3 ms·
I know, yes, but did not want to digress. Thanks though. Actually, it is the "extension" to K[e] with e^2=0, as "usual".
by pfortuny 7mo ago
I know, yes, but did not want to digress. Thanks though. Actually, it is the "extension" to K[e] with e^2=0, as "usual".
- measurablefunc 7mo agoThe infinitesimally thickened point.