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.
Best Answer
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.