Timeline for Quotient of $Z[x_1,...,x_n]$ by a maximal ideal is a finite field [duplicate]
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| May 20, 2018 at 12:58 | comment | added | Todd Trimble | The link is irreparable since Stallings's Berkeley page seems to be long gone (he died in 2008). But let's leave it up in case someone has any idea where those notes can be found (maybe they downloaded a copy). | |
| Aug 3, 2017 at 10:55 | comment | added | Benjamin Steinberg | @Arrow, I don't know where they can be found now. | |
| Aug 3, 2017 at 9:40 | comment | added | Arrow | @BenjaminSteinberg the link you posted is broken. Where can I find those notes? | |
| Oct 12, 2013 at 13:03 | vote | accept | Hugo Rafael Oliveira Ribeiro | ||
| Oct 12, 2013 at 8:26 | history | closed |
Martin Brandenburg Chris Godsil Evan Jenkins Ricardo Andrade Carlo Beenakker |
Duplicate of A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite | |
| Oct 11, 2013 at 23:16 | comment | added | Benjamin Steinberg | @Margaux, what's in a name :) | |
| Oct 11, 2013 at 23:11 | review | Close votes | |||
| Oct 12, 2013 at 8:26 | |||||
| Oct 11, 2013 at 22:03 | comment | added | Marguax | @Benjamin Steinberg: Sorry, I don't know anything about mathematical logic. Also, perhaps it's a matter of convention for the names of results, but when working over a general field I always thought that the meaning of "Nullstellensatz" is exactly what is called "Zariski's Lemma" by you and Wikipedia (corporations are people, my friend). | |
| Oct 11, 2013 at 15:22 | comment | added | Benjamin Steinberg | @Marguax, what you really seem to be using is Zariski's lemma en.wikipedia.org/wiki/Zariski%27s_lemma. It seems to me that this is not really any easier than first order logic, although it has the advantage of being part of Commutative Algebra. The logic proof just uses ultraproducts and that two algebraically closed fields of the same characteristic and cardinality are isomorphic. It takes a couple of pages in Marker's book. | |
| Oct 11, 2013 at 2:55 | comment | added | Marguax | Let $K$ be a field finitely generated as a $\mathbf{Z}$-algebra. Assume $\mathbf{Q}\subset K$, so $K$ is finitely generated as a $\mathbf{Q}$-algebra, so $\mathbf{Q}$-finite by the Nullstellensatz over fields. A finite set of $\mathbf{Z}$-algebra generators of $K=\mathbf{Q}\otimes_{\mathbf{Z}}O_K$ lies in $O_K[1/N]$ for a sufficiently divisible integer $N>0$, yet $O_K[1/N]$ is a ring, so $K=O_K[1/N]$, absurd (Euclid!). Thus, $p:={\rm{char}}(K)>0$, so $K$ is a finitely generated $\mathbf{F}_p$-algebra, so $\mathbf{F}_p$-finite (Nullstellensatz again). No Jacobson rings, no first-order logic. :) | |
| S Oct 11, 2013 at 2:50 | history | suggested | ctype.h | CC BY-SA 3.0 |
Improved grammar, link description
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| Oct 11, 2013 at 2:40 | comment | added | Benjamin Steinberg | A very short proof due to Shalen can be found in Lemma 10.3 and Thm 10.4 of math.berkeley.edu/~stall/ch3oz.pdf | |
| Oct 11, 2013 at 2:16 | review | Suggested edits | |||
| S Oct 11, 2013 at 2:50 | |||||
| Oct 11, 2013 at 1:39 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Oct 11, 2013 at 0:30 | answer | added | Steven Landsburg | timeline score: 7 | |
| Oct 11, 2013 at 0:01 | review | First posts | |||
| Oct 11, 2013 at 0:03 | |||||
| Oct 10, 2013 at 23:41 | history | asked | Hugo Rafael Oliveira Ribeiro | CC BY-SA 3.0 |