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The claim on this page that the second part of the definition of elementary substructure can be expressed by stating simply that ${\displaystyle \mathrm {Th} _{L(M)}(M)=\mathrm {Th} _{L(M)}(N)}$ is incorrect if ${\displaystyle \mathrm {Th} _{L(M)}(M)}$ refers to the set of first-order ${\displaystyle L}$-sentences satisfied by ${\displaystyle M}$. In this case, the latter is a statement only of elementary equivalence, and it is possible for a structure to have elementarily equivalent substructures that are not elementary substructures. As an example, consider the structure ${\displaystyle A=(\mathbb {N} ,<)}$ consisting of the naturals with the usual ordering. It is easily seen that the substructure consisting of the evens with the induced ordering is isomorphic, and hence elementarily equivalent, to ${\displaystyle A}$; but it is not an elementary substructure by Tarski's criterion, since every natural is definable in ${\displaystyle A}$ (the natural ${\displaystyle n}$ by the formula stating 'there exist exactly ${\displaystyle n-1}$ elements less than me').

On the other hand, the author may be using a slightly nonstandard notation, where ${\displaystyle \mathrm {Th} _{L(M)}(M)}$ refers to something else.

This is notation I've seen before, ${\displaystyle L(M)}$ refers to the language ${\displaystyle L}$ extended with constant symbols ${\displaystyle a_{n}}$ for every element ${\displaystyle n}$ in ${\displaystyle M}$. So ${\displaystyle \mathrm {Th} _{L(M)}(M)}$ is the set of first order sentences satisfied by ${\displaystyle M}$ when these constants are included, i.e. the theory of ${\displaystyle M}$ considered as a structure for language ${\displaystyle L(M)}$ (called the elementary diagram). This gives ${\displaystyle \mathrm {Th} _{L(M)}(M)=\mathrm {Th} _{L(M)}(N)}$ as a statement equivalent to the second part of the definiton holds, but using constants symbols rather than parameters ${\displaystyle a}$. This works because ${\displaystyle \mathrm {Th} _{L(M)}(M)}$ contains sentences stating that no two constants are equal!

I hope this makes sense, the only decent definiton/proof I can find online is on page 13 of http://www.maths.ox.ac.uk/~zilber/lect.pdf ---Rsimmonds01 15:16, 2 June 2006 (UTC)[reply]

Thanks for the info. That indeed makes sense, since any formula whose free variables are interpreted as elements of ${\displaystyle M}$ will be equivalent to a sentence using the appropriate introduced constants.

It might be a good idea to define that on the page (or reference a definition, if one exists), unless you feel that the notation is common enough.--Blargoner 09:49, 3 June 2006 (UTC)[reply]

Good idea, I'll do that. --Richard CHS 12:51, 7 June 2006 (UTC)[reply]

"More generally, any first-order theory has non-isomorphic, elementary equivalent models, which can be obtained via the Löwenheim–Skolem theorem. Thus, for example, there are non-standard models of Peano arithmetic, which contain other objects than just the numbers 0, 1, 2, etc., and yet are elementarily equivalent to the standard model"

Isn't this claim false? Don't we need the condition that the first-order theory have infinite models? (I.e. doesn't the theory that (forall x)(forall y)(x = y) have a single model?) Or are we not assuming that each model interprets '=' as the diagonal relation? — Preceding unsigned comment added by 128.135.100.105 (talk) 19:24, 26 November 2012 (UTC)[reply]

The current version states:

An elementary embedding of a structure N into a structure M of the same signature σ is a map h: N → M such that for every first-order σ-formula φ(x₁, …, x_n) and all elements a₁, …, a_n of N,

N ${\displaystyle \models }$ φ(a₁, …, a_n) if and only if M ${\displaystyle \models }$ φ(h(a₁), …, h(a_n)).

Every elementary embedding is a strong homomorphism, and its image is an elementary substructure.

The previous version had "implies" in place of "if and only if". It does not really make any difference but just for the purposes of understanding the issue, it seems to me that "implies" is also accurate. This is because we quantify over all φ, so in particular the negation of it will give the opposite implication. Just wondering if this is right. Tkuvho (talk) 14:36, 12 June 2013 (UTC)[reply]

Yes, that seems correct to me. The key point is that e.g. ${\displaystyle N\not \models \phi }$ if and only if ${\displaystyle N\models \lnot \phi }$. — Carl (CBM · talk) 15:20, 12 June 2013 (UTC)[reply]

Then I think the definition would be clearer with "implies" instead of "if and only if", because that would seem to be the more significant direction in the context of embeddings. Tkuvho (talk) 12:16, 13 June 2013 (UTC)[reply]

I don't have any strong preference, but if there is some potential for confusion, maybe the "if and only if" is better, just because it's more direct. — Carl (CBM · talk) 17:48, 13 June 2013 (UTC)[reply]

It would we useful to add (to this article) that

A theory T is said to have definable Skolem functions if for every formula ϕ(x;y) there is a definable function f(y) such that whenever M⊨T, b∈M, and ϕ(M;b) is non-empty, then f(b)∈ϕ(M;b). This is a weaker condition than having Skolem functions.

An equivalent condition to having definable skolem functions is that every definably closed subset of a model is an elementary substructure.

(For now, just copy-pasted from modeltheory.wikia.com.)
A source: "A note on definable Skolem functions" by D. A. Anapolitanos.
Boris Tsirelson (talk) 13:52, 5 December 2018 (UTC)[reply]