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Let $\mathcal{L}$ be a first-order language, and $M$ and $N$ be two $\mathcal{L}$-structures. We say that $M$ and $N$ are elementarily equivalent (write $M \approx N$) if they satisfy the same first-order formulas in the language $\mathcal{L}$. It is clear that if $M$ and $N$ are isomorphic (write $M \simeq N$), then $M \approx N$, but the converse does not holds (see below).

It is an important problem in algebra, given an algebraic structure $M$, to determine what algebraic properties are determined by its elementary equivalence class. This is related to the problem of determining which algebraic properties can be expressed in first-order logic, and has some classical problems, such as Tarski's Problem As a trivial exemple, given two fields $K$ and $L$, if $K \approx L$, then they have the same characteristic; as a less trivial example, if $K \approx L$ then $K$ is algebraically closed if and only iff $L$ is algebraically closed.

As an example of two elementarily equivalent structures which are not isomorphic, we can consider the algebraic closures $K$ and $L$ of $\mathbb{Q}(x)$ and $\mathbb{Q}(x,y)$, respectively. Since the theory of algebraically closed fields of zero characteristic is complete, clearly $K \approx L$; however, the fields are not isomorphic, as $K$ has transcendece degree $1$ over its prime field; and $L$ has transcendence degree $2$.

The theory of elementary equivalence of fields is a mature subject with many references (for one which I found very interesting, we have this)

In looking for references for the theory of elementary equivalence of rings. I know some classical results (for example, if the rings $R$ and $S$ are elementarily equivalent, $R$ is Jacobson-semisimple if and only if $S$ is Jacobson-semisimple; if $R$ and $S$ are elementarily equivalent commutative Noetherian rings and $R$ has finite Krull dimension $n$, then $S$ also has Krull dimension $n$; the first-order theory of rings is undecidable; etc), but I could not find many papers dealing specifically with this topic.

An exception is the very interesting paper by Kharlampovich and Myasnikov about the analogue of Tarski's Problem for the free associative algebras (the paper can be found here).

So, I'm interested in surveys and important papers about the elementary theory of rings.

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    $\begingroup$ One issue with a 'complete' development of elementary equivalence for rings is that many algebraically tame rings are model-theoretically wild. The full first-order theory of the integers is badly not computable, for instance. I think a lot more is understood about the model theory of modules over a fixed ring. $\endgroup$ Commented Nov 27, 2023 at 15:41
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    $\begingroup$ I recommend the books Model Theoretic Algebra by Jensen and Lenzing and The Use of Ultraproducts in Commutative Algebra by Hans Schoutens. Both should contains lots of relevant information. $\endgroup$ Commented Nov 27, 2023 at 16:10
  • $\begingroup$ For the special class of rings called Boolean algebras the (countably many) elementary types were described by Tarski. I don't believe Tarski published this work, but there is a 1964 exposition by Yu. L. Ershov in Russian. By now it is probably covered in some English text. $\endgroup$
    – bof
    Commented Nov 28, 2023 at 0:11
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    $\begingroup$ Elementary equivalence of rings subsumes elementary equivalence of models of arithmetic (Peano arithmetic and its fragments), as any such model has a canonical extension with negative elements to become a ring. There is tons of literature on models of arithmetic. In any case, this should tell you that determining the elementary equivalence of two given rings is in general very hard, and there is no simple set of invariants like the characteristic and Krull dimension. $\endgroup$ Commented Nov 28, 2023 at 7:37
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    $\begingroup$ Expanding on Emil's comment a bit, the point is that the arithmetic of integers proves (via Lagrange's theorem) that the non-negative integers are exactly the sums of four squares. This means that the ring of "integers" obtained from a model of arithmetic already knows all the arithmetic. Moreover, since integers are also quantifier-free definable in the field of rationals, even just the model theory of (arbitrary) fields already subsumes arithmetic, never mind arbitrary rings. $\endgroup$
    – tomasz
    Commented Nov 28, 2023 at 17:22

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