If $f$ is a prime polynomial, then $f$ and $f'$ have no common root

Asked Apr 17 '23 at 19:41

Active Apr 17 '23 at 19:41

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Problem: If $f$ is a prime polynomial from $\mathbb Q[x]$, then $f$ and $f'$ have no common root.

My attempt: Suppose $f$ and $f'$ have the root $c$ in common. Divide $f$ by $f'$ to get $f = f'\cdot q + r$. Then $r = f - f'\cdot q$ and so $r(c) = 0$.

Question: I know that the degree of $r$ is less than that of $f$, but have I found a contradiction somehow? I just don't see it.

asked Apr 17 '23 at 19:41

Ungar Linski

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What you've done is the first step of the Euclidean algorithm. That's a hint how to continue. – Chad K Apr 17 '23 at 19:56
The difference between this question and the link is that this is a prime polynomial. – Ungar Linski Apr 17 '23 at 20:00
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By the linked dupe, if $f,f'$ had a common root $r$ then $,f = (x-r)^2 g,,$ which is reducible so not prime. – Bill Dubuque Apr 18 '23 at 19:26
@BillDubuque Should I assume that the root is in $\mathbb Q$ as well? – Ungar Linski Apr 19 '23 at 01:49

If $f$ is a prime polynomial, then $f$ and $f'$ have no common root

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