My question is the following Lemma.
Throughout this question, $R$ always denotes a prime ring with involution ${*}$ and an automorphism $\sigma$, which is not commutative. An additive map $D: R\rightarrow R$ is called a Jordan $({*}, \sigma)$-derivation if $ D(x^2) = D(x)x^{*} + x^{\sigma}D(x)$ for all $x \in R.$
LEMMA: Suppose that there exists $q \in Q_{ml}(R)$ such that $ D(x) = x^{\sigma}q - qx^{*}$ for all $x \in R$. Then $q \in Q_{ms}(R)$. Here $Q_{ml}(R)$ and $Q_{ms}(R)$ are the maximal left quotient ring and maximal symmetric ring of quotient respectively.
PROOF: The definition of $Q_{ms}(R)$ is the set consisting of the elements $x$ such that $xA\subseteq R$ and $Bx\subseteq R$ for some dense right ideal $A$ of $R$ and some dense left ideal $B$ of $R$.
Now, $ D(y) = y^{\sigma}q - qy^{*}$ for all $y \in R$. Since $q \in Q_{ml}(R)$, $\lambda q \subseteq R$ for some dense left ideal $\lambda$ of $R$. Note that $\lambda^{\sigma}$ is a dense left ideal and $\lambda^{*}$ is a dense right ideal of R. For $y \in \lambda$, we have to prove $ qy^{*} = y^{\sigma}q - D(y) \in R$ ? But how could I show $y^{\sigma}q \in R$ ? That is, $\lambda^{\sigma}q \subseteq R$ ? Or we have to find a dense right ideal $J$ of $R$ such that $qJ \subseteq R$. Then this will proves that $q \in Q_{ms}(R)$, but I was unable to find such an ideal. Please give me some hint. How could I prove it?