# Can one embed two division rings in a common one?

I could not get an answer to this question in MathStackExchange, so I dare ask it here.

Given any two fields, $\rm F_1,F_2$ over the same prime subfield $\rm F$, the quotient $\rm \mathbf F=F_1\otimes_F F_2/\mathcal M$ of the tensor product $\rm F_1\otimes_F F_2$ by a maximal proper ideal $\mathcal M$ provides a field with two embeddings $$\rm F_1\rightarrow\mathbf F,\ f\mapsto f\otimes 1 +\mathcal M\quad\text{and}\quad F_2\rightarrow\mathbf F,\ f\mapsto 1\otimes f +\mathcal M.$$

Question. Given two division rings $\rm R_1,R_2$ having the same characteristic, is there a way to find a division ring $\rm R$ with two embedings $\rm R_1⊂R$ and $\rm R_2⊂R$ ?

Yes, this is possible. PM Cohn first showed that the amalgamated product $R_1 * R_2$ over a common subfield is a "fir" (free ideal ring), and then in
that one can adjoin inverses to get a division ring containing $R_1 * R_2$.
• @Drike What I mean by amalgamated product is also called a free product (in fact this is what Cohn calls it). Say $F$ is a common subfield $R_1$ and $R_2$, and look at generators and relations for $R_1$ and $R_2$. A free product $R_1$ and $R_2$ with $F$ amalgamated is a ring $R$ with generators and relations unions of those for $R_1$ and $R_2$ such that inside $R$, $R_1 \cap R_2 = F$. Conditions for the existence of free products were given in Cohn (1959, Math Z). – Kimball Sep 25 '17 at 1:12