2
$\begingroup$

I would like to show that

The Kernel of a coring morphism $\phi:C\rightarrow C'$ between two $R$-corings $(C,\Delta,\epsilon)$ and $(C',\Delta',\epsilon')$ is a coideal.

The only point that I can't show is that $$\Delta(\text{Ker}(\phi))\subset \text{Ker}(\pi\otimes\pi)$$

This should be straightforward but I can't show it.

EDIT:

$\pi$ is the natural projection of $C$ unto $C/\text{Ker}(\phi)$

The definition I use are

A coring morphism from an $R$-coring $(C,\Delta,\epsilon)$ to another $R$-coring $(C',\Delta',\epsilon')$ is a $R$-bimodule morphism $\phi$ such that $\epsilon'\circ\phi = \epsilon$ and $(\phi\otimes\phi)\circ\Delta = \Delta'\circ\phi$.

A coideal $I$ of the $R$-coring $C$ is a $R$-subbimodule of $C$ such that $I\subset \text{Ker}(\epsilon)$ and \begin{align*} \Delta(I) \subset \text{Ker}(\pi\otimes\pi) \end{align*} where $\pi:C\rightarrow C/I$ is the cannonical projection map.

$\endgroup$
  • $\begingroup$ What is $\pi$ here? $\endgroup$ – Alex Clark Apr 6 '18 at 14:03
  • $\begingroup$ Sorry, $\pi$ is the natural projection of $C$ unto $C/\text{Ker}(\phi)$ $\endgroup$ – tomak Apr 6 '18 at 14:17
  • $\begingroup$ Since you are speaking generally about corings (and not especially about coalgebras) could you please make explicit the definition of the notion of co-ideal you are using ? $\endgroup$ – KonKan Apr 8 '18 at 21:51
2
$\begingroup$

The statement:

The Kernel of a coring morphism $\phi:C\rightarrow C'$ between two $R$-coring's $(C,\Delta,\epsilon)$ and $(C',\Delta',\epsilon')$ is a coideal.

is not true in general:

  • It is true if one considers the special case of coalgebras over a field $k$ (a proof of this, can be found at Hopf algebras: an introduction, Dascalescu et al, p. 25-26, proposition 1.4.9),
  • However, in the general case that we have an $A$-coring $C$ instead of a coalgebra over a field $k$ (now the linear structure of $C$ is that of an $A$-bimodule ${}_AC_A$, where $A$ is an arbitrary -that is: non-commutative in general- algebra) the statement is not valid as it is: In this case, the coideal is by definition the kernel of a surjective $A$-coring morphism (and not of any $A$-coring morphism). For this last case, see the proof and futher details at: Corings and Comodules, Brzezinski et al, proposition 17.14, p. 177-178.
    The same situation holds in the case of coalgebras over a commutative ring $R$ (for a proof of the case where the "scalars" are elements of a commutative ring $R$, see: Corings and Comodules, Brzezinski et al., proposition 2.4, p. 9-10)
$\endgroup$
  • $\begingroup$ Thanks for the references. I am taking my statement from a paper by Sweedler "The prequel to the Jacobson-Bourbaki theorem" where it is clearly stated that the kernel of a coring morphism is a coideal. I'll put the definitions as an edit $\endgroup$ – tomak Apr 10 '18 at 8:14
  • $\begingroup$ @tomak: thank you for your edit. It contributes to clarifying the problem. However, i tried a little search on the Sweedler's paper you are citing and i was not able to find it online. Could you add some link to it? $\endgroup$ – KonKan Apr 10 '18 at 20:03
  • $\begingroup$ I got the pdf from my teacher. Any way I could send it to you? $\endgroup$ – tomak Apr 11 '18 at 9:00
  • $\begingroup$ Could you possibly use an upload-a-file-to-share facility, such as for example: expirebox.com and share the corresponding link here? $\endgroup$ – KonKan Apr 11 '18 at 15:27
  • $\begingroup$ @tomak: you could also send me an email. I d' be glad to take a look at the paper you are citing. Of course you should also ask your teacher ;) $\endgroup$ – KonKan Apr 16 '18 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.