MathAction #253 factor returns wrong result

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I just ran across the following astonishing bug:

fricas

s :=-x^3+1/6*(-2*sqrt(6)+2*sqrt(3)+3*sqrt(2))*x^2+1/6*((sqrt(3)+sqrt(2))*sqrt(6)-2*sqrt(2)*sqrt(3))*x-sqrt(2)*sqrt(3)*sqrt(6)/6

$\label{eq1}\begin{array}{@{}l} \displaystyle -{{x}^{3}}+{{{-{2 \ {\sqrt{6}}}+{2 \ {\sqrt{3}}}+{3 \ {\sqrt{2}}}}\over 6}\ {{x}^{2}}}+ \ \ \displaystyle {{{{{\left({\sqrt{3}}+{\sqrt{2}}\right)}\ {\sqrt{6}}}-{2 \ {\sqrt{2}}\ {\sqrt{3}}}}\over 6}\ x}-{{{\sqrt{2}}\ {\sqrt{3}}\ {\sqrt{6}}}\over 6}$

(1)

Type: Polynomial(AlgebraicNumber?)

fricas

factor s

$\label{eq2}- 1$

(2)

Type: Factored(Polynomial(AlgebraicNumber?))

This is the same problem as 191ExquoAndThereforeGcdCannotHandleUPXEXPRINT. Namely, the roots sqrt(2), sqrt(3) and sqrt(6) are dependent, which cause problems because sqrt(6)^2 = 6 = 23 = sqrt(2)^2sqrt(3)^2 but we do not know if sqrt(6) = sqrt(2)sqrt(3) or sqrt(6) = -sqrt(2)sqrt(3). This effectively creates ring with zero divisors, while factoring/GCD routines assume a field.

There are several things to notice, in fact:

The factorisation is nonsense
I think that AlgebraicNumber? should be able to simplify $\sqrt{2}\sqrt{3}\sqrt{6}$ to $\sqrt{36}$
shouldn't be simplified to ? Usually, sqrt denotes the positive square root.

Martin

... --kratt6, Wed, 25 Jan 2006 04:50:05 -0600 reply

In fact, the problem shows already with

fricas

s :=x^2-sqrt(2)*sqrt(3)*sqrt(6)

$\label{eq3}{{x}^{2}}-{{\sqrt{2}}\ {\sqrt{3}}\ {\sqrt{6}}}$

(3)

Type: Polynomial(AlgebraicNumber?)

and it seems to occur in 'InnerAlgFactor?':

  \begin{axiom}
  )tr InnerAlgFactor  )ma
  factor(s)
  \end{axiom}

... --test1, Fri, 19 Jun 2015 15:52:44 +0000 reply

Status: open => duplicate

Subject: (replying) Be Bold !!

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