Quartics and Quadratics

I've been playing with polynomials a bit recently, and so I expect this post will be the first of several, explaining my explorations. The question I'd like to look at today is: When is a quartic (degree 4) polynomial the composite of two quadratics (degree 2)? Can every quartic be written as such a composite? Uniquely?

I wouldn't be surprised to find somebody has written this up before. In fact, they've probably dealt with more general questions, in an easier manner. All the same, this is how I spent my day, so I thought I'd share.

You might notice that 4 is the smallest degree polynomial that can be written as a composite of lower degree polynomials. If $p_i$ is a polynomial of degree $d_i$ (for $i=1,2$), then $p_1\circ p_2$ has degree $d_1\cdot d_2$. So if we wanted a cubic as a composite of lower degree polynomials, we'd have trouble doing it, and similarly for quadratics. However, these same considerations help us out a little bit. Suppose $f$ is a quartic that can be written as the composite $p_1\circ p_2$ of quadratics, and that T is a (non-constant) linear function. Then $T\circ f=T\circ p_1\circ p_2=(T\circ p_1)\circ p_2$, showing that $T\circ f$ is still the composite of two quadratics. The same argument works in reverse: if $T\circ f=p_1\circ p_2$, then $f=T^{-1}\circ p_1\circ p_2$ (and $T^{-1}$ exists, since $T$ was assumed linear), which is still the composite of quadratics.

What's the point of this linear function $T$? Well, suppose

$f=ax^4+bx^3+cx^2+dx+e.$

Let $T=\frac{1}{a}(x-e)$, and then notice that $T\circ f$ is monic (leading coefficient 1) and has 0 constant term. If the polynomial $T\circ f$ can be written as a composition of quadratics, then by the previous paragraph, $f$ can be written similarly. So, we can make our lives easier, and assume that $a=1$ and $e=0$.

Let $p_1=a_1x^2+b_1x+c_1$ and $p_2=a_2x^2+b_2x+c_2$. We'd like to know if we can write our $f$ above as $p_1\circ p_2$. If you're following along at home, I'll wait while you write out the composite $p_1\circ p_2$. To check your work (and mine), when you want $f=p_1\circ p_2$, you set the corresponding coefficients equal, and obtain the following system of equations:

$a=a_1a_2^2$
$b=2a_1a_2b_2$
$c=2a_1a_2c_2+a_1b_2^2+a_2b_1$
$d=2a_1b_2c_2+b_1b_2$
$e=a_1c_2^2+b_1c_2+c_1$

Our simplifications let us write $a=1$ and $e=0$. Let's also go ahead and assume $a_1=a_2=1$ (that is, our quadratics are monic), to make our calculations easier. It's also believable (I hope) that we haven't lost anything with this assumption, but perhaps I should think about this some more. I had convinced myself of it in the office this afternoon. Anyway, this makes our equations

$b=2b_2$
$c=2c_2+b_2^2+b_1$
$d=2b_2c_2+b_1b_2$
$0=c_2^2+b_1c_2+c_1$

Remember that we're assuming $b$, $c$, and $d$ were fixed, and are hoping to find $b_1,b_2,c_1$, and $c_2$. The first line clearly makes it easy to find $b_2=b/2$, and we can substitute that in the remaining lines:

$c=2c_2+b_1+(b^2/4)$, or $c-b^2/4=2c_2+b_1$
$d=(b/2)(2c_2+b_1)$, or $2d/b=2c_2+b_1$
$0=c_2^2+b_1c_2+c_1$

Now we're making progress. But first, we've made an assumption in the line $2d/b=2c_2+b_1$. What is it? Well, to divide by $b$, we assume $b\neq 0$. If, alternatively, $b=0$, then also $d=0$ (since $d$ is a multiple of $b$) and $b_2=0$. With all of these values being 0, the above system has 2 equations (the first and third) and 3 unknowns ($b_1$, $c_1$, and $c_2$). We can solve for $b_1$ and then $c_1$ in terms of a free variable $c_2$. Thus we obtain

Proposition: A quartic $x^4+cx^2+dx$ with $d\neq 0$ can not be written as the composition of two quadratics. The quartic $x^4+cx^2$ can be written as the composite $p_1\circ p_2$ where

$p_1=x^2+(c-2c_2)x+(c_2^2-cc_2)$
$p_2=x^2+c_2$

and $c_2$ can be any value.

Ok, so, assuming $b\neq 0$, what happens? The equations

$c-b^2/4=2c_2+b_1$
$2d/b=2c_2+b_1$

mean that $c-b^2/4=2d/b$, which we chose to re-write as $8d=4bc-b^3$. This is a requirement on the coefficients of the original quartic. However, if this is satisfied, then we can pick

$b_1=2d/b-2c_2$
$c_1=c_2^2-2dc_2/b$

in terms of a free variable $c_2$. We conclude by stating the final proposition, which can be checked by brute force (like everything else above):
Proposition: If $b\neq 0$ in the quartic $f=x^4+bx^3+cx^2+dx$, then $f$ can be written as the composite of quadratics only if $8d=4bc-b^3$ - in which case $f=p_1\circ p_2$ for

$p_1=x^2+(2d/b-2c_2)x+(c_2^2-2dc_2/b)$
$p_2=x^2+b/2 x+c_2$

where $c_2$ can be any value.

So there you have it. Not all quartics can be written as composites of quadratics. For those that can, there are a 1-parameter family of choices for monic quadratic factorizations.

Update 20081115: I've been thinking about my reduction that made the quadratics monic. I've been trying to decide if it was necessary. Playing around with the equations some more, it's not too hard to tell that you can't get rid of the relation $8d=4bc-b^3$ on the coefficients of the quartic, even if you don't assume your quadratics are monic. If you allow non-monic quadratics (but still want the composition to be monic), you have a free choice of value for $a_1$ (as long as it's positive) and for $c_2$ still, and the other coefficients can be solved for in terms of these two values (and the coefficients for the quartic).

Also, the two propositions could be combined. The relation $8d=4bc-b^3$ is present in both propositions, and there's no reason to assume $b\neq 0$ in that relation. Of course, we do end up dividing by $b$ to solve for coefficients in $p_1$ and $p_2$, so it's something to keep in mind.

Ok, I better go do some real work. I've still got other polynomial questions to look at, but they'll have to wait.