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I'm looking to come up with a $\mathcal{L}_{ring}{-sentence}$ which have constant symbols $0, 1$ and binary functions $+, \times$.

I'd like to find a sentence using this structure that compares $\mathbb{Z}$ and $\mathbb{C}$ such that it satisfies one but not the other.

I understand the differences, I'm just having a hard time expressing it in terms of logic..

To give an example using $(\mathbb{Z}, \mathbb{Q})$ we can define $\phi = \forall x \forall y\exists z(y\times z =x).$ This statement only satisfies the rationals, but does not necessarily for integers.

How do I come up with something along these lines for the complex numbers?


Would something like

$\forall x ( x\times x=(0-1))$ work?

The only issue is that I don't have a subtraction operator.


$\exists x ( x\times x + 1 = 0)$

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  • $\begingroup$ Hint: what's the defining property of $i$? $\endgroup$ – Steven Stadnicki Apr 9 '18 at 15:58
  • $\begingroup$ The only thing I can think of is $i=\sqrt{-1}$ $\endgroup$ – 1011011010010100011 Apr 9 '18 at 15:59
  • $\begingroup$ Do you mean to compare $\mathbb{Q}$ or $\mathbb{R}$ to $\mathbb{C}$? Because otherwise your original $\phi$ (with a small fix, which is needed in the original case: think about $y=0$ ...) does the job. $\endgroup$ – Noah Schweber Apr 9 '18 at 15:59
  • $\begingroup$ I'd like to compare $\mathbb{Z}$ to $\mathbb{C}$ $\endgroup$ – 1011011010010100011 Apr 9 '18 at 16:02
  • $\begingroup$ I have edited the above. Would this make sense? $\endgroup$ – 1011011010010100011 Apr 9 '18 at 16:20
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In $\mathbb{C}$ the formula $\exists x: x \times x + 1 = 0$ holds, which is expressed in terms of $0,1, +, \times$. This does not hold in $\mathbb{Z}$, nor in $\mathbb{Q}$ or $\mathbb{R}$.

But maybe easier : $\forall x : \lnot (x = 0) \to (\exists y: x \times y =1)$ does not hold in $\mathbb{Z}$ (as it is not a field), but it does hold in $\mathbb{R}, \mathbb{C}, \mathbb{Q}$ or any other field.

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    $\begingroup$ And (conversely) $\exists x. (x\neq 0 \land (\forall y.xy \neq 1))$ holds in $\Bbb{Z}$ but not in any field. $\endgroup$ – Rob Arthan Apr 9 '18 at 19:38

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