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Oct 29, 2014 at 12:17 comment added XL _At_Here_There Joel, your answer or the definition by model theory is so universal, I am wondering whether there is any linkage between transcendental number theory and such a definition.
Apr 4, 2013 at 3:31 vote accept XL _At_Here_There
Mar 8, 2013 at 1:47 comment added Yemon Choi Joel: stubby fingers have undone me again :) (It was meant to be "informative", as you no doubt guessed
Mar 8, 2013 at 0:32 vote accept XL _At_Here_There
Apr 4, 2013 at 3:31
Mar 8, 2013 at 0:31 vote accept XL _At_Here_There
Mar 8, 2013 at 0:32
Mar 8, 2013 at 0:24 comment added Joel David Hamkins Oh, I'm very sorry if my answer has caused any looseness! :-) I probably agree with you, though, since I am unsure myself what kind of answer the OP may have had in mind. Meanwhile, I took the opportunity to mention a notion of algebraicity that I suspect is not so widely known outside logic, but which I find to be very interesting and easy to understand (and which answers the question).
Mar 7, 2013 at 18:58 comment added Yemon Choi This is very unfirmative, but I can't help feeling it answers a different question from the one intended by the OP
Mar 7, 2013 at 18:37 comment added Joel David Hamkins In any sufficiently saturated structure, an object is algebraic if and only if it has finite orbit under the action of the automorphism group. But these concepts are not equivalent in an arbitrary structure, for non-algebraic elements can nevertheless be fixed by every automorphism, as they are in $\langle\mathbb{R}, {+},{\cdot},0,1,\lt\rangle$. But meanwhile, an element is algebraic if and only if it has finite orbit in every elementary extension of the structure.
Mar 7, 2013 at 17:15 history edited Joel David Hamkins CC BY-SA 3.0
Improved exposition; added 167 characters in body
Mar 7, 2013 at 12:32 history answered Joel David Hamkins CC BY-SA 3.0