$R$ is a division ring and $a \in R$ then is $N(a):=\{x\in R : xa=ax \}$ a division ring?

Question

If $R$ is a division ring and $a \in R$ then is $N(a):=\{x\in R : xa=ax \}$ a division ring ?

Answer 1

It's very easy to show that if $x,y\in N(a)$ then $x-y$ and $xy$ are in $N(a)$, and this shows it is at least a subring. (Work out all the details, if necessary.)

Finally, if $x$ is in $N(a)$, we have $xa=ax$. If $a$ is nonzero, we can multiply on the left and then on the right with $x^{-1}$ to get $ax^{-1}=x^{-1}a$, so every element of $N(a)$ has an inverse in $N(a)$. If $a=0$, I trust you can see the solution yourself.

Answer 2

Yes it is. One can show quite easily that $N(a)$ is a subring of $R$.

It remains to show that $\forall x \in N(a), x\neq 0$ we have $x^{-1}\in N(a)$. But now

$$ax^{-1}x = a = x^{-1}xa = x^{-1}ax$$ so multiplying by $x^{-1}$ on the right we get $ax^{-1} = x^{-1}a$.