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Is there a function to obtain the monic form of a multivariables polynomial?

My polynomial is:

f = -2 x^2 - x^3 - 3 y

sorted with a lexicographic order.

The monic polynomial that I want to obtain is the polynomial divided by the coefficient of the monial of the higher rank.

f = -2/3 x^2 - 1/3 x^3 - y

In other words, polynomials whose leading coefficients are 1 are called monic.

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closed as unclear what you're asking by yohbs, MarcoB, happy fish, mikado, Young May 2 '17 at 15:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What does this question mean? What kinds of transformations do you allow? $\endgroup$ – Igor Rivin Apr 28 '17 at 20:05
  • $\begingroup$ It's not clear what you're asking. Can you give the explicit answer you're looking for, perhaps for a few different examples? $\endgroup$ – yohbs Apr 28 '17 at 20:26
  • $\begingroup$ This does what you want for the example cited: Expand[f/3]. Not sure what you are looking for in general. $\endgroup$ – bill s Apr 28 '17 at 22:03
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Perhaps

poly = -2 x^2 - x^3 - 3 y;
f = poly/(Abs[CoefficientList[poly, y]] // Last) // Expand

-((2 x^2)/3) - x^3/3 - y

This can be generalized into a function.

monic[poly_, var_] := poly/(Abs[CoefficientList[poly, var]] // Last) // Expand

Then

monic[-2 x^2 - 3 x^3 - 3 x y + 7 y^3, y]

-((2 x^2)/7) - (3 x^3)/7 - (3 x y)/7 + y^3

and

monic[-2 x^2 - 3 x^3 - 3 x y + 7 y^3, x]

-((2 x^2)/3) - x^3 - x y + (7 y^3)/3

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    $\begingroup$ Is Mathematica's ordering total degree? Is this documented anywhere? $\endgroup$ – Igor Rivin Apr 29 '17 at 0:25
  • $\begingroup$ Thank you for your help. But, i think that it doesn't work with f = -2 x^2 - x^3 - 3 y + 4*xy. The coefficient that should be found is 4. The leading power product ordered lexicographically is xy. $\endgroup$ – Bendesarts Apr 29 '17 at 6:35
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    $\begingroup$ What makes x*y the leading term? (Hint: there is an unstated assumption on the ordering being made.) $\endgroup$ – Daniel Lichtblau Apr 29 '17 at 15:51
  • $\begingroup$ @DanielLichtblau in fact, i want to use the lexigraphic order and consequently, i would like to find the coefficient of the leading term with this order. $\endgroup$ – Bendesarts Apr 29 '17 at 20:02

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