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?
Best Answer
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 $\geq 0$.