \otimes\ f}>>D\otimes D \end{CD} $$ Now, we would like to determine which $k$-subspaces of $C$ are in the kernel of the coalgebra map $f:C\to D$.
Since $\Delta_D\big(f(Ker f)\big)=0$, by the comm. diagram above, we get that $\big(f\otimes f\big)\big(\Delta_C(Ker f)\big)=0$ and thus $$ \Delta_C(Ker f)\subseteq Ker(f\otimes f) $$ Consequently, the description of the kernel of $f$ and its behaviour under $\Delta$ are related to the description of $Ker(f\otimes f)$.
Now, recall that from linear algebra: $$ Ker(f\otimes f)=Ker f\otimes C+C\otimes Ker f $$ Since we want our notion of coideal to describe something which behaves similarly to kernels of coalgebra maps, the standard definition seems reasonable, in view of the relation $$ \Delta_C(Ker f)\subseteq Ker f\otimes C+C\otimes Ker f $$ On the other hand, the second relation which is implied by the fact that $f:C\to D$ is a coalgebra morphism is that $$ \varepsilon_D\circ f=\varepsilon_C $$ thus: $\varepsilon_D\big(f(Ker f)\big)=0=\varepsilon_C(Ker f)$ and consequently: $$ \varepsilon_C(Ker f)=0 $$ which justifies the second relation of the definition of the coideal.

I hope the above description, will shed some light into the ... coidea behind the definition of a coideal (for coalgebras over fields).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .