How to interpret this strange property of the resultant for $f$ and $1-f$?

Question

What I know about the resultant (e. g. from Wikipedia) is its, more or less, defining property: $\operatorname{Res}(f,g)$ is a polynomial expression in the coefficients of $f$ and $g$ which vanishes if and only if $f$ and $g$ have a common root in some extension field.

I now compute (with Mathematica) that $$ \operatorname{Res}\left(a_0+a_1x+...+a_nx^n,1-\left(a_0+a_1x+...+a_nx^n\right)\right)=(-a_n)^n. $$

How to interpret this? I mean, $f$ and $1-f$ cannot have a common root unless $0=1$. Besides, if $a_n$ vanishes, then we get a polynomial one degree smaller, and the story repeats. But if it vanishes, there should already be a common root! Something's wrong. I am confused. Maybe it tells something about rings where $a_n$ can be a nontrivial nilpotent?

Answer 1

Let’s assume that $f$ is not a constant polynomial, or you get a $0\times0$ matrix when you compute the resultant.

The formula you get should be read: the resultant of $f$ and $1-f$ is the negative of the leading coefficient of $f$ raised to the degree of $f$.

The leading coefficient of a nonzero polynomial is nonzero by definition.

Nilpotent elements are out of the question, since you need a field to state the theorem on the resultant. Of course the resultant can be used in a more general context, but outside fields you cannot say that vanishing of the resultant means having a common root somewhere.

Answer 2

The formula you wrote is only valid for $a_n \neq 0$.

Very often, when you see a formula or a theorem for polynomials of particular degrees — in this case, that $f$ is a polynomial of degree $n$ — it only applies when polynomials really do have that degree.

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Incidentally, for some purposes there is a useful way to interpret $f$ as a polynomial of degree $n$ even when $a_n = 0$, and in this case $\infty$ is a root of the polynomial.

More precisely, you can homogenize the polynomial:

$$ F(Z,X) = a_0 Z^n + a_1 Z^{n-1} X + \ldots + a_{n-1} Z X^{n-1} + a_n X^n $$

The variables are related by $x = X/Z$; that is, these polynomials satisfy the identity

$$ F(Z, X) = Z^n f(X/Z)$$

In the case that $a_n = 0$, $Z$ will be a factor of $F(Z, X)$. Since $x = X/Z$, $Z=0$ in some sense corresponds to $x = \infty$. This can be made more precise via the arithmetic of the projective line.