# Embeddings $A → B → A$, but $A \not\cong B$?

Are there any nice examples of structures (groups, modules, rings, fields) $A$ and $B$ such that there are embeddings $A → B → A$ while $A \not\cong B$? I would especially like to see an example for modules $A$, $B$. Or is it even true that the existence of such embeddings implies $A \cong B$?

Background: I’m correcting exercises and I wanted to give a counterexample to a failing argument. (Well, I’m not certain it fails, but I’m pretty sure it does and it’s not sufficiently justified at least.)

• See here and also the link given there to the analogous question on MO. – Andrés E. Caicedo May 20 '14 at 20:22
• mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold – Asaf Karagila May 20 '14 at 20:22
• You can have a look to Counterexamples to Cantor-Schroeder-Bernstein here. – Watson Aug 17 '16 at 12:55
• For rings, $k[t^2,t^3] \subset k[t] \hookrightarrow k[t,y]/(t^2-y^3) \cong k[x^2,x^3]$, but one is a PID while the other is not even integrally closed because $(t/y)^3 \equiv t$ (so it is not a UFD, so it is not a PID), or because $k[t^2,t^3]$ has Krull dimension $2$, since $k[t,y]$ is integral over it. – Watson Jan 28 '17 at 17:17

Modules, rings: $A=\Bbb Q^{\oplus\omega}$, $B=A\oplus\Bbb Z$. To see $A\not\cong B$ consider additive divisibility.

Fields: For every char $p\ge0$ and cardinal $\kappa\ge{\frak c}$ there exists a unique algebraically closed field of characteristic $p$ and cardinality $\kappa$. If $F$ is an infinite field then $|\overline{F}|=|F|$. Let $F$ be an algebraically closed field of cardinality $|F|\ge{\frak c}$. Then $\overline{F(T)}\cong F$ which yields $F(T)\hookrightarrow F$. Thus we achieve a sequence $F(T)\hookrightarrow F\hookrightarrow F(T)$. To see why $F\not\cong F(T)$, note $F(T)$ is not algebraically closed.

Linear (hence lattice, partial) orders: $A=(0,1)$, $B=[0,1)$. To see $A\not\cong B$ consider minima.

The above is also an example for topological spaces: $B$ can be written as a disjoint union of a singleton and a connected subset, while $A$ cannot.

• Thanks! A very nice example! That’ll do. – k.stm May 20 '14 at 20:22

For groups, you may consider $$\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots.$$

They are not isomorphic: in the second group, any element of order two is divisible by 2.

Another example, but finitely-generated, is (where $\mathbb{F}_n$ denotes the free group of rank $n$): $$\mathbb{F}_2 \hookrightarrow \mathbb{F}_3 \hookrightarrow \mathbb{F}_2.$$

See for example here for the existence of the monomorphisms. To prove that $\mathbb{F}_2$ and $\mathbb{F}_3$ are not isomorphic, notice that their abelianizations are respectively $\mathbb{Z}^2$ and $\mathbb{Z}^3$.

Shapes up to scale: a triangle embeds into a square embeds into a bigger triangle.

• Thanks. What would you consider as structure preserving maps? – k.stm May 21 '14 at 18:19
• Compositions of isometric embeddings and scalings. (You can think of the objects of the category as metric spaces and the morphisms as maps $f : X \to Y$ such that there exists a constant $k$ such that $d(f(x), f(y)) = k d(x, y)$.) – Qiaochu Yuan May 21 '14 at 18:19