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.