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Nov 23, 2021 at 21:26 vote accept Martin Brandenburg
Nov 23, 2021 at 21:26 comment added Martin Brandenburg What is implicit here: To arrive at a categorical characterization, we also need to remark that $C_K$ is the kernel pair of $R \to R/K$ in $\mathbf{CRing}$, and that $R \to R/K$ is the coequalizer of its kernel pair --- so this explains why we can work with congruences instead of quotients here.
Nov 23, 2021 at 21:23 comment added Martin Brandenburg I only see this answer now (I don't check notifications anymore here). Awesome, finally this problem gets solved :). Thanks a lot anonymous user.
Jan 3, 2017 at 18:12 comment added user13113 @R.vanDobbendeBruyn: Right; the overall idea was inspired by Martin's suggested approach by unitalization.
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Jan 3, 2017 at 16:54 comment added R. van Dobben de Bruyn And I suppose that the point of this answer is that $C_K$ and all other objects involved are in fact rings (rather than mere modules).
Jan 3, 2017 at 15:45 history edited user13113 CC BY-SA 3.0
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Jan 3, 2017 at 15:41 comment added user13113 @R.vanDobbendeBruyn: Hrm. In my scratchwork I had used $R \oplus K$ throughout, and only changed to $C_K$ in the final revision, so the choices were indicated by the notation. In the end they only matter up to isomorphism, so I don't think it necessary to bog it down with the details of a particular choice, but I suppose it's short enough that it shouldn't hurt much to fill in more details.
Jan 3, 2017 at 15:06 comment added R. van Dobben de Bruyn Your notation is very sloppy. Which of the two projections $p_i \colon C_J \to R$ do you use for $\pi_0$? How do you view $R$ as a subring of $C_J$ in the definition of $T$? Which 'evident isomorphism' $C_K \cong R \oplus K$ do you choose? You could take $(p_1,p_1-p_2)$, or $(p_2,p_1-p_2)$, or a sign variation thereof.
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Jan 3, 2017 at 5:56 history answered user13113 CC BY-SA 3.0