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From: Kent Holing on 26 Jul 2010 05:34
Given a monic quartic equation Q(x) = x^4 + a x^3 + b x^2 + c x + d = 0 with integer coefficients with the cubic (Lagrange) resolvent

R(t) = t^3 – b t^2 + (ac – 4d) t + 4bd – a^2d – c^2 = 0.

Can it happen that Q(x) = 0 and R(x) = 0 have a common root?

If so, Q(x) must be reducible; otherwise Q(x)|R(x), which is of course not possible.

The only possible case when R(x) is irreducible is that Q(x) = (x – r) R(x) for r an integer root of Q(x).

(Q(x) = (x – r)q3(x) for q3(x) irreducible cubic. Since R(x) is irreducible R(r) /= 0, R(x) = 0 and q3(x) = 0 have a common root, so q3(x) == R(x).)

(The other case Q(x) = p(x) q(x) for p, q 2 irreducible polynomials of degree 2 cannot happen:

Either p or q have a common root with R(x), but if so, p(x) == R(x) or q(x)== R(x). This is of course not possible.)

It can further be proved that (1) r =-(a+b) and that (2) R(r) = +-1.

Now, (1) follows since r + x1 + x2 + x3 = -a (the roots of the quartic are x1, x2, x3 and x4 = r) and the roots of the resolvent are x1, x2 and x3, so r + b = - a.

Proving (2): discriminant of Q = R(r)^2 discriminant of R, discriminant of Q = discriminant of R, both discriminants /= 0, so R(r)^2 = 1.

In addition to the above if R(x) is reducible (and Q(x) is reducible), can we then have a common root for Q(x) = 0 and R(x) = 0?

Any comments are welcome.

Kent
From: José Carlos Santos on 26 Jul 2010 09:43
On 26-07-2010 14:34, Kent Holing wrote:

> Given a monic quartic equation Q(x) = x^4 + a x^3 + b x^2 + c x + d = 0 with integer coefficients with the cubic (Lagrange) resolvent
>
> R(t) = t^3 � b t^2 + (ac � 4d) t + 4bd � a^2d � c^2 = 0.
>
> Can it happen that Q(x) = 0 and R(x) = 0 have a common root?

Yes. Take a = b = c = d = 0. Then 0 is a solution of both equations.

Best regards,

Jose Carlos Santos
From: Robert Israel on 26 Jul 2010 13:21

> On 26-07-2010 14:34, Kent Holing wrote:
>
> > Given a monic quartic equation Q(x) = x^4 + a x^3 + b x^2 + c x + d = 0
> > with integer coefficients with the cubic (Lagrange) resolvent
> >
> > R(t) = t^3 � b t^2 + (ac � 4d) t + 4bd � a^2d � c^2 = 0.
> >
> > Can it happen that Q(x) = 0 and R(x) = 0 have a common root?
>
> Yes. Take a = b = c = d = 0. Then 0 is a solution of both equations.

Or for a somewhat less trivial case, a=-3, b=4, c=-3, d=1.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Kent Holing on 26 Jul 2010 11:26
Hi all! What I really ask is: Can you give examples where the resolvent R(x) = 0 is irreducible having a root common with its quartic Q(x) = 0? (Note that then all roots of the resolvent are roots of the quartic since R|Q.)

The given examples are all with reducible R-s. (Note Q must be reducible.)

Kent
From: Ken Pledger on 26 Jul 2010 17:29
In article
<1470860066.2550.1280151278590.JavaMail.root(a)gallium.mathforum.org>,
Kent Holing <KHO(a)statoil.com> wrote:

> Given a monic quartic equation Q(x) = x^4 + a x^3 + b x^2 + c x + d = 0 with
> integer coefficients with the cubic (Lagrange) resolvent
>
> R(t) = t^3 � b t^2 + (ac � 4d) t + 4bd � a^2d � c^2 = 0.
>
> Can it happen that Q(x) = 0 and R(x) = 0 have a common root?
> ....


I'd attack this by looking at expressions for the roots. If
Q(x) = 0 has roots x_1, x_2, x_3, x_4, then IIRC R(t) = 0 has
roots x_1.x_2 + x_3.x_4, x_1.x_3 + x_2.x_4, x_1.x_4 + x_2.x_3. Try
equating one of those to x_1, etc.

Ken Pledger.
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