1. Overview
Our goal is to prove the following conjecture when $ F $ has degree $ n=p^r $ for a prime $ p $.
Casas-Alvero Conjecture. Let $ F(X) \in k[X] $ be monic of degree $ n $, where $ k $ has characteristic 0. If $ F $ shares a root with $ F^{(i)} $ for every $ 1 \leq i \leq n-1 $, then $ F(X) = (X - \alpha)^n $ for some $ \alpha \in k $.
Example 1.1. This is from [2]. Consider $ F(X) = X^4 + X^2 + 2X $ in $ \FF_7[X] $. Then $ F'(X) = 4X^3 + 2X + 2 $ shares the root $ 4 $ with $ F $, $ F''(X) = 5X^2 + 2 $ shares the root $ 6 $, and $ F'''(X) = 3X $ shares the root $ 0 $. But $ F(X) $ is not of the desired form.
We can assume the following: (1) Dividing by the leading coefficient, we can assume $ F $ is monic. Hence,
\[F(X) = \sum a_i X^i = X^n + a_{n-1} X^{n-1} + \cdots + a_1 X + a_0.\](2) The $ i \text{th} $ Hasse derivative is given by $ H^i F(X) = F^{(i)}(X)/i! $. Since our field is of characteristic 0, the Casas-Alvero hypothesis is equivalent to assuming that $ F $ shares a root with $ H^i F $ for each $ 1 \leq i \leq n-1 $. We can expand this as
\[H^i F(X) = {n \choose i} X^{n-i} + \cdots + {i+1 \choose i} a_{i+1} X + a_i.\](3) Because $ H^{n-1} F $ is linear, we observe $ F $ has a zero $ \alpha $ in $ k $. Translating $ \alpha $ to 0 we can further assume $ a_0 = 0 $. Thus, we need to show that $ a_1 = \cdots = a_{n-1} = 0 $, allowing us to state our target theorem as follows:
Theorem 1.2. Let $ F \in k[X] $ be a monic polynomial of degree $ n = p^r $, where $ p $ is prime and $ k $ has characteristic zero. Suppose $ a_0=0 $ and $ F $ shares a root with $ H^iF $ for every $ 1 \leq i \leq n-1 $. Then $ a_1 = \cdots = a_{n-1} = 0 $.
2. Proof
In the sequel, we will write $ k $ for a field of characteristic zero. Let $ F,G\in k[X] $, with $ F(X)=\sum a_iX^i $, $ G(X)=\sum b_iX^i $, $ \deg F=m $, and $ \deg G=n $. Their resultant is
\[\Res(F,G)=\det\operatorname{Syl}(F,G),\]where $ \operatorname{Syl}(F,G) $ is the $ (m+n)\times(m+n) $ Sylvester matrix: its first $ n $ rows are the coefficient vector of $ F $ and its successive shifts, and its last $ m $ rows are the coefficient vector of $ G $ and its successive shifts. If $ F $ has roots $ \alpha_1,\dots,\alpha_m $ and $ G $ has roots $ \beta_1,\dots,\beta_n $, then
\[\Res(F,G) =a_m^n b_n^m\prod_{i=1}^m\prod_{j=1}^n(\alpha_i-\beta_j).\]In particular, $ F $ sharing a root with $ H^i F $ is equivalent to $ \Res(F, H^i F) = 0 $.
Setting
\[F = X^n + a_{n-1} X^{n-1} + \cdots + a_1 X,\]the resultant $ \Res_X(F,H^iF) $ is an element of $ \ZZ[a_1,\dots,a_{n-1}] $. Our problem reduces to studying the common zero locus of these resultant polynomials. We can rephrase this using a weighted projective scheme.
Let $ (w_1,\dots,w_r) $ be an $ r $-tuple of positive integers and $ R $ a commutative ring. Grade $ R[x_1,\dots,x_r] $ by setting $ x_i $ to have weight $ w_i $. We define weighted projective space as
\[\PP_R(w_1,\dots,w_r)=\operatorname{Proj}(R[x_1,\dots,x_r]).\]Over an algebraically closed field $ K $, its points may be described by nonzero tuples modulo
\[(x_1,\dots,x_r)\sim(\lambda^{w_1}x_1,\dots,\lambda^{w_r}x_r),\]with $ \lambda\in K^\times $. For example, $ x_1^2x_3+x_2 $ is homogeneous of weighted degree $ 4 $ for the weights $ (1,4,2) $.
Give the coefficient $ a_j $ of $ X^j $ weight $ n-j $; thus $ (a_{n-1},\dots,a_1) $ have weights $ (1,\dots,n-1) $. Then $ \Res(F,H^iF) $ is weighted homogeneous of degree $ n(n-i) $. Hence, the ideal
\[I_n = \left< \Res(F, H^iF) \mid 1 \leq i \leq n-1 \right>\]is a homogeneous ideal in the weighted coordinate ring. We define
\[X_n=\operatorname{Proj}\left( \ZZ[a_{n-1},\dots,a_1]/I_n \right) \subseteq\PP_{\ZZ}(1,\dots,n-1).\]Over an algebraically closed field $ k $, the points $ X_n(k) $ correspond to nontrivial normalized coefficient tuples for which $ F $ shares a root with each Hasse derivative. Since the desired tuple $ a_1=\cdots=a_{n-1}=0 $ is not a point of projective space, we have
Lemma 2.1. For an algebraically closed field $ k $, the normalized Hasse-derivative Casas-Alvero condition in degree $ n $ has only the pure-power solution if and only if $ X_n(k) $ is empty. In characteristic zero this is the ordinary Casas-Alvero property.
The following lemma lets us pass from an empty geometric fiber in characteristic $ p $ to characteristic zero.
Lemma 2.2. If $ X_n(\overline{\FF}_p) $ is empty for some prime $ p $, then $ X_n(k) $ is empty for every field $ k $ of characteristic zero.
Proof. Consider the structure morphism $ \phi_n\colon X_n\to\Spec\ZZ $. Since $ X_n $ is weighted projective over $ \ZZ $, this morphism is proper, and hence its image is closed. The hypothesis \(X_n(\overline{\FF}_p)=\varnothing\) is equivalent to the fiber \(X_n\times_{\ZZ}\Spec\FF_p\) being empty. Therefore the open subset
\[U=\Spec\ZZ\setminus\phi_n(X_n)\]contains the point $ (p) $. Every open neighborhood of a closed point of $ \Spec\ZZ $ contains the generic point $ (0) $, so the generic fiber $ X_n\times_{\ZZ}\Spec\QQ $ is empty. Base change to any field $ k $ of characteristic zero remains empty, and therefore $ X_n(k)=\varnothing $. $ \blacksquare $
Lemma 2.3. Let $ n=p^r $ for $ p $ prime. Then for $ 1\leq i\leq n-1 $ we have $ {n\choose i}\equiv0 $ modulo $ p $.
Proof. In $ \FF_p[T] $, the Frobenius identity gives
\[(1+T)^{p^r}=1+T^{p^r}.\]Comparing coefficients of $ T^i $ for $ 1\leq i\leq p^r-1 $ proves the claim. $ \blacksquare $
We now have the tools needed to prove Theorem 1.2.
Proof (Theorem 1.2). By Lemmas 2.1 and 2.2, it is enough to prove that $ X_n(\overline{\FF}_p) $ is empty. Let $ F(X)\in\overline{\FF}_p[X] $ have degree $ n=p^r $ and satisfy the normalized Casas-Alvero conditions. Recall
\[H^i F(X) = {n \choose i} X^{n-i} + \cdots + {i+1 \choose i} a_{i+1} X + a_i.\]In particular, $ F $ shares a root with
\[H^{n-1} F(X) = {n \choose n-1} X + a_{n-1} = n X + a_{n-1} = a_{n-1},\]but this is only possible if $ a_{n-1}=0 $. Suppose inductively that $ a_{s+1}=\cdots=a_{n-1}=0 $ for some $ 1\leq s<n $. Then
\[H^sF(X) = {n\choose s}X^{n-s} +\sum_{j=s}^{n-1}{j\choose s}a_jX^{j-s} =a_s,\]because Lemma 2.3 kills the leading term and the induction hypothesis kills every term with $ j>s $. A nonzero constant polynomial has no root, so the common-root hypothesis forces $ a_s=0 $. Descending induction gives $ a_1=\cdots=a_{n-1}=0 $. Hence the geometric fiber is empty, and Lemma 2.2 completes the proof in characteristic zero. $ \blacksquare $
3. References
- Eduardo Casas-Alvero, "Higher order polar germs", Journal of Algebra 240 (2001), no. 1, 326–337.
- Jan Draisma and Johan P. de Jong, "On the Casas-Alvero conjecture", Newsletter of the European Mathematical Society 80 (2011), 29–33.
- H.C. Graf von Bothmer, O. Labs, J. Schicho, and C. van de Woestijne, "The casas-alvero conjecture for infinitely many degrees", Journal of Algebra 316 (2007), no. 1, 224–230.