I am studying non commuatitve ring theory and I'm having problems with the following exercise:
Let $R$ be a ring and $(\mathcal{D},\varphi)$ an $R$-division ring. Then, $(\mathcal{D},\varphi)$ is epic iff the division closure of $\varphi(R)$ in $\mathcal{D}$ is $\mathcal{D}$.
First of all let us fix some definitions and notations:
Let $R$ be a ring. An $R$-ring is a pair $(S,\varphi)$, where $S$ is a ring and $\varphi:R\to S$ is a ring homomorphism. Moreover, if $S$ is a division ring we say that it is a $R$-division ring.
Let $R$ be a ring. An $R$-ring $(S,\varphi)$ is epic if for any ring $S'$ and maps $\psi,\psi':S\to S'$ such that $\psi\circ\varphi=\psi'\circ\varphi$ then $\psi=\psi'$.
Let $R$ be a ring and $(S,\varphi)$ an $R$-ring. The domain of $\varphi$ is defined as: $$\text{dom}(\varphi)=\lbrace s\in S\mid \forall\text{ ring }S'\text{ and }\psi,\psi':S\to S'\text{ such that }\psi\circ\varphi=\psi'\circ\varphi\text{ then }\psi(s)=\psi'(s)\rbrace$$
Hence it is immediate to notice that $\text{dom}(\varphi)=S$ iff $(S,\varphi)$ is epic.
Now, I have the following lemma:
Let $R$ be a ring and $(S,\varphi)$ an $R$-ring, the following are equivalent:
- $x\in\text{dom}(\varphi)$.
- For every $S$-bimodule $M$ and $\forall~y\in M$ such that $ry=yr$ for all $r\in\varphi(R)$ it holds that $xy=yx$.
- $1\otimes_{R} x=x \otimes_{R} 1$.
- For every right $S$-bimodules $M$ and $N$ and $\psi\in\text{Hom}_R(M,N)$ it holds that $\psi(yx)=\psi(y)x~\forall~ y\in M$.
What I want to do is to use this lemma to solve the exercise.
I was able to do one of the implications:
If the division closure of $\varphi(R)$ is $\mathcal{D}$ and $0\neq s\in \text{dom}(\varphi)$ then: $$s^{-1} \otimes_{R} 1=s^{-1} \otimes_{R} ss^{-1}=s^{-1}s \otimes_{R} s^{-1}=1 \otimes_{R} s^{-1}$$ and since by the previous lema we have: $$\text{dom}(\varphi)=\lbrace s\in S\mid 1\otimes_{R} s=s \otimes_{R} 1\rbrace$$ this implies that $\text{dom}(\varphi)=\mathcal{D}$ and so $(\mathcal{D},\varphi)$ is epic.
But I dont know how to do the other implication. If you can check whether my first implication is good and help me with the other one I would be very grateful
Let $T$ denote the division closure of $\varphi(R)$ in $S$. In order to complete the exercise, assuming the given lemma, you must show that if $x\notin T$, then $x\otimes_R 1\neq 1\otimes_R x$ in $S\otimes_R S$.
Suppose $a,b,c\in S$ satisfy for all $s\in S$: $$sa\otimes_R s'=s\otimes_R as',\qquad sb\otimes_R s'=s\otimes_R bs',\qquad sc\otimes_R s'=s\otimes_R cs',$$ and $c\neq 0$.
\begin{eqnarray*} sab\otimes_R s'&=&sa\otimes_R bs'=s\otimes_R abs',\\ s(a+b)\otimes_R s'&=&sa\otimes_R s'+sb\otimes_R s'=s\otimes_R as'+s\otimes_R bs'=s\otimes_R (a+b)s',\\ sc^{-1}\otimes_Rs'&=&sc^{-1}\otimes_R cc^{-1}s'=sc^{-1}c\otimes_R c^{-1}s'=s\otimes_R c^{-1}s'. \end{eqnarray*}
For any $r\in \varphi(R)$ we have $sr\otimes_R s'=s\otimes_R rs'$ and this property is preserved by addition, multiplication, and taking inverses, so for any $t\in T$ we have $st\otimes_R s'=s\otimes_R ts'$.
We conclude $$S\otimes_R S=S\otimes_T S.$$
Now $S$ has a basis $\{e_i|\,\, i\in I\}$ as a right module over $T$ and a basis $\{f_j|\,\, j\in J\}$ as a left module over $T$, where we may assume that for some $i_0\in I$ and $j_0\in J$ we have $e_i=f_j=1$. As $T$ is by construction a division ring, the proof is identical to the usual proof that every vector space has a basis via Zorn's lemma.
We next define a function $$\Theta\colon S\otimes_T S\to {\rm Hom}_{\rm Set}(I\times J, T),$$ by $$\Theta\left(\sum_k s_k\otimes_T s'_k\right)(i,j) = \sum_k e^i (s_k)f^j( s'_k), $$ where $\{e^r|\,\,r\in I\}$ is the dual set of right $T$-linear maps $S\to T$ to the basis $\{e_r|\,\,r\in I\}$ and similarly $\{f^r|\,\,r\in J\}$ is the dual set of left $T$-linear maps $S\to T$ to the basis $\{f_r|\,\,r\in I\}$.
We must check that $\Theta$ is well defined. At present it is only well defined if we abuse notation and regard the sum of tensor products as a set of pairs of elements of $S$. For $s,s_1,s_2,s',s_1',s_2'\in S$ and $t\in T$ we have: \begin{eqnarray*} \Theta((s_1+s_2)\otimes_T s')(i,j)&=&\Theta(s_1\otimes_T s')(i,j)+\Theta(s_2\otimes_T s')(i,j)=\Theta(s_1\otimes_T s'+s_2\otimes_T s')(i,j),\\\\ \Theta(s\otimes_T (s_1'+s_2'))(i,j)&=&\Theta(s\otimes_T s_1')(i,j)+\Theta(s\otimes_T s_2')(i,j)=\Theta(s\otimes_T s_1'+s\otimes_T s_2')(i,j),\\\\ \Theta(st\otimes_T s')(i,j)&=&e^i(st)f^j(s')=e^i(s)tf^j(s')=e^i(s)f^j(ts')=\Theta(s\otimes_T ts'). \end{eqnarray*} Thus $\Theta$ is well defined on $S\otimes_T S$, and we may stop abusing notation.
For $w\in S\otimes_T S$, let the support of $\Theta(w)$ be the set: $$\{(i,j)\in I\times J|\,\, \Theta(w)(i,j)\neq 0\}.$$
Now suppose $x\notin T$. Then the support of $\Theta(x\otimes_T 1)$ consists only of elements of the form $(i,j_0)$ including ones where $i\neq i_0$. On the other hand the support of $\Theta(1\otimes_T x)$ consists only of elements of the form $(i_0,j)$ including ones where $j\neq j_0$.
Thus we have $$x\otimes_T 1\neq 1\otimes_T x,$$ and $$x\otimes_R 1\neq 1\otimes_R x,$$ as required.