# Solving an equation in a noncommutative ring.

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.

• What do you mean by a "prime" in a non-commutative ring? See here. – Dietrich Burde May 8 '19 at 19:28
• 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. – user530316 May 8 '19 at 21:31
• A right Ore domain does not have to be Noetherian on either side... where’d you get that? – rschwieb May 9 '19 at 2:30
• 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. – rschwieb May 9 '19 at 2:37
• 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. – user530316 May 9 '19 at 3:02