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 at 13:32 |
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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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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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