Proposition 3

— Let f be a non-zero polynomial in K[X] and let Ω be an algebraically closed extension of K. The following conditions are equivalent: a) The polynomial f is relatively prime to its derivative f' in K[X]. b) Either deg(f) = 0, or deg(f) > 0 and dis(f) ≠ 0 (IV, p. 83). c) There exists an extension L of K such that f splits in L[X] into a product of distinct polynomials of degree ≤ 1. d) The roots off in Ω are simple. e) The K-algebra K[X]/(f) is etale (V, p. 28, Def. 1).

a) ⇒ d): Under the hypothesis a) there exist two polynomials g and h in K[X] such that fg + f'h = 1 (IV, p. 12). Let a be a root of fin Ω; we have

whence $f'(a) \neq 0$; therefore $a$ is a simple root of fin Ω (IV, p. 17, Prop. 7).

d) ⇒ c): If d) holds, f splits in Ω[X] into a product of distinct factors of degree ≤ 1.

c) ⇒ b): Under the hypothesis c) there exist an element $A \neq 0$ in L and distinct elements $a_1, ..., a_n$ of L such that $f(X) = \lambda (X - a_1) ... (X - a_n)$. If $\deg(f) > 0$, we have (IV, p. 83, Prop. 11),

b) ⇒ a): Let c be the leading coefficient and D the discriminant of f; the resultant of $f'$ and f is equal to ± cD (IV, p. 84, Formula (54)), hence is non-zero; therefore (IV, p. 78, Cor. 2) the polynomials f and $f'$ are relatively prime in K[X].

a) ⇒ e): Let A be the K-algebra K[X]/(f) and x the image of X in A; by III, p. 573, Prop. 22, the A-module $\Omega_K(A)$ is generated by the elements $dx$, subject to the single relation $f'(x)\ dx = 0$. By V, p. 33, Th. 3, the K-algebra A is thus etale if and only if $f'(x)$ is an invertible element in A, which means that f and $f'$ are relatively prime in K[X].