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Consider the first-order language of ring theory (here rings are defined with 1): We have variables $x_1,x_2,...$, constant symbols $0,1$, the binary function symbols $+, *$ and the unary function symbol $-$ (used to assign an element $a$ with its additive inverse $-a$).

One possible structure for this language could be, for example: $\mathcal U=(\mathbb R,0,1,+,\cdot,-)$ (all defined in the usual way).

We can create a sentence $\varphi$ such that $\mathcal U\models\varphi$ iff $\cal U$ is a ring.

My question is, is it possible to, in this language, to "define" an Ideal of a ring?

My guess is no, because if we have a structure $\cal U\models\varphi$ ($\cal U$ is a ring), then an ideal would be a subset of the universe of $\cal U$, and first-order logic does not have the ability to do operations such as quantification with subsets of the universe of $\cal U $, but I'm not sure about this. Maybe I need to go to second-order logic or monadic second-order logic?

Sorry if this post sounds a little vague, but I just started learning model theory, and I'm still getting used to some of these concepts.


Edit: For some background on this question, I was taking some non-commutative algebra and I wanted to formalize the notion that if I have some property expressed as a sentence $\phi$ that is true for all ideals in a commutative ring (hence the need to quantify over ideals) such that the proof does not use the commutativity of $\cdot$, then it must also be valid of all Ideals inside a non-commutative ring. But I'm having some trouble defining an ideal in the language of rings.

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  • $\begingroup$ You could add a unary predicate to the first-order language signifying membership in $I$, and add axioms related to that unary predicate. $\endgroup$ Commented Sep 27, 2023 at 21:54
  • $\begingroup$ On the other hand, I think it's probably not possible to write a first-order sentence in the language of rings expressing the property that the ring is Noetherian, for example. $\endgroup$ Commented Sep 27, 2023 at 21:57
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    $\begingroup$ But that wouldn't allow me to quantify over ideals of a ring, for example, right? @DanielSchepler $\endgroup$ Commented Sep 27, 2023 at 21:57
  • $\begingroup$ I don't know if it helps to better understand what kind of solution I'm looking for @DanielSchepler, but check the edit $\endgroup$ Commented Sep 27, 2023 at 22:03
  • $\begingroup$ A subset $I$ is an ideal if and only if it is the equivalence class of $0$ in a congruence on the ring $R$ (an equivalence relation on $R$ that is a subring of $R\times R$). Could this be leveraged? $\endgroup$ Commented Sep 27, 2023 at 22:07

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It depends exactly what you want to do.

  • We can write a sentence in the language of rings together with a new unary predicate symbol $U$ such that whenever $\mathcal{R}$ is the expansion of a ring $\mathcal{R}_0$ by interpreting $U$ as a set $X$, we have $\mathcal{R}\models\varphi$ iff $U^\mathcal{R}$ is an ideal of $\mathcal{R}_0$. This isn't hard: it's just $$U(0)\wedge \forall x,y(U(x)\rightarrow U(xy))\wedge \forall x,y(U(x)\wedge U(y)\rightarrow U(x+y)).$$

  • However, this trick won't let us quantify over ideals in any way. We could fix this by looking at a larger structure (essentially $\mathcal{R}_0\sqcup\mathcal{P}(\mathcal{R}_0)$), but general ideals are as complicated as general sets in the following sense: if we expand first-order logic by the ability to quantify over ideals of (interpretations) of rings, we get something as powerful (in a somewhat coarse sense admittedly) as second-order logic.

  • That said, ideals are often still analyzable in a first-order way by reducing to the principal case. For example, suppose $\mathcal{A}$ is a simple ring and $\mathcal{B}\preccurlyeq\mathcal{A}$. Then I claim $\mathcal{B}$ is simple as well! This is because $(i)$ every nontrivial ideal contains a nontrivial principal ideal, and $(ii)$ quantifying over principal ideals is no harder than quantifying over elements. (A similar argument shows that elementary substructures of simple groups are simple groups.)

So whether or not ideal-ness is out of reach of first-order logic depends on exactly what you mean by that in the first place.

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I wanted to formalize the notion that if I have some property expressed as a sentence $\phi$ that is true for all ideals in a commutative ring (hence the need to quantify over ideals) such that the proof does not use the commutativity of $\cdot$, then it must also be valid of all Ideals inside a non-commutative ring.

This doesn't require formalization. If you have a proof, and it doesn't use commutativity, then it's valid in the noncommutative case, by hypothesis!

Of course you need to be very careful about whether your proof in fact does not use commutativity; e.g. commutativity is already used in the use of the word "ideal" without qualifying whether you mean left, right, or two-sided ideals, so if at any point in the proof you only check that something is a left ideal but then quotient by it to get a quotient ring as if it were a two-sided ideal, then you've used commutativity.

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