# Under what conditions two cyclic modules are isomorphic?

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 $$\frac RI \cong \frac RJ$$?

• @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 ! – Max Sep 19 at 19:12
• @Max Oops, you're correct. My bad. – Luiz Cordeiro Sep 19 at 19:14
• MO crosspost: mathoverflow.net/questions/342407 – YCor Sep 25 at 9:52
• 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! – 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 $$c,d$$.

Equivalences:

(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.

• Remark: in (b), the condition $L_a^{-1}(J)=I$ means the same as [$aI\subset J$ and $a(R-I)\subset R-J$]. – YCor Sep 25 at 12:13
• Thank you. I really liked the condition c. – Sara.T Sep 25 at 16:21
• 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. – 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.

• @Luc Guyot: thanks! Feels much nicer now :) – 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 $$R/I$$.