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If $f(x) \in F[x]$ is irreducible, then
1. If the characteristic of $F$ is 0, then $f(x)$ has no multiple roots.
2. If the characteristics of $F$ is $p \neq 0$ then $f(x)$ has multiple roots if it is of form $f(x)=f(x^p)$.

In my book,

For the first one, since if $f(x)$ has multiple root, then $f'(x)$ and $f(x)$ have non trivial common root. Thus $f(x)|f'(x)$ which forces $f'(x) = 0$ or $f(x) $ is constant.

My question is how come $f(x)|f'(x)$ if they have common factor? shouldn't there be some irreducible factor of $f(x)$ that remains at denominator? Also what does it have to do with characteristic $0$?

for second the book writes,

in characteristic $p \neq 0$, this forces $f(x) = g(x^p)$
I don't understand how ??

I am using I.N. Herstein Topics in Algebra second edition.

Mula Ko Saag
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1 Answers1

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If $a$ is a multiple root that $f(X)=(X-a)^2g(X)$ (in an extension) and $f'(X)=2(X-a)g(X)+(X-a)^2g'(X)$ have (at least) $X-a$ as common factor. But if $f$ is already irreducible(!), there is no such thing as a proper factor of $f$. That is, $f$ itself is a factor of $f'$. But as $\deg f'<\deg f$, this implies $f'=0$. So far, this works for all characteristics. However, in characteristic $p>0$ we may have $f'=0$ even for nonconstant $f$. For example the derivative of $X^p$ is $0$. Looking at the derivative of $\sum_{k=0}^n a_kX^k$ we note that it is zero iff $ka_k=0$ for all $k$. For $k\ne 0$ (which means $p\not\mid k$ if we work in characteristic $p$) this means $a_k=0$; but for $p\mid k$, no condition is obtained. Thus all coefficients of $X^{ip}$ are allowed to be arbitrary. This precisely says that we have a polynomimal in $X^p$.

S L
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Hagen von Eitzen
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    Is the question about multiple roots over F or over an extension of F? In the first case the answer is too convoluted, in the latter case it is incorrect: "But if is already irreducible(!)" -- it is irreducible over F, but not over an extension of F. – Adam Jan 20 '23 at 20:49