Infinitesimal calculus and the introduction of derivatives is often linked to Newton and Leibniz.

I was wondering, when and why the idea of studying formal derivatives (e.g., of a formal polynomial) emerged?

For instance, what were the motivations to introduce derivatives to study things over finite fields and their extensions (where the visual intuition of infinitesimal as linked to Newton and Leibniz is lost)?

  • 1
    $\begingroup$ Of course formal derivatives for finite fields is much more recent, since finite fields only began in the 1800s $\endgroup$ Commented Mar 13, 2021 at 15:16
  • 3
    $\begingroup$ Finite fields aside, the idea of detecting double roots of polynomials by using (formal) derivatives predates infinitesimal calculus. It was introduced by Descartes in La Geometrie (1637) and developed by Hudde, see Is there a 'lost calculus'? A point of tangency can be interpreted as a double root of a system of a polynomial and a line equation, and the corresponding coefficient of the line equation is the derivative. In the algebraic case one does not need limits or infinitesimals even with visual intuition. $\endgroup$
    – Conifold
    Commented Mar 13, 2021 at 17:05
  • 2
    $\begingroup$ @Conifold, the idea of detecting a double root is earlier than Descartes. It was introduced by Fermat as part of his technique of adequality. Some historians hold that there were two techniques involved here. At any rate, detecting the double root so as to identify a maximum/minumum was certainly part of the package. Fermat's discoveries date from the late 1620s. They became known among the Parisian mathematicians around 1635 when Fermat sent his text to Mersenne. $\endgroup$ Commented Jun 20, 2023 at 10:31
  • $\begingroup$ Babylonian mathematics are worth mention here as they worked with quadratics and square roots. en.wikipedia.org/wiki/Babylonian_mathematics That said, my understanding is that the first formal proofs in the history of mathematics were Euclid's. $\endgroup$
    – DJohnson
    Commented Jul 23, 2023 at 13:10

1 Answer 1


The motivation for applying derivatives to polynomials over general fields is their use in detecting multiple roots: if $K$ is a general field, a polynomial $f(x)$ in $K[x]$ has no repeated roots (edit: in a field extension of $K$ over which $f(x)$ splits completely) if and only if $f(x)$ is relatively prime to $f'(x)$ in $K[x]$.

Formal derivatives on polynomials in $K[x]$ for a general field $K$ were introduced by Steinitz in Section 9 of his big 1910 paper on the algebraic theory of fields and used in Section 10 in order to determine when a polynomial (especially an irreducible polynomial) over a general field has a multiple root. (Perfect fields are defined in Section 11 and separable field extensions are defined in Section 13 as extensions of the "first kind".)

Bourbaki writes in the Historical Note after the section Commutative Fields in his book Algebra II that in the 19th century, Dedekind "sensed" the general notion of a derivation (look at his paper here and see the footnote by Emmy Noether on the 2nd page) and that Galois "is fully aware of the formal character" of differentiation of polynomials and used this to show an irreducible polynomial in $\mathbf F_p[x]$ has no multiple roots in his development of finite fields. So if you are looking for the very first use of derivatives outside the setting of real or complex polynomials, then it goes back to Galois for the case of polynomials over a finite field.

Going back even farther, the use of derivatives as a formal operation (no limits) to detect multiple roots of real polynomials in some sense precedes calculus by Newton and Leibniz! This is called Hudde’s rule and you can read about it here.

  • $\begingroup$ Shouldn't the statement be: A polynomial $f$ has no repeated root in the algebraic closure of $K$ iff f is relatively prime to $f'$ in $K[x]$? $\endgroup$
    – Weier
    Commented Jun 19, 2023 at 11:08
  • 1
    $\begingroup$ Sure, and I did not say the roots were supposed to be in $K$. They can be taken in a splitting field over $K$ too. $\endgroup$
    – KCd
    Commented Jun 20, 2023 at 7:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.