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Any other definition for algebraic number than the root of algebraic equation?

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closed as not a real question by Dan Petersen, Steven Landsburg, ACL, Chandan Singh Dalawat, Chris Godsil Mar 7 '13 at 13:32

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What do you mean? Do you want a definition equivalent to the one you give or you want to ask if there are some texts where algebraic number means something else? – Jérémy Blanc Mar 7 '13 at 12:25
Perhaps the OP is looking for some other characterisation of algebraic numbers than "is the root of a monic rational polynomial"? – Ketil Tveiten Mar 7 '13 at 12:29
@Gerald: You might want to eliminate the first syllable of "independent". – Lee Mosher Mar 7 '13 at 12:59
Eigenvalues of a rational matrices. – Name Mar 7 '13 at 13:03
@Lee: I noticed the error, but have to wait for 2.0 to edit comments. – Gerald Edgar Mar 7 '13 at 13:18
up vote 12 down vote accepted

In model theory, an object is algebraic in a structure $M$ if it satisfies a property that only finitely many other objects in $M$ exhibit, where by "property" here we mean one that is expressible in the first-order language of the structure. This is a weakening of definability, since $a$ is definable in $M$ if it satisfies a property that no objects other than $a$ have.

More generally, we say that $a$ is algebraic in $M$ over $A$, where $A\subset M$, to mean that one may use parameters from $A$ in describing the property. Thus, $a$ is algebraic in $M$ over $A$ if there is some first-order formula $\varphi$ and parameters $\vec c$ in $A$ such that $M\models \varphi(a,\vec c)$ and $\{b\in M\mid M\models\varphi(b,\vec c)\}$ is finite.

This general notion of an algebraic object is often illuminating in diverse mathematical structures, such as graphs, digraphs, orders, lattices and so on, where one has no polynomials.

Meanwhile, it agrees with the usual notion in the structure $\langle\mathbb{R},{+},{\cdot},0,1\rangle$, for example, and similarly for $\langle\mathbb{C},{+},{\cdot},0,1\rangle$, where the algebraic elements are precisely the solutions of (nondegenerate) integer polynomial equations. Of course, any such solution set is finite, but conversely, it follows from Tarski's deep theorem on elimination of quantifiers in real-closed fields, that every finite definable set in these structures turns out to be describable as a solution set of such polynomials. It is quite remarkable that quantification is so powerless in real-closed fields, and polynomial equations already capture the full expressive power of the language.

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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. – Joel David Hamkins Mar 7 '13 at 18:37
This is very unfirmative, but I can't help feeling it answers a different question from the one intended by the OP – Yemon Choi Mar 7 '13 at 18:58
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). – Joel David Hamkins Mar 8 '13 at 0:24
Joel: stubby fingers have undone me again :) (It was meant to be "informative", as you no doubt guessed – Yemon Choi Mar 8 '13 at 1:47
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. – XL _at_China Oct 29 '14 at 12:17

Assuming ZFC, one can characterize the algebraic numbers in $\mathbb{C}$ as those numbers which lie in finite orbits under the action of ${\rm Aut}(\mathbb{C})$ on $\mathbb{C}$.

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Thank you Stefan,I think I should accept your post as an answer either,but it is only possible to accept one. – XL _at_China Mar 8 '13 at 0:34

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