Meaning for a structure $\mathcal{M}$ to be said a model of a theory $T$ if the language of $\mathcal{M}$ is the same as the language of $T$?

My question come from this

"A structure $${\mathcal {M}}$$ is said to be a model of a theory T if the language of $$\mathcal {M}$$ is the same as the language of T and every sentence in T is satisfied by $$\mathcal {M}$$. Thus, for example, a "ring" is a structure for the language of rings that satisfies each of the ring axioms, and a model of ZFC set theory is a structure in the language of set theory that satisfies each of the ZFC axioms." https://www.wikiwand.com/en/Structure_(mathematical_logic)

I don't get

"if the language of $$\mathcal {M}$$ is the same as the language of T"

Then they set an example

"Thus, for example, a "ring" is a structure for the language of rings that satisfies each of the ring axioms"

I don't know what the language of ring is : for me rings is a formal theory constructed over a first order logic language (hence a formal system ), a language which can have different formal symbols (provided by its alphabet) from one of its model : if one takes the set of integer, one can substitutes some formal symbols of the theory with integers and addition symbol and multiplication symbol (if they were not used already in the theory), though the syntax remains, one changed some symbols, therefore the language used by the model is not the same as in the theory, where am I wrong?

• The language of rings requires, at a minimum, two distinct binary function symbols. You might also use a constant symbol (to represent $0$) but that's not necessary because you can use axioms to define $0$. The language of groups, in contrast, requires only one binary function symbol. Commented Apr 18, 2021 at 20:18

Suppose I have a ring $$\mathcal{R}=(R; +^\mathcal{R},\times^\mathcal{R}, 0^\mathcal{R},1^\mathcal{R})$$. There are two first-order languages which are naturally associated to $$\mathcal{R}$$:

• The "nothing-taken-for-granted" language $$\Sigma_{rings}$$ which just has the binary function symbols $$+$$ and $$\times$$ and the constant symbols $$0$$ and $$1$$. This is the "language of rings."

• The "all-about-$$\mathcal{R}$$" language $$\Sigma_{\mathcal{R}}$$ which has, in addition to the four symbols above, a constant symbol $$c_r$$ for every $$r\in R$$.

There's a crucial subtlety here: despite the tight connection between $$\mathcal{R}$$ and $$\Sigma_\mathcal{R}$$, the structure $$\mathcal{R}$$ is not a $$\Sigma_\mathcal{R}$$-structure but merely a $$\Sigma_{rings}$$-structure; this was indicated when I wrote $$\mathcal{R}=(R; +^\mathcal{R},\times^\mathcal{R}, 0^\mathcal{R},1^\mathcal{R})$$ at the beginning of this answer. What is true is that $$\mathcal{R}$$ has a natural expansion $$\mathcal{R}_{nameeverything}$$ to $$\Sigma_\mathcal{R}$$. We often conflate $$\mathcal{R}$$ and $$\mathcal{R}_{nameeverything}$$ and e.g. write "$$\mathcal{R}\models r+s=t$$" in place of "$$\mathcal{R}_{nameeverything}\models c_r+c_s=c_t$$," but strictly speaking they are different objects.

• Thank you, I have few questions about what you said : · the Σ𝑟𝑖𝑛𝑔𝑠 is like a signature isn't it? · $\Sigma_\mathcal{R}$ seems to me denoting the language of the theory of rings (particularly because of " 𝑐𝑟 for every 𝑟∈𝑅"), I mean does those 𝑐𝑟 are uses as the arguments of the function interpretation for other sets than 𝑅, like $N$? · you use the convention "$+^\mathcal{R}$" which I saw elsewhere, does it means something like "+ is the binary function symbol of the structure $\mathcal{R}$"? Is there a particular reason why the $\mathcal{R}$ symbol is placed as an exponent? Commented Apr 22, 2021 at 15:35
• what you describe when "we conflate $\mathcal{R}$ and $\mathcal{R}$𝑛𝑎𝑚𝑒𝑒𝑣𝑒𝑟𝑦𝑡ℎ𝑖𝑛𝑔" is an homomorphism, isn't it? Commented Apr 22, 2021 at 15:42
• @GuilhemEscudéro The $\Sigma_{rings}$ is a signature ("signature" and "language" are synonymous). $\Sigma_\mathcal{R}$ is not the language of the theory of rings - for different rings $\mathcal{R},\mathcal{S}$ we'll generally have $\Sigma_\mathcal{R}\not=\Sigma_\mathcal{S}$! The language of rings should be the language common to all rings - and this is exactly $\Sigma_{rings}$. Commented Apr 22, 2021 at 17:54
• Yes, "$+$" is a binary function symbol and "$+^\mathcal{R}$" is its interpretation in $\mathcal{R}$. As to where this exponent-like notation came from, I have no idea but it's standard at this point. Finally, that conflation is not a homomorphism - homomorphisms are maps between structures in the same language, and $\mathcal{R}$ has a different language than $\mathcal{R}_{nameeverything}$! The language of the former is the language of rings $\Sigma_{rings}$, while the language of the latter is $\Sigma_\mathcal{R}$, a particular extension of the language of rings special to $\mathcal{R}$. Commented Apr 22, 2021 at 17:54
• Thanks @noahschweber ! So to summarize (under your rating if you don't mind) : · $\mathcal{R}_\text{nameeverything}$ and $\mathcal{R}$ are algebraic structures · $\Sigma_{\mathcal{R}}$ and Σ𝑟𝑖𝑛𝑔𝑠 are signatures · $\Sigma_{\mathcal{R}}$ is defined by a "natural expansion of $\mathcal{R}$ : from $\mathcal{R}_\text{nameeverything}$ to $\Sigma_{\mathcal{R}}$ · it is meant to conflate $\mathcal{R}$ and $\mathcal{R}$𝑛𝑎𝑚𝑒𝑒𝑣𝑒𝑟𝑦𝑡ℎ𝑖𝑛𝑔 so we can write "$\mathcal{R}$ ⊨ 𝑟+𝑠=𝑡" in place of "$\mathcal{R}$𝑛𝑎𝑚𝑒𝑒𝑣𝑒𝑟𝑦𝑡ℎ𝑖𝑛𝑔 ⊨ 𝑐𝑟+𝑐𝑠=𝑐𝑡" Commented Apr 24, 2021 at 9:17