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Jul 30, 2022 at 1:25 comment added Joe Shipman @AlexKruckman , you don’t need odd degree, you only need odd prime degree, as I proved in 2007. — Joe Shipman
Sep 16, 2021 at 8:07 history edited Oscar Cunningham CC BY-SA 4.0
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Sep 13, 2021 at 17:23 vote accept Oscar Cunningham
Sep 13, 2021 at 16:59 comment added Emil Jeřábek Ok, done. -----
Sep 13, 2021 at 16:55 answer added Emil Jeřábek timeline score: 10
Sep 8, 2021 at 20:08 comment added Oscar Cunningham @EmilJeřábek Thanks. It looks like your comment has the community support. If you add it as an answer I'll accept it.
Sep 8, 2021 at 20:07 comment added Oscar Cunningham @AlexKruckman Thanks, I've fixed it.
Sep 8, 2021 at 11:22 history edited Oscar Cunningham CC BY-SA 4.0
added 200 characters in body; edited title
Sep 8, 2021 at 5:32 comment added Emil Jeřábek While we are talking axiomatizations, $\mathrm{Th}(\mathbb C,0,1,+,\cdot,\bar{})$ is axiomatized by algebraically closed fields + $\bar z$ is an automorphism of order $2$.
Sep 7, 2021 at 20:28 comment added Alex Kruckman To axiomatize RCF, in addition to saying that $-1$ is not a sum of squares and every odd-degree polynomial has a root, you need to say that for all $a$, $a$ or $-a$ is a square.
Sep 7, 2021 at 19:04 comment added Asaf Karagila @Emil: Imagine how much money we can make if we start charging a fee for constants. We can fund whole research centres with minimal bureaucracy!
Sep 7, 2021 at 16:25 comment added Emil Jeřábek $\mathrm{Th}(\mathbb R,0,1,+,\cdot)$ is bi-interpretable with $\mathrm{Th}(\mathbb C,0,1,i,+,\cdot,\bar{})$ (you don’t really need $0$ and $1$ in either structure, they are definable). As you say, bi-interpretations preserve the automorphism group, so the result is not true without the $i$ constant. The thing is, model theorists are generally used to take constants from the model for free.
Sep 7, 2021 at 16:08 history edited Oscar Cunningham CC BY-SA 4.0
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Sep 7, 2021 at 15:54 history asked Oscar Cunningham CC BY-SA 4.0