## Jategaonkar’s observation

Posted: December 26, 2009 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Jategaonkar’s lemma. Let $R$ be a ring and $S \neq (0)$ a subring of $R.$ If $a,b$ are in the centralizer of $S$ in $R$ and $a,b$ are left or right linearly independent over $R,$ then $S \langle a,b \rangle \cong S \langle x,y \rangle.$

Proof. I will assume that $a,b$ are right linearly independent over $R.$ The argument for the “left” version is identical. We basically need to prove that the set of all monomials in $a,b$ is an $S$-basis for the ring generated by $a,b$ over $S.$ So suppose that the claim is false. Then there exists a non-zero $f \in S \langle x,y \rangle$ of minimum total degree such that $f(a,b)=0.$ Write

$f(x,y)=s + xg(x,y) + yh(x,y),$

where $s \in S$ and $g, \ h \in S \langle x,y \rangle$ with $g \neq 0.$ Now

$0=f(a,b)b=bs + ag(a,b)b + bh(a,b)b=ag(a,b)b + b(s + h(a,b)b).$

Thus $g(a,b)b=0$ because $a,b$ are right linearly independent over $R.$ Again, we can write

$g(x,y)=s' + xg'(x,y)+yh'(x,y),$

where $s' \in S$ and $g',h' \in S \langle x,y \rangle.$ Then

$0=g(a,b)b=ag'(a,b)b + b(s' + h'(a,b)b)$

and so, since $a,b$ are right linearly independent over $R,$ we get $g'(a,b)b=0.$ But that contradicts the minimality of the total degree of $f$ because

$\deg g'(x,y)y \leq \deg g(x,y) < \deg f(x,y). \ \Box$