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$\mathbb C$ is an algebraic closed field with characteristic $0$, hence $Th(\mathbb C)$ is a recursive satisfiable complete theory, thus recursive axiomatizable. So if $\mathbb N$ is definable in $\mathbb C$, then $+$,$\cdot$ are definable, hence $0$,$1$,$<$,$S$(successor),$E$(power) are definable too.

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Without exponentiation, it is known that $Th(\mathbb C,+,\cdot)$ is decidable, therefore $\mathbb N$ cannot be definable. –  sdcvvc Nov 29 '12 at 12:28
    
@sdcvvc I'm not quite clear why the undefinability of $\mathbb N$ can be immediately concluded by $Th((\mathbb C,+,\cdot))$ is decidable? –  Popopo Nov 29 '12 at 14:02
    
You could recursively transform formulas speaking about $(\mathbb N, +, \cdot)$ into formulas speaking about $(\mathbb C, +, \cdot)$, by relativizing the universe to $\mathbb C$. However, $Th(\mathbb N, +, \cdot)$ is undecidable. –  sdcvvc Nov 29 '12 at 16:40
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Also, if you have $\exp$ and $\pi$ note that you can define integers as solutions of $\exp (2 \pi n)=1$, and define $\mathbb N$ in $\mathbb Z$, e.g. with Lagrange's four squares theorem. –  sdcvvc Nov 29 '12 at 16:41
    
You don't need $\pi$, it's enough to have the exponential function, because $\mathbb{Z}$ is precisely the set of elements in $\mathbb{C}$ which fixes the set of solutions to $exp(z)=1$ under multiplication. –  Marcel T. Aug 3 '13 at 23:27

1 Answer 1

up vote 2 down vote accepted

The theory of algebraically closed fields of characteristics zero is strongly minimal.

This means that every definable set [in every model] is finite or co-finite.

I took this fact from: A Guide to Classical and Modern Model Theory, p.77.

If you search for a reference about minimal and strongly minimal theories I am sure you can find more about this, including various proofs if you are interested in them.

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I see, $Th(\mathbb C)$ is strongly minimal, hence $\mathbb N$ is not definable due to it is neither finite nor co-finite. –  Popopo Nov 29 '12 at 12:48
    
@Popopo: Exactly. –  Asaf Karagila Nov 29 '12 at 12:51
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@GEdgar: Well, there is no real reason to expect that it would be. Definable essentially means expressible by polynomials, and there is no polynomial whose zeros are just $\mathbb R$... –  Asaf Karagila Nov 29 '12 at 13:13
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Well, we can add the extra operation of complex conjugation, as we often do. Then the reals can be defined, but the integers still cannot. –  GEdgar Nov 29 '12 at 13:18
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@GEdgar: If you talk about "as we often do" then we can also use second-order formulas and then the integers can be defined... –  Asaf Karagila Nov 29 '12 at 13:20

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