Multiplicity of Zeroes and Derivatives

Question

Does it hold that:

If $\alpha \in K^{alg}$, $K$ a field and $f \in K\left[ X \right]$ then it holds:

$\alpha$ is a zero of $f$ with multiplicity greater than one iff $f(\alpha) = f(\alpha)' = 0$.

Answer 1

It is true.

Suppose that $\alpha$ is a zero of $f$ with multiplicity at least $2$ -- that is, $f(X) = \left( X -\alpha \right)^{2} g(X)$ for some polynomial $g \in K^{\text{alg}}[X]$. We have $f^{\prime}(X) = 2 \left( X -\alpha \right) g(X) +\left( X -\alpha \right)^{2} g^{\prime}(X)$. Therefore, $f(\alpha) = f^{\prime}(\alpha) = 0$.

Conversely, suppose that $f(\alpha) = f^{\prime}(\alpha) = 0$. The remainder of the Euclidean division of $f(X)$ by $X -\alpha$ is $f(\alpha) = 0$, and hence $f(X) = \left( X -\alpha \right) g(X)$ for some polynomial $g \in K^{\text{alg}}[X]$. We have $f^{\prime}(X) = g(X) +\left( X -\alpha \right) g^{\prime}(X)$, and hence $g(\alpha) = f^{\prime}(\alpha) = 0$. It follows that $g(X) = \left( X -\alpha \right) h(X)$ for some polynomial $h \in K^{\text{alg}}[X]$. Thus, we have $f(X) = \left( X -\alpha \right)^{2} h(X)$, which proves that $\alpha$ is a zero of $f$ with multiplicity at least $2$.

Note that the natural generalization to higher multiplicities is true when the characteristic of $K$ is $0$ but not true in general. For example, $0$ is a zero of $f(X) = X^{2} \in \mathbb{F}_{2}[X]$ with multiplicity $2$ but $f(0) = f^{\prime}(0) = f^{\prime \prime}(0) = 0$.