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

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?

• The determinant of the Sylvester matrix, or this, is taken as the definition. It is only in an integral domain that you get the property that the resultant is zero iff there is a common root in an algebraically closed field containing the coefficients.
– user550929
Commented Apr 12, 2018 at 6:44

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.

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.

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.

• Interesting what you say last. Is there a way to define resultant of two finite subsets of the projective line? Commented Apr 12, 2018 at 10:12