Clarification of: $f$ has no multiple root in any field extension of $F$ unless $f'$ is the zero polynomial

Question

In Artin's Algebra, Proposition 15.6.8 is: Let $f$ be an irreducible polynomial in $F[x]$.

$f$ has no multiple root in any field extension of $F$ unless $f'$ is the zero polynomial.

Then, it is written that in a field $F$ of characteristic $5$, since derivative of $f(x)=x^{15}+ax^{10}+bx^5+c$ is zero, its roots will be multiple roots in any extension.

How does this follow from the statement given?

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The conditional statement $q$ unless $\neg p$ is equivalent to if $p$ then $q$. So,

$f$ has no multiple root in any field extension of $F$ unless $f'$ is the zero polynomial.

is equivalent to

If $f'$ is not the zero polynomial, then $f$ has a multiple root in any extension of $F$.

and by contraposition, this is equivalent to

If $f$ has no multiple root in any field extension of $F$, then $f'$ is the zero polynomial.

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So, how does the derivative equivalent to zero imply $f$ has a multiple root in any extension?

Answer 1

You're right that the conclusion doesn't follow from the statement you quoted first. But it is reasonably easy to prove from first principles:

Let $f(x)=x^{15}+ax^{10}+bx^5+c$ in some field of characteristic $5$, and suppose $\xi$ is a root of $f$. Now consider the function $$ g(y)=f(y+\xi)=(y+\xi)^{15}+a(y+\xi)^{10}+b(y+\xi)^5+c$$ This is a polynomial in $y$. If we use the binomial theorem to expand $(y+\xi)^{5n}$, the only terms whose binomial coefficient don't vanish modulo $5$ are those where the power of $y$ is a multiple of $5$. So $g$ must actually have the form

$$ g(y)=y^{15}+a'y^{10}+b'y^5+c'$$ and since $g(0)=0$ (because $\xi$ was a root of $f$) we have that $c'=0$. Thus $$ g(y) = p(y)y^5 $$ for some polynomial $p$. But then $f(x)=g(x-\xi) = p(x-\xi)\cdot(x-\xi)^5$ which by definition means that $\xi$ is an at least fivefold root of $f$.

Answer 2

The multiple roots of a polynomial $f$ are the roots of $\gcd(f, f')$ (see https://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor). If $f' = 0$, $\gcd(f, f') = f$, so all the roots of $f$ are multiple roots.

I am not familiar with Artin's book, but it must surely include the above well-known characterisation of multiple roots. I think it is the above characterisation rather than the proposition you quote that Artin has in mind when he states that the polynomial in $x^5$ will have multiple roots in any extension field.