Is the field $\mathbb{Q}(\pi)$ elementarily equivalent to $\mathbb{Q}$?

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?

• Is your formula $\sigma$ correct? It does hold in $\Bbb Q$: $y=z=2$: $2\cdot 2=2+2$. Sep 18, 2020 at 0:57
• 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. Sep 18, 2020 at 0:57
• $0, 1$ 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 $\forall z a \cdot z=z$. Sep 18, 2020 at 1:00
• 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. Sep 18, 2020 at 1:03
• @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 Sep 18, 2020 at 1:04

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 $$\geq 0$$.