Timeline for Rational numbers as an extension of the field with one element?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2010 at 1:52 | comment | added | JBorger | at times field-specific concepts in descent theory do not translate so well. In retrospect, this is not so surprising, because the importance of fields, as opposed to rings, is mostly of a psychological or historical nature (IMHO). | |
| Mar 24, 2010 at 1:46 | comment | added | JBorger | the relative Galois theory. This is in contrast to usual field theory, where for any extension $L/K$, the functor $L\otimes_K -$ from $K$-algebras to $L$-algebras always has the descent property; but it is similar to general ring theory where, for instance, the functor $\mathbf{Q}\otimes_{\mathbf{Z}}-$ does not have the descent property. The moral of the story is that, from the $\Lambda$-ring point of view, $\mathbf{F}_1$ fails to have lots of properties that usual fields (as opposed to rings) have. So it's a bit safer to think of this $\mathbf{F}_1$ as a generalized ring, and so... | |
| Mar 24, 2010 at 1:41 | comment | added | JBorger | Thanks for the kind words. I'd just like to emphasize (not to suggest that you meant or said otherwise) that these analogies ($\mathrm{Gal}(\bar{\mathbf{Q}}/\mathbf{F}_1)$ and so on) should be taken with a grain of salt. For instance, it is good to think of the functor $\mathbf{Q}\otimes -$ from $\Lambda$-rings to $\mathbf{Q}$-algebras as the functor $\mathbf{Q}\otimes_{\mathbf{F}_1} -$, but this functor does not have the descent property, which is what Galois theory is really used for. So while it is reasonable to call $\mathbf{Q}$ an $\mathbf{F}_1$-algebra, it's a bit abusive to speak of... | |
| Mar 23, 2010 at 20:03 | history | edited | lieven lebruyn | CC BY-SA 2.5 |
added 45 characters in body; added 45 characters in body
|
| Mar 23, 2010 at 19:38 | history | answered | lieven lebruyn | CC BY-SA 2.5 |