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Let $R$ be a non-commutative integral domain with unity which is also a right Noetherian ring. By integral domain I mean that the product of nonzero elements is always nonzero. I am trying to show the following easy thing: Let $a,b\in R$ be two nonzero elements.

I want to show that there exists $p,q \in R$ such that $aq=bp\neq 0$

I am searching for a down to earth proof. (not by any famous theorem) Thanks in advance for any help.

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The trick (used by Goldie) is to show that if $aR\cap bR=\{0\}$, then the elements of $\{a^ib\mid i\in \Bbb N\}$ are $R$-linearly independent on the right.

If $\sum_{i=0}^n a^ibr_i=0$, then $-br_0=\sum_{i=1}^n a^ibr_i$, but $a$ factors to the left side of this last expression. Therefore both sides are zero, so $r_0=0$, and $\sum_{i=1}^n a^{i-1}br_i=0$. Doing this repeatedly, you show all the $r_i$ are zero.

Since this set exists, it means that $\bigoplus_{i\in \Bbb N}a^ibR$ is a submodule of $R$. But this submodule clearly contains infinite, strictly ascending chains. If $R$ is right Noetherian, this is clearly a contradiction.

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