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Dear John, your help is greatly appreciated :) One further question : What i want to do is exactly what you have descibed at the end, but with an expression of the form $\frac{f(a)}{g(a)}$ where $f,g$ are polynomials in $a$ (and with coefficients in $Z[m])$ instead of just a polynomial $f(a)$. So when I define say f=((a^2-m^2)^2)(a^3-m) what Sage notebook gives is

1/256a^13/(a^2 - 4m)^2 - 1/8a^11m/(a^2 - 4m)^2 + 7/4a^9m^2/(a^2 - 4m)^2 - 14a^7m^3/(a^2 - 4m)^2 + 70a^5m^4/(a^2 - 4m)^2 - 1/2a^7m/(a^2 - 4m)^2 - 224a^3m^5/(a^2 - 4m)^2 + 8a^5m^2/(a^2 - 4m)^2 + 448am^6/(a^2 - 4m)^2 - 48a^3m^3/(a^2 - 4m)^2 - 512m^7/((a^2 - 4m)^2a) + 128am^4/(a^2 - 4m)^2 + 256m^8/((a^2 - 4m)^2a^3) - 128m^5/((a^2 - 4m)^2a) + 16am^2/(a^2 - 4m)^2

That seems rather incomprehensible as it computes the final expression in disctinct fractions. Is there any way to format this in a nice expression of the form ''polynomial over (other)polynomial'' where both polynomials will be factored ?

Dear John, your help is greatly appreciated :) One further question : What i want to do is exactly what you have descibed at the end, but with an expression of the form $\frac{f(a)}{g(a)}$ where $f,g$ are polynomials in $a$ (and with coefficients in $Z[m])$ instead of just a polynomial $f(a)$. So when I define say f=((a^2-m^2)^2)(a^3-m) $$f=\frac{(a^2-m^2)^2}{a^3-m}$$ what Sage notebook gives is

1/256a^13/(a^2 - 4m)^2 - 1/8a^11m/(a^2 - 4m)^2 + 7/4a^9m^2/(a^2 - 4m)^2 - 14a^7m^3/(a^2 - 4m)^2 + 70a^5m^4/(a^2 - 4m)^2 - 1/2a^7m/(a^2 - 4m)^2 - 224a^3m^5/(a^2 - 4m)^2 + 8a^5m^2/(a^2 - 4m)^2 + 448am^6/(a^2 - 4m)^2 - 48a^3m^3/(a^2 - 4m)^2 - 512m^7/((a^2 - 4m)^2a) + 128am^4/(a^2 - 4m)^2 + 256m^8/((a^2 - 4m)^2a^3) - 128m^5/((a^2 - 4m)^2a) + 16am^2/(a^2 - 4m)^2

That seems rather incomprehensible as it computes the final expression in disctinct fractions. Is there any way to format this in a nice latexed expression of the form ''polynomial over (other)polynomial'' where both polynomials will be factored ?