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The following result about polynomials is known:

Proposition: Let $K$ be a subfield of $\mathbb{C}$, $f(x) \in K[x]$ a polynomial with degree $n \geq 1$ and $\alpha \in \mathbb{C}$ a root of $f(x)$. Then

a) $\alpha$ is a simple root of $f(x)$ $\iff$ $f(\alpha) = 0$ and $f'(\alpha) = 0$;

b) if $f(x)$ is irredutible over $K$ then all the roots of $f(x)$ are simple.

Is this result true if we exchange $\mathbb{C}$ for any algebraically closed field?

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I believe you want to say that $\alpha$ is a simple root implies that $f(\alpha)=0, f'(\alpha)\neq 0$. This is not true is the characteristic is $p$, $X^p-a=(X-a)^p$ and $a$ is simple.

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  • $\begingroup$ It is a subfield of $\mathbb{C}$ Char=0 $\endgroup$
    – IrbidMath
    Commented Jul 1, 2020 at 23:54
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    $\begingroup$ an algebraically closed field is not always a subfield of $\mathbb{C}$. $\endgroup$
    – Tsemo Aristide
    Commented Jul 1, 2020 at 23:55
  • $\begingroup$ I thought you missed read that $K$ is subfield of $\mathbb{C}$ but it seems I missed read the questions $\endgroup$
    – IrbidMath
    Commented Jul 2, 2020 at 0:17
  • $\begingroup$ Thus the item-b) is not true too. Right? $\endgroup$
    – Croos
    Commented Jul 2, 2020 at 2:23

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