Suppose $R$ is noncommutative ring with unit and has the properties necessary for a right and left skew field of fractions to exist (i.e. $R$ has no zero divisors and satisfies the left and right Ore condition). Let $x,y\in R$ be nonzero elements in $R$ so that $x,y$ are prime and $xy=yx$ in $R$.

I am interested in showing that the solutions to the equation $$xp+yq=0 $$ are exactly $p=yt, q=-xt$ for all $t\in R$.

Here is my reasoning:

$p=q=0$ is a solution. Now consider all other solutions.

If we assume one of $p$ or $q$ is nonzero, then since there are no zero divisors in $R$, we know both $p$ and $q$ are nonzero. In the skew field of fractions we can rewrite the equation as $$p=-x^{-1}yq $$ but since $x$ and $y$ commute, we have $$p=-yx^{-1}q. $$ Since the left side resides in $R$ if seems like $q=xt$ for some $t\in R$, since $y$ can not contain a factor of $x$ by the primeness assumption. By a similar argument, we can conclude that $p=ys$ for some $s\in R$. Hence the equation becomes $$xys+yxt=0 .$$ Multiplying on the left by $x^{-1}y^{-1}$ yields $s=-t$ Proving the result.

I feel I may be making a mistake. Some of the steps may not work in such a general setting? Thanks.

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    $\begingroup$ What do you mean by a "prime" in a non-commutative ring? See here. $\endgroup$ – Dietrich Burde May 8 '19 at 19:28
  • $\begingroup$ I'm using the notion of prime given in Cohn's paper which appears in your link, i.e. an element is prime if it cannot be written as the product of two nonunits. $\endgroup$ – user530316 May 8 '19 at 21:31
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    $\begingroup$ A right Ore domain does not have to be Noetherian on either side... where’d you get that? $\endgroup$ – rschwieb May 9 '19 at 2:30
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    $\begingroup$ I don’t understand why you’re rejecting $p=q=0$ as a solution... when you say “since there are no zero divisors.” And you didn’t use your primeness assumption? Fishy. $\endgroup$ – rschwieb May 9 '19 at 2:37
  • $\begingroup$ The left (right) Noetherian was overkill. I recall that left (right) Noetherian implies the left (right) Ore condition is satisfied. In my case I'm interested in $R$ a skew Laurent polynomial ring, in which case I believe $R$ satisfies these stronger conditions. My apologies. I should have mentioned that above. $\endgroup$ – user530316 May 9 '19 at 3:02

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