Let $R$ be a ring with identity and let $I$ and $J$ be right ideals of $R$. I know that If $R$ is COMMUTATIVE then the right $R$-modules $\frac RI$ and $\frac RJ$ are isomorphic if and only if $I=J$.

What happens if $R$ is not Commutative?, i.e., is there any necessary and sufficient condition (in terms of $I$ and $J$) to force that $\frac RI \cong \frac RJ$?

  • $\begingroup$ @LuizCordeiro : they're not isomorphic as $R$-modules. The claim is true. I think the claim might also be true for noncommutative rings but with a more subtle proof. In any case, it's a good question ! $\endgroup$ – Max Sep 19 at 19:12
  • $\begingroup$ @Max Oops, you're correct. My bad. $\endgroup$ – Luiz Cordeiro Sep 19 at 19:14
  • $\begingroup$ MO crosspost: mathoverflow.net/questions/342407 $\endgroup$ – YCor Sep 25 at 9:52
  • $\begingroup$ Just to say it: one can look for a counterexample in the realm of lie algebras, by taking the enveloping algebra. In some cases one has a good understanding of cyclic modules, like in the case of highest weight modules. It's been a while I don't study this kind of stuff, so if someone is fresher maybe can have an idea. Also the case of group representations can give some constructions! $\endgroup$ – Andrea Marino Sep 25 at 11:38

(I assume $R$ is associative, guessing it's implicit.) Write $L_c(r)=cr$: then $L_c$ is an endomorphism of $R$ as right $R$-module, and $L_{cd}=L_c\circ f_d$ for all $c,d$.


(a) $R/I$ and $R/J$ are isomorphic right $R$-modules.

(b) there exists $a\in R$ such that $L_a^{-1}(J)=I$ and $aR+J=R$.

(c) there exist $a,b\in R$ such that (c1) $aI\subset J$ (c2) $bJ\subset I$ (c3) $ba-1\in I$ (c4) $ab-1\in J$.

If (a) holds, consider an isomorphism $q:R/I\to R/J$, and lift the image of $1$ as an element $a\in R$. Then $q$ is induced by $L_a$ and (b) follows. Also, choose a lift $b$ of the image of $1$ by the $q^{-1}:R/J\to R/I$. Then (c) holds.

Suppose that $a,b$ as in (c) exist. Then by (c1) $L_a$ induces an homomorphism $q:R/I\to R_J$, by (c2) $L_b$ induces an homomorphism $q':R/J\to R_I$. So $L_{ba}$ and $L_{ab}$ induce endomorphisms of $R/I$ and $R/J$, which by (c3) and (c4) are the identity. Hence $q$ and $q'$ are inverse to each other. So (a) holds.

Similarly if $a$ exists as in (b) then by the first half of (b), $L_a$ induces an injective homomorphism $R/I\to R/J$, which is surjective by the last part.

  • $\begingroup$ Remark: in (b), the condition $L_a^{-1}(J)=I$ means the same as [$aI\subset J$ and $a(R-I)\subset R-J$]. $\endgroup$ – YCor Sep 25 at 12:13
  • $\begingroup$ Thank you. I really liked the condition c. $\endgroup$ – Sara.T Sep 25 at 16:21
  • $\begingroup$ A sufficient condition is the existence of invertible $a$ such that $aI=J$. To get a concrete example (but I switch left to right: otherwise take the opposite ring), take in dimension $\ge 2$, $R=\mathrm{End}(V)$ a matrix algebra, and $I_v=\{f:fv=0\}$: this is a left ideal, and for $g$ invertible, $I_vg=I_{g^{-1}v}$. In particular all $R/I_v$ for nonzero $v$ are isomorphic left $R$-modules. $\endgroup$ – YCor Sep 25 at 17:36

The criterion given by YCor also yields a counterexample to the fact that $I,J$ must always coincide. The idea is to take all the letters implied in the condition and build a "free" counterexample, where one straight impose all conditions.

Explicitly, take $R = \mathbb{C} \langle a,b\rangle$, the free $\mathbb{C}$-algebra over $\{a, b\}$, and the right ideals: $$I = (b^{n_1} a^{m_1} \ldots b^{n_r}a^{m_r}(ba-1), r\ge 0, m_i, n_i >0 ) + (b^{n_1} a^{m_1} \ldots b^{n_r}a^{m_r}b^k(ab-1), r\ge 0, n_i, m_i,k >0 )$$ Note the slight difference between the first and the second kind of generators: in the second, the degree is at least three.

$$J= (a^{n_1} b^{m_1} \ldots a^{n_r}b^{m_r}(ab-1), r\ge 0, m_i, n_i >0 ) + (a^{n_1} b^{m_1} \ldots a^{n_r}b^{m_r}a^k(ba-1), r\ge 0, n_i, m_i,k >0 )$$

This ideal is analogous to the first, with $a,b$ swapped.

Firstly, note that they satisfy the criterion (c). Indeed, $ba - 1 \in I, ab-1 \in J$, and a simple calculation yields $aI \subset J$: indeed, $a*(\cdot)$ takes first kind generators of $I$ in second kind generators of $J$, and second kind to first kind.

Finally, note that they are different. Observe that $I$ is contained in the bilateral ideal $K$ generated by $ba-1$: in fact the second kind of generators has the form $xb(ab-1) = x(ba-1)b$. We now show that $ab - 1 \in J$ does not map to zero in the quotient ring $S = R/K$. Establishing this fact will complete the proof.

The ring $S$ is generated as a $\mathbb{C}$-algebra by $a$ and $b$ subject to the relations $ba = 1$. So, it is isomorphic to the monoid $\mathbb{C}$-algebra $\mathbb{C}[M]$ of the bicyclic monoid $M \Doteq \langle a, b \, \vert \, ba = 1 \rangle$. Since $ab\neq 1$ in $M$, we have $ab \neq 1$ in $S$, which concludes the proof.

The last fancy part can be substituted by a more direct computation, but I couldn't do the latter in a clean way and I chose for this version.

  • $\begingroup$ @Luc Guyot: thanks! Feels much nicer now :) $\endgroup$ – Andrea Marino Sep 29 at 16:18

We shall give an example of a non-commutative PID $R$ with two distinct right principal ideals $I$ and $J$ such that $R/I$ and $R/J$ are isomorphic as right $R$-modules.

We begin with a trivial remark.

Let $\text{ann}(M) \Doteq \{ r \in R \,\vert \, mr = 0 \}$ denote the annihilator of a right $R$-module $M$. For a right ideal $I$ of a unital and associative ring $R$, we have more specifically $$\text{ann}(R/I) = (R : I) \Doteq \{ r \in R \,\vert \, Rr \subseteq I \} \subseteq I.$$

Thus, if $I$ is two-sided, then $\text{ann}(R/I) = I$. Therefore the following is immediate.

Claim 1. Let $R$ be an associative unital ring and let $I$ and $J$ be two-sided ideals of $R$. Then the following are equivalent:

  • $R/I$ and $R/J$ are isomorphic as right $R$-modules;

  • $I = J$.

In general, we cannot conclude that $I = J$. A counter-example will follow from:

Claim 2 [Theorem 1.(2)]. Let $k$ be a field, $\sigma$ an automorphism of $k$ and let $R = k[X;\sigma]$ be the univariate skew polynomial ring defined via $aX = X \sigma(a)$ for every $a \in k$. Let $\beta \in k \setminus \{0\}$ and set $\alpha \Doteq \frac{\beta}{\sigma(\beta)}$. Then the right $R$-modules $R/(X - 1)R$ and $R/(X - \alpha)R$ are isomorphic.

Proof. By assumption, the map $1 + (X - 1)R \mapsto \beta + (X - \alpha)R$ induces an $R$-homomorphism which is easily seen to be surjective. As $R/(X - 1)R$ is isomorphic to $k$ as a $k$-algebra, the previous homomorphism is an isomorphism of right $R$-modules.

If we specialize $k$ in Claim 2 to the finite field with $4$ elements and $\sigma$ to the Frobenius automorphism for instance, we obtain the desired counter-example.

Take $k = \mathbb{C}$ and $\sigma$ to be the complex conjugation, then we get uncountably many pairwise distinct right principal ideals $I = (X - \alpha)R$ for $\alpha \in \mathbb{S}^1$ with $R$-isomorphic quotient $R/I$.


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