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In model theory, we say that two structures are elementarily equivalent if they satisfy the same first-order sentences. For instance, in the language $\mathcal{L}=\{+,\cdot\}$, the fields $(\mathbb{Q},+,\cdot)$ and $(\mathbb{Q}(\sqrt{2}),+,\cdot)$ are not elementarily equivalent because the $\mathcal{L}$-sentence

$$\sigma: \exists y\,\exists z\,\forall x\,(x\cdot z=x \wedge y\cdot y=z+z.)$$

holds in the second field, but not in the first. (Basically, the sentence states that "there is an element whose square is equal to $1+1$")

However, is there an easy way to show that $(\mathbb{Q},+,\cdot)$ and $(\mathbb{Q}(\pi),+,\cdot)$ are (or are not) elementarily equivalent?

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  • $\begingroup$ Is your formula $\sigma$ correct? It does hold in $\Bbb Q$: $y=z=2$: $2\cdot 2=2+2$. $\endgroup$
    – markvs
    Sep 18, 2020 at 0:57
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    $\begingroup$ Surely you need $xz = x$ if you want $z$ to be the identity? But even needing to do this seems very strange to me. To me the first-order language of fields includes $0, 1$ as constants. $\endgroup$ Sep 18, 2020 at 0:57
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    $\begingroup$ $0, 1$ are not needed in the definition of the first order theory of fields. $0$ is given by the formula $\forall z: z+a=z$ and $1$ is given by the formula $\forall z a \cdot z=z$. $\endgroup$
    – markvs
    Sep 18, 2020 at 1:00
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    $\begingroup$ Yes, I get that, but it seems perverse to me. You’re just making sentences harder to write with extra quantifiers for no reason that I can see. $\endgroup$ Sep 18, 2020 at 1:03
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    $\begingroup$ @Dario: apparently first-order sentences can detect transcendence degree although the construction is quite nontrivial: math.upenn.edu/~pop/Research/files-Res/AWS03-fin.pdf $\endgroup$ Sep 18, 2020 at 1:04

1 Answer 1

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For every rational number $x$, either $x$ or $-x$ is a sum of four squares in $\mathbb{Q}$. But neither $\pi$ nor $-\pi$ is a sum of four squares in $\mathbb{Q}(\pi)$. One quick way to see this is that there is an automorphism of $\mathbb{Q}(\pi)$ swapping $\pi$ and $-\pi$, so $\mathbb{Q}(\pi)$ admits an ordering in which $-\pi<0$ and another in which $\pi<0$, but in an ordered field any sum of squares is $\geq 0$.

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    $\begingroup$ Thank you Alex! $\endgroup$
    – Darío G
    Sep 18, 2020 at 2:26

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