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 $\exists x$
  • $\begingroup$ Would "is finitely generated" work? There are rings that are Noetherian on one side but not the other, so e.g., every left ideal is finitely generated, but there are right ideals that are not. $\endgroup$ Feb 12 at 1:53
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    $\begingroup$ @ArturoMagidin That's definitely not first-order. $\endgroup$ Feb 12 at 2:39
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    $\begingroup$ Incidentally, a question of vaguely-related flavor (to which I recall seeing a positive answer on MO, but I can't verify that at the moment): is there a ring which is not elementarily equivalent to its opposite ring? $\endgroup$ Feb 12 at 2:45

1 Answer 1


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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    $\begingroup$ That's a wonderful example. Thank you. May I just ask what you mean by $\exists^{=p^2}$? I don't remember seeing this notation $\endgroup$ Feb 12 at 18:39
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    $\begingroup$ @EduardoMagalhães my pleasure, happy to help! given a natural number $n$ and a formula $\psi(x)$, one may write $\exists^{\geqslant n}x\psi(x)$ to abbreviate the formula $\exists x_1\dots\exists x_n\left[\bigwedge_i\psi(x_i)\wedge\bigwedge_{i\neq j}x_i\neq x_j\right]$. then $\exists^{<n}x\psi(x)$ abbreviates $\neg\exists^{\geqslant n}x\psi(x)$, and $\exists^{=n}x\psi(x)$ abbreviates $\exists^{\geqslant n}x\psi(x)\wedge\exists^{<n+1}x\psi(x)$. it is often a convenient shorthand :) $\endgroup$ Feb 12 at 19:17

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