My understanding of the remainder theorem for one variable is that for $$f(x)=(x-a)q(x)+r(x)$$$\qquad$ if $x=a\implies f(a)=0\times q(a)+r(a)$ so $f(a)=r(a)$ Is this correct for a multivariate function? $$f(x,y,z)=(x-g(y,z))(y-h(x,z))(z-i(x,y))q(x,y,z)+r(x,y,z)$$so $x=g(y,z)\implies f(g(y,z))=r(x,y,z)$ thanks for any help.
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a set of polynomialsand the rest of the division is unique only if the set of polynomials is ‘well chosen’ in a certain sense, leading to the concept of aGröbner basis. – Bernard Jan 05 '15 at 21:09similar to the integers. Anyway, I guess the question is now whether said algorithm interacts in any special way with substitution? – GPerez Jan 05 '15 at 22:07