Finding a property that is true for every left ideal but not for right ideals

Question

I'm trying to find (or prove that it cannot exist) a property that is true for all left ideals of a ring (with unity) but fails for some right ideal.

To rephrase this more rigorously:

Consider the first-order languange $\mathcal{L} = (0,+,\cdot,-)$. Given a fixed ring $R$ and a left (right) ideal $I$, we can define the $\mathcal L$-Structure $(I,0,+,\cdot,-)$ in the obvius way. I want to find a sentence $\varphi$ (or prove that one cannot exist) such that:

1. For any left ideal $I_l$, we have $I_l\models \varphi$
2. $I_r \not \models \varphi$ for some right ideal $I_r$

If there were no restrictions on the property, then this would be an easy task, but I have no idea if one exists if we restrain the property to ones that can be expressed by a FOL sentence in this language.

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I'll try to rephrase the problem without using any logical concepts so that people well-versed in ring theory but without any logical background may be able to provide help.

We want the property to be expressed by a logical formula $\varphi$ satisfying all of the following:

1. The only logical symbols allowed on the formula are $\forall, \exists, \implies, \wedge, \vee, \sim$
2. We can only quantify over elements of the ideal, and not over subsets, functions, or relations. For example, $\forall J \subseteq I$ is not allowed.
3. Quantification of elements of the ideal must always range throughout the whole ideal, so for example, you can't write $\forall x \in J$ for some $J\subseteq I$. All quantification must be of the form $\forall x \in I$ or $\exists x \in I$.
4. The property must be internal, meaning that you fix an ideal $I$, and you can only work inside that ideal.
5. The formula cannot contain the multiplicative identity of the ring.
6. The formula cannot have free variables, meaning that if $x$ appears in the formula, it should be under the scope of some $\forall x$ or $\exists x$

Answer 1

Here is an example. Fix a prime $p>0$ and let $\mathbb{F}_p$ denote the field with $p$ elements, and consider the ring $R=\begin{bmatrix} \mathbb{Z} &\mathbb{F}_p\times\mathbb{F}_p \\ 0 & \mathbb{F}_p \end{bmatrix} $. Then the subset $\begin{bmatrix} 0 &\mathbb{F}_p\times 0 \\ 0 & \mathbb{F}_p \end{bmatrix} $ is a right ideal of $R$ of size $p^2$ in which there are two elements that do not commute (eg $\begin{bmatrix} 0 &(1,0) \\ 0 & 0 \end{bmatrix} $ and $\begin{bmatrix} 0 &(1,0) \\ 0 & 1 \end{bmatrix} $). On the other hand, using the criterion from rschwieb's answer here, the only left ideal of $R$ of size $p^2$ is $\begin{bmatrix} 0 &\mathbb{F}_p\times\mathbb{F}_p \\ 0 & 0 \end{bmatrix} $, in which any two elements commute. (To see this, note that any finite subgroup of $\mathbb{Z}\oplus(\mathbb{F}_p\times\mathbb{F}_p)$ is contained in $0\oplus(\mathbb{F}_p\times\mathbb{F}_p)$.)

Thus, if $\phi$ is the sentence $\exists^{=p^2}x(x=x)\to\forall x\forall y(xy=yx)$, then $\phi$ holds in every left ideal of $R$ but not in every right ideal of $R$.

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Note that this also gives an example to the question Noah posed in the comments, of a ring that is not elementarily equivalent to its opposite, since it is first-order expressible to say there is a right ideal of size $p^2$ in which there are two non-commuting elements.