Ring Theory – Property True for Every Left Ideal but Not Right Ideals

first-order-logicmodel-theoryring-theory

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.

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 $\exists x$

Best Answer

Here is an example. Fix a prime $p>0$ and consider the ring $R=\begin{bmatrix} \mathbb{Z} &\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z} \\ 0 & \mathbb{Z}/p\mathbb{Z} \end{bmatrix} $. Then the subset $\begin{bmatrix} 0 &\mathbb{Z}/p\mathbb{Z}\times 0 \\ 0 & \mathbb{Z}/p\mathbb{Z} \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 &(\bar{p},0) \\ 0 & \bar{0} \end{bmatrix} $ and $\begin{bmatrix} 0 &(\bar{p},0) \\ 0 & \bar{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{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z} \\ 0 & 0 \end{bmatrix} $, in which any two elements commute. (To see this, note that any finite subgroup of $\mathbb{Z}\oplus\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ is contained in $0\oplus\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.)

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 $R$.

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.

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