Invertible elements of subrings Suppose that a ring $R$ with unity is free as a left module over a subring $H$ and let $B$ be a basis of $R$ on $H$. If an element $x$ of $H$ has a right inverse $y$ in $R$, is it true that $y\in H$? 
This is clear to me if $1\in B$, but what happens in general? Thanks.  
 A: It is not necessarily true that every right inverse $y \in R$ of $x \in H$ is already contained in $H$:
Let $H'$ be a ring (with unity) such that there exists elements $f, g, h \in H'$ with $fg = 1$ but $fh = 0$ while $h \neq 0$, i.e. $f$ has a right inverse but is a left zero divisor.
One can take, for example, the endomorphism ring $H' = \operatorname{End}_{\mathbb{R}}(\mathbb{R}^\mathbb{N})$ and consider the left shift operator $f((a_n)_{n \geq 0}) = (a_{n+1})_{n \geq 0}$, the right shift operator $g((a_n)_{n \geq 0}) = (b_n)_{n \geq 0}$ with $b_n = a_{n-1}$ for $n \geq 1$ and $b_0 = 0$, and the trunction operator $h((a_n)_{n \geq 0}) = (b_n)_{n \geq 0}$ with $b_0 = a_0$ and $b_n = 0$ for $n \geq 1$.
Then $R = \operatorname{M}_2(H')$ is free as an $H'$-module (with basis given by the usual $E_{ij}$-matrices), and therefore also free as an $H$-module for the subring
$$
      H
  :=  \left\{
        \begin{bmatrix}
          a & 0 \\
          0 & a
        \end{bmatrix}
      \,\middle|\,
        a \in H'
      \right\}.
$$
(One can think of $R$ as non-central $H'$-algebra.)
The element
$$
      x
  =   \begin{bmatrix}
        f & 0  \\
        0 & f
      \end{bmatrix}
  \in H
$$
has the right inverse
$$
      y
  =   \begin{bmatrix}
        g & h  \\
        0 & g
      \end{bmatrix}
  \in R,
$$
which is not contained in $H$.
(The statement is true if $R$ is commutative, see below.)

What is true is that if $x \in H$ has a right inverse $y \in R$, then it already has a right inverse in $H$:
We will only consider the case $B \neq \emptyset $ because otherwise $R = H = 0$.
The map $R \to R$, $r \mapsto xr$ is surjective because $xR \supseteq xyR = R$.
This shows that the left module action of $x$ on $R$ is surjective.
We have that $R \cong H^{\oplus B}$ as left $H$-modules, so it follows that the left module action of $x$ on $H^{\oplus B}$ is surjective.
It further follows from the decomposition $H^{\oplus B} = \bigoplus_{b \in B} Hb$ into left $H$-modules, that the left module action of $x$ on every direct summmand $Hb$ is surjective.
It follows from $B \neq \emptyset$ that there exists such a direct summand $Hb$, for which we then have $Hb \cong H$ as left $H$-modules.
It follows that the left module action of $x$ on $H$ itself is surjective.
This shows that for every $z \in H$ there exists some $y' \in H$ with $xy' = z$;
this holds in particular for $z = 1$.

If $H$ is commutative, then the original statement is true, i.e. it follows that $y \in H$:
It follows from the above that $x$ has a right inverse $y' \in H$.
Since $H$ is commutative, it follows that $y'$ is a two-sided inverse of $x$, so that $x$ is a unit in $H$.
Then the left module action of $x$ on $H$ is bijective, and it follows that the same goes for $H^{\oplus B}$ and thus also for $R$.
Then $y, y' \in R$ are both right inverse to $x$, so that $xy = 1 = xy'$.
Since the left module action of $x$ on $R$ is injective, it follows that $y = y' \in H$.
