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{\bf Math 222: Cubic Equations and Algebraic Expressions}

{\bf Example.} We claimed in class that none of the solutions to the equation
$8x^3 - 6x - 1 = 0$ ha a degree two algebraic expression in the integers, $\zz$
\dots from which we concluded that one could not trisect an arbitrary angle
with
ruler and compass. How would one rigorously prove this claim about the
solutions?

Let's step back a bit.  Consider quadratic equations: $ax^2 + bx + c = 0$,
where
the coefficients $a, b, c \in \zz$. We know that every solution is
degree two algebraic in  $\zz$ - via the quadratic formula. But, sometimes the
expression is {\em rational} in $\zz$.  That is we don't need to take a
square root.  For example, $x^2 - 4 = 0$ clearly has solutions $x = \pm 2$.

{\bf Note:} A number has a rational expression in the integers precisely if
it is a
rational number, i.e.\ a quotient of integers.

{\bf Question:} Is there a way to ``characterize'' when the solutions are
actually rational?

{\bf Answer:} Yes!  The point is that we can always {\bf factor} the quadratic.
e.g.\ $x^2 - 4 = (x - 2)(x + 2)$ or $x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2})$.
What is different about these two factorizations?  In the first case, the
factors
are linear polynomials with rational (in fact integer) coefficients, whereas in
the second case, the coefficients are not all rational numbers (e.g. $\sqrt{2}$
is not a rational number.  In summary, can you prove the following fact?

{\bf Proposition 1.} The solutions to a quadratic equation $ax^2 + bx + c = 0$
with $a, b, c \in \mathbb{Z}$ are rational if and only if $ax^2 + bx + c$
factors
as $(dx + e)(fx + g)$ where each of $d, e, f, g$ are rational numbers.

{\bf Question:} Do you think a similar sort of result holds for cubic
polynomials?

The answer is again yes, but life is a little more complicated.  From
Section 3.1,
we now know that the solutions to any cubic (with integer coefficients)
have degree
three algebraic expressions in the integers.  However, they might be
rational or
degree two algebraic in the integers.  How do we tell what happens?  Well,
this is
more complicated to prove, but what turns out to be true is the following.

{\bf Proposition 2.} The solutions to a cubic equation $ax^3 + bx^2 + cx +
d = 0$
with $a, b, c, d, \in \mathbb{Z}$ are all degree two algebraic in the
integers if
and only if the polynomial factors as\\
$(ex + f)(gx^2 + hx + j)$ for some rational numbers $e, f, g, h, j$.

{\bf Question.} What is the condition for all the solutions to in
fact be rational numbers?

{\bf Example.} We have seen that the cube roots of unity in fact have
degree two
algebraic expressions in $\mathbb{Z}$.  Of course, these are solutions to the
equation $x^3 - 1 = 0$.  Notice that $x^3 - 1 = (x - 1)(x^2 + x + 1)$.  The
first
factor has solution $x = 1$ of course, and the solutions of the second
factor are precisely $\omega$ and $\omega^2$.

{\bf Example.} Let's return now to our original example of $8x^3 - 6x - 1 = 0$.
We want to show that the solutions are not degree two algebraic in
$\mathbb{Z}$.
>From Proposition 2, it suffices to show that the polynomial does not factor
over
the rational numbers.  Hmm... how to do this. The key is to notice that if it
factors, then one of the factors is of the form $ex + f$ and so one
solution is $x
= -f/e$, a rational number!!  Hence, all the solutions will be degree two
algebraic
in $\mathbb{Z}$ if and only if at least one of the solutions is a rational
number.

Now, how can we figure out if that is so or not?  Obviously we can't check
every rational number.  Well, there is a very beautiful little fact which
narrows
things down for us. Indeed, you may have seen this before.

{\bf Proposition 3 - Eisenstein's Criterion.} Let $ax^3 + bx^2 + cx + d =
0$ be a
cubic equation with $a, b, c, d \in \mathbb{Z}$ (with $a \neq 0$).  Suppose
that
$x = \frac{u}{v}$ is a rational solution.  That is $u, v \in \mathbb{Z}$.
Assume
further that the fraction is reduced so that $u$ and $v$ have no common
factors.
Then $u$ must divide $d$ and $v$ must divide $a$.

{\em Proof.} We are assuming that $\frac{u}{v}$ is a solution so
$$
a\left(\frac{u}{v}\right)^3 + b\left(\frac{u}{v}\right)^2 +
c\left(\frac{u}{v}\right) + d = 0.
$$
Multiplying through by $v^3$ (notice that $v \neq 0$), we get
$$
au^3 + bu^2v + cuv^2 + dv^3 = 0.
$$
By appropriate subtractions and factoring, we get
$$
(1) \hskip 0.7in au^3 = -(bu^2v + cuv^2 + dv^3) = -(bu^2 + cu^2v + dv^2)v
$$
and
$$
(2) \hskip 0.7in u(au^2 + buv + cv^2) = au^3 + bu^2v + cuv^2 = -dv^3.
$$
>From (1) we see that $v$ must divide $au^3$, but $u$ and $v$ have no
common factors so $v$ must divide $a$.  Similarly, from (2) we see that
$u$ must divide $dv^3$, and so $u$ must divide $d$ as claimed.

{\bf Note.} This fact holds for polynomials of any degree. Can you state
what's true in general?

{\bf Example.} Let's apply Eisenstein's Criterion to the equation $8x^3 -
6x - 1 =
0$.  Here $a = 8$ and $d = -1$. If a rational number $u/v$ is a solution,
then $u$
must divide $-1$, so $u = \pm 1$.  And $v$ must divide $8$, so $v = \pm 1,
\pm 2,
\pm 4, \pm 8$.  In summary, the only rational numbers which could possibly be
solutions are $\pm 1, \pm\frac12, \pm\frac14,\pm\frac18$.  Now, one just
explicitly checks that none of these eight possibilities works.  Hence the
polynomial does not factor over the rationals and so none of the solutions can
be degree two algebraic over $\mathbb{Z}$.

{\bf Final Thoughts:} You may or may not have noticed a variety of subtleties
in the above discussion about factoring polynomials. Factoring polynomials is
something we will discuss in more detail later in the semester.  One point to
note for now is that when factoring, one has to be very clear about the world
in which one is living, i.e.\ what sort of numbers (integers, rational, real,
complex ??) you allow as coefficients when you factor.

{\bf Final Question:}  Is there a similar characterization in terms of
factoring for higher degree polynomials?  Notice that the solutions to
$x^4 - 2 = 0$ are all degree two algebraic in $\zz$, but this does not factor
into quadratics over the rational numbers.
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