If the option 'symbolic'=true is given and a RootOf whose minimal polynomial factors nontrivially is detected, then Squarefree will reduce it to a RootOf of lower degree by picking one of the factors arbitrarily. This will eliminate the possibility of a "reducible RootOf detected" error. The default is 'symbolic'=false.

If the option 'expanded'=true is given, then all polynomials in the factorization with the same multiplicity will be combined and expanded, producing a single polynomial factor of higher degree, but the same multiplicity.  In this case, all $e_i$'s in the output will be distinct. The default is 'expanded'=false.

If the option 'characteristic'=p is given, where p is a non-negative integer, the square-free factorization is computed over an extension of the ring ${\mathrm{\mathbb{Z}}}_p$ of integers modulo p. The default is 'characteristic'=0 and means that the square-free factorization is computed over an extension of the rational numbers.

In positive characteristic, if the input contains a RootOf with a reducible minimal polynomial, then squarefree may not return the correct squarefree factorization or may raise an error.  As well, correctness of the result is not guaranteed if the algebraic numbers in the input do not form an independent set of extensions.

Note that if p is positive but not a prime, then ${\mathrm{\mathbb{Z}}}_p$ is not a field, so Squarefree may not be able to compute a square-free factorization of the input.  If it does not succeed because it encounters an integer that has no inverse modulo p, it issues the error "zero divisor modulo p detected".

If the option 'makeindependent'=true is given, then Squarefree will always try to find a field representation for algebraic numbers in the input, regardless of how many algebraic objects the input contains. If the input contains many RootOfs, then this can be a very expensive calculation. If 'makeindependent'=false is given, then no independence checking is performed. The default is 'makeindependent'=FAIL, in which case algebraic dependencies will only be checked for if there are $4$ or fewer algebraic objects in the input.

The Squarefree function computes a square-free factorization of the multivariate polynomial or rational function p in the variable(s) v over an algebraic number field.

The square-free factorization is returned as a nested list of the form $\left[u\,\left[\left[f_1\,e_1\right]\,\mathrm{...}\,\left[f_n\,e_n\right]\right]\right]$ such that $p=u{f}_1^{e_1}\mathrm{...}{f}_n^{e_n}$ where $f_i$ is a monic and square-free polynomial in the variable(s) v. Formally, GreatestCommonDivisor $\left(f_i\,\frac{\partial }{\partial v}{f}_i\right)=1$ for all $i$ and GreatestCommonDivisor $\left(f_i\,f_j\right)=1$ for all $i\ne j$.  The coefficient $u$ is an element of the coefficients field of p and is free of v. The $f_i$'s are collected with respect to the variable(s) v.

Algebraic numbers may be represented by radicals or with the RootOf notation.  Usually, they will be returned in the same representation as they were received. (see type,algnum, type,radnum).

Non-algebraic sub-expressions such as $\sin \left(x\right)$ that are neither variables, rational numbers, or algebraic objects are frozen and temporarily replaced by new local variables, which are not considered to be constant in what follows.

The argument p is considered as a polynomial or a rational function in the variable(s) v. Other names are considered as elements of the coefficient field. If the optional argument v is omitted, then all the names in p are used (including non-algebraic sub-expressions).

Functions such as $\sin \left(x\right)$ can also be included in v and treated as variables of p.  However, functions that are also of type AlgebraicObject, such as $\sin \left(\frac{\mathrm{\pi }}{3}\right)$ will be converted to algebraic numbers before proceeding, so they cannot be included. Proceed with caution when including functions in v, as treating some functions as variables may produce mathematically unsound results.

The $e_i$'s will always be integers.  If p is a polynomial in v, then they will all be positive integers.  If p is a rational function in v, then factors in the denominator that cannot be cancelled will be returned with negative corresponding $e_i$'s.

By default, the $e_i$'s may not be distinct, because partial factorizations in the input are preserved.  If the option 'expanded'=true is specified (see below), then factorizations of polynomials with the same multiplicity are not preserved, so all $e_i$'s will be distinct. In any case, the $e_i$'s are returned sorted by non-decreasing value.

Algebraic numbers occurring in the result are reduced modulo their minimal polynomial (see Reduce).

If p contains functions, their arguments will be normalized recursively (see Normal).

Terms will be sorted according to a block-lexicographic ordering of the variables, where all global variables are considered larger than all local ones. If all variables have different names, then this ordering is session-independent.

If the set of radicals and RootOfs in the input cannot be embedded into a field algebraically, then Squarefree may not be able to compute a square-free factorization. Squarefree will try to find a field representation if there are at most $4$ algebraic objects in the input (unless option 'makeindependent' is given; see below), and otherwise attempt to proceed anyway. If unsuccessful, a "reducible RootOf detected" error will be returned (unless the option 'symbolic'=true is given; see below).

This function does not support input containing floats or radical functions such as $\sqrt{x}$.

$\mathrm{with}\left(\mathrm{Algebraic}\right)\:$

$\mathrm{Squarefree}\left(2x^2+4\mathrm{sqrt}\left(2\right)xy+4y^2\right)$

$\left[2\,\left[\left[\sqrt{2}y+x\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\frac{x^4{y}^5}{z^2}\right)$

$\left[1\,\left[\left[z\,\mathrm{-2}\right]\,\left[x\,4\right]\,\left[y\,5\right]\right]\right]$

$\mathrm{Squarefree}\left(\frac{\left(x^2-2\right)\left(x+\mathrm{sqrt}\left(2\right)\right)}{{\left(x-\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\,\mathrm{index}=1\right)\right)}^2}\right)$

$\left[1\,\left[\left[x-\sqrt{2}\,\mathrm{-1}\right]\,\left[x+\sqrt{2}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+\mathrm{sqrt}\left(2\right)\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\,\mathrm{index}=1\right)\right)\right)$

$\left[1\,\left[\left[x+\sqrt{2}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x^2+1\right)\left(2\mathrm{I}x+2\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\,\mathrm{index}=1\right)\right)\right)$

$\left[2\mathrm{I}\,\left[\left[x-\mathrm{I}\,2\right]\,\left[x+\mathrm{I}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(4x^2+4x+1\right)$

$\left[4\,\left[\left[x+\frac{1}{2}\,2\right]\right]\right]$

$\mathrm{p1}≔\mathrm{expand}\left(\left(x^2+{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\right)}^{3}x+2\right)\left(2z+3\right)\right)$

$\mathrm{p1}≔2x^{2}z+3x^2+2{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\right)}^{3}xz+3{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\right)}^{3}x+4z+6$

$\mathrm{Squarefree}\left(\mathrm{p1}\right)$

$\left[2\,\left[\left[z+\frac{3}{2}\,1\right]\,\left[x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\right)\,2\right]\right]\right]$

$\mathrm{p2}≔3x^2{y}^4+12x{y}^4+12y^4$

$\mathrm{p2}≔3x^2{y}^4+12x{y}^4+12y^4$

$\mathrm{Squarefree}\left(\mathrm{p2}\right)$

$\left[3\,\left[\left[x+2\,2\right]\,\left[y\,4\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{p2}\,\left[x\,y\right]\right)$

$\left[3\,\left[\left[x+2\,2\right]\,\left[y\,4\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{p2}\,\left[x\right]\right)$

$\left[3y^4\,\left[\left[x+2\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{p2}\,\left[y\right]\right)$

$\left[3{\left(x+2\right)}^2\,\left[\left[y\,4\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{p2}\,\left[z\right]\right)$

$\left[3x^2{y}^4+12x{y}^4+12y^4\,\left[\right]\right]$

$\mathrm{Squarefree}\left(x^2+3x+2\right)$

$\left[1\,\left[\left[x^2+3x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+2\right)\right)$

$\left[1\,\left[\left[x+1\,1\right]\,\left[x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+2\right)\left(\mathrm{sqrt}\left(2\right)x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-8\,\mathrm{index}=1\right)\right)\right)$

$\left[\sqrt{2}\,\left[\left[x+1\,1\right]\,\left[x+2\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+2\right)\,'\mathrm{expanded}'=\mathrm{true}\right)$

$\left[1\,\left[\left[x^2+3x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x^2{y}^2{z}^3\,'\mathrm{expanded}'=\mathrm{true}\right)$

$\left[1\,\left[\left[xy\,2\right]\,\left[z\,3\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+\tan \left(\frac{\mathrm{\pi }}{8}\right)\right)\left(x+\mathrm{RootOf}\left(x^2-2\,\mathrm{index}=1\right)-1\right)\right)$

$\left[1\,\left[\left[x+\sqrt{2}-1\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x^2+1\right)\left(x+\exp \left(\frac{\mathrm{\pi }\mathrm{I}}{2}\right)\right)\right)$

$\left[1\,\left[\left[x-\mathrm{I}\,1\right]\,\left[x+\mathrm{I}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x-5\mathrm{sqrt}\left(5\right)\right)\left(x+{\mathrm{RootOf}\left(Z^2-5\,\mathrm{index}=2\right)}^3\right)\right)$

$\left[1\,\left[\left[x-5\sqrt{5}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(x^3{\mathrm{RootOf}\left({\mathrm{\_Z}}^2+\mathrm{\_Z}+2\right)}^3\right)$

$\left[-\mathrm{RootOf}\left({\mathrm{\_Z}}^2+\mathrm{\_Z}+2\right)+2\,\left[\left[x\,3\right]\right]\right]$

$\mathrm{Squarefree}\left(x\sin \left(x\right)+2\sin \left(x\right)\right)$

$\left[1\,\left[\left[\sin \left(x\right)\,1\right]\,\left[x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x\sin \left(x\right)+2\sin \left(x\right)\,x\right)$

$\left[\sin \left(x\right)\,\left[\left[x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x\sin \left(x\right)+2\sin \left(x\right)\,\sin \left(x\right)\right)$

$\left[x+2\,\left[\left[\sin \left(x\right)\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(\exp \left(x\right)+\exp \left(\frac{\mathrm{\pi }\mathrm{I}}{2}\right)\right)\left({\left(\exp \left(x\right)\right)}^2+\cos \left(0\right)\right)\right)$

$\left[1\,\left[\left[{\ⅇ}^x-\mathrm{I}\,1\right]\,\left[{\ⅇ}^x+\mathrm{I}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(\sin \left(\frac{x^2-1}{x-1}\right)+2\right)\left({\sin \left(x+1\right)}^2-4\right)\right)$

$\left[1\,\left[\left[\sin \left(x+1\right)-2\,1\right]\,\left[\sin \left(x+1\right)+2\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\sin \left(x+\frac{\mathrm{\pi }}{2}\right)\cos \left(x\right)\right)$

$\left[1\,\left[\left[\cos \left(x\right)\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(x^2+\cos \left(\frac{\mathrm{\pi }y+\mathrm{\pi }}{y+1}\right)x+\frac{1}{4}\,x\right)$

$\left[1\,\left[\left[x-\frac{1}{2}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(x^2+2x\mathrm{sqrt}\left(y\right)+y\right)$

Error, (in Algebraic:-Normal) ratpoly(s) with radalgnum coefficients expected

$\mathrm{Squarefree}\left(x^2-x+0.25\right)$

Error, (in Algebraic:-Squarefree) ratpoly(s) with radalgnum coefficients expected

$\mathrm{Squarefree}\left(x^2+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-4\right)x+1\right)$

$\left[1\,\left[\left[x+\frac{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-4\right)}{2}\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(x^2+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-4\,\mathrm{index}=2\right)x+1\right)$

$\left[1\,\left[\left[x-1\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\,\mathrm{index}=1\right)\right)\right)$

$\left[1\,\left[\left[x+1\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\,\mathrm{index}=2\right)\right)\right)$

$\left[1\,\left[\left[x+1\,1\right]\,\left[x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\right)\right)\right)$

Error, (in Algebraic:-Squarefree) reducible RootOf detected. Substitutions are {RootOf(_Z^2-3*_Z+2) = 1, RootOf(_Z^2-3*_Z+2) = 2}

$\mathrm{Squarefree}\left(\left(x+1\right)\left(x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\right)\right)\,'\mathrm{symbolic}'=\mathrm{true}\right)$

$\left[1\,\left[\left[x+1\,1\right]\,\left[x+2\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\right)x^2+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\right)x^4\right)$

$\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\mathrm{\_Z}+2\right)\,\left[\left[x^2+1\,1\right]\,\left[x\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{\_Z}\right)x^2+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{\_Z}\right)x^4\right)$

Error, (in Algebraic:-Squarefree) reducible RootOf detected. Substitutions are {RootOf(_Z^2-_Z) = 0, RootOf(_Z^2-_Z) = 1}

$\mathrm{Squarefree}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{\_Z}\right)x^2+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{\_Z}\right)x^4\,'\mathrm{symbolic}'=\mathrm{true}\right)$

$\left[1\,\left[\left[x^2+1\,1\right]\,\left[x\,2\right]\right]\right]$

$\mathrm{Squarefree}\left(x^2+1\,'\mathrm{characteristic}'=2\right)$

$\left[1\,\left[\left[x+1\,2\right]\right]\right]$

$a≔\mathrm{Expand}\left({\left(x+y\right)}^3{\left(2x+y\right)}^5{x}^2\,'\mathrm{characteristic}'=3\right)$

$a≔2x^{10}+2y{x}^9+2y^2{x}^8+y^6{x}^4+y^7{x}^3+y^8{x}^2$

$\mathrm{Squarefree}\left(a\,'\mathrm{characteristic}'=3\right)$

$\left[2\,\left[\left[x\,2\right]\,\left[x+y\,3\right]\,\left[x+2y\,5\right]\right]\right]$

$b≔\mathrm{Reduce}\left(\mathrm{eval}\left(a\,y=\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)\right)\,'\mathrm{characteristic}'=3\right)$

$b≔2x^8{\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)}^2+2x^9\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)+2x^{10}+x^4{\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)}^2+x^3{\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)}^2+x^4\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)+2x^2{\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)}^2+2x^3\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)+x^4+2x^3+2x^2$

$\mathrm{Squarefree}\left(b\,'\mathrm{characteristic}'=3\right)$

$\left[2\,\left[\left[x\,2\right]\,\left[x+\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)\,3\right]\,\left[x+2\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\mathrm{\_Z}+1\right)\,5\right]\right]\right]$

$\mathrm{CubeRootOf}\left[-4\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3+4\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_{\mathrm{-4}}≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3+4\,\mathrm{index}=1\right)$

$\mathrm{CubeRootOf}\left[-2\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_{\mathrm{-2}}≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\,\mathrm{index}=1\right)$

$\mathrm{CubeRootOf}\left[2\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_2≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=1\right)$

$\mathrm{CubeRootOf}\left[3\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-3\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_3≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-3\,\mathrm{index}=1\right)$

$\mathrm{CubeRootOf}\left[4\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-4\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_4≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-4\,\mathrm{index}=1\right)$

$\mathrm{CubeRootOf}\left[6\right]≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-6\,\mathrm{index}=1\right)$

${\mathrm{CubeRootOf}}_6≔\mathrm{RootOf}\left({\mathrm{\_Z}}^3-6\,\mathrm{index}=1\right)$

$\mathrm{Squarefree}\left(x+\mathrm{CubeRootOf}\left[4\right]\mathrm{CubeRootOf}\left[-2\right]\mathrm{CubeRootOf}\left[3\right]-\mathrm{CubeRootOf}\left[-4\right]\mathrm{CubeRootOf}\left[6\right]\,x\right)$

$\left[1\,\left[\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^3+2\,\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^3-3\,\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^3-4\,\mathrm{index}=1\right)-\mathrm{RootOf}\left({\mathrm{\_Z}}^3+4\,\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^3-6\,\mathrm{index}=1\right)+x\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x+\mathrm{CubeRootOf}\left[4\right]\mathrm{CubeRootOf}\left[-2\right]\mathrm{CubeRootOf}\left[3\right]-\mathrm{CubeRootOf}\left[-4\right]\mathrm{CubeRootOf}\left[6\right]\,x\,'\mathrm{makeindependent}'=\mathrm{true}\right)$

$\left[1\,\left[\left[x\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x+\mathrm{CubeRootOf}\left[2\right]\mathrm{CubeRootOf}\left[3\right]-\mathrm{CubeRootOf}\left[6\right]\,x\right)$

$\left[1\,\left[\left[x\,1\right]\right]\right]$

$\mathrm{Squarefree}\left(x+\mathrm{CubeRootOf}\left[2\right]\mathrm{CubeRootOf}\left[3\right]-\mathrm{CubeRootOf}\left[6\right]\,x\,'\mathrm{makeindependent}'=\mathrm{false}\right)$

$\left[1\,\left[\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^3-3\,\mathrm{index}=1\right)+x-\mathrm{RootOf}\left({\mathrm{\_Z}}^3-6\,\mathrm{index}=1\right)\,1\right]\right]\right]$