# Question on existence of finite-index ideals in non-commutative rings

If $$R$$ is a commutative ring with a subring $$S$$ of finite index, it is not hard to show that there is an ideal $$I$$ of $$R$$ with $$[R:I]$$ finite and $$I\subset S$$. The idea is to take $$I = \{s\in S : rs\in S, \ \forall r\in R \}$$, and then observe that $$I$$ is the kernel of the natural map $$S\to End(R/S)$$ taking $$s\mapsto ([r]\mapsto [rs])$$, where $$R/S$$ is just an abelian group and $$End(R/S)$$ is its endomorphism ring. Note that $$S$$ doesn't have to contain a multiplicative unit; we just need that $$S$$ is closed under addition and multiplication.

I wanted to know if the same result actually holds true for non-commutative rings: if $$R$$ is a (unital if necessary) non-commutative ring with a subring $$S$$ of finite index, is there a 2-sided ideal $$I$$ in $$S$$ whose index in $$R$$ is finite?

The idea I had had was to define $$I_R = \{s\in S: rs\in S, \ \forall r\in R \}$$ and similarly $$I_L = \{s\in S: sr\in S, \ \forall r\in R \}$$; these are 1-sided ideals in $$R$$ contained in $$S$$, of finite index in $$R$$, by a similar argument to the above. We could then say that $$I_R\cap I_L$$ is finite index in $$R$$, but the intersection of left ideal and right ideal is not in general a 2-sided (or even 1-sided) ideal. Perhaps instead I should use the set $$J=\{ s\in S: rsr'\in S \ \forall r, r'\in R\}$$, which is a 2-sided ideal contained in $$S$$, but I'm uncertain it needs to have finite index.