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  

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