There is the well-known proposition (for a specific reference: see section 14.1 of Dummit/Foote):

Proposition. If $\sigma \in \text{Aut}(K/F)$ and $f(x) \in F[x]$ has $\alpha \in K$ as a root, then $f(x)$ also has $\sigma(\alpha)$ as a root.

This got me wondering about the following scenario:

Suppose there exists distinct $\sigma_1, \sigma_2 \in \text{Aut}(K/F)$ with the property that $\sigma_1(\alpha) = \sigma_2(\alpha)$, where $\alpha \in K$ is some root of $f(x)$. Then does $f(x)$ have $\sigma_1(\alpha)$ as a root with multiplicity $2$? (i.e. one multiplicity "comes from" $\sigma_1$ and one multiplicity "comes from" $\sigma_2$)

After some thought, I think this is false. For example, consider a Galois extension $K/F$ and some irreducible polynomial $p(x) \in F[x]$ with $\alpha$ as a root. If it so happens that there are distinct $\sigma_1, \sigma_2 \in \text{Aut}(K/F)$ with $\sigma_1(\alpha) = \sigma_2(\alpha)$, then it cannot be that $p(x)$ has $\sigma_1(\alpha)$ as a multiple root, since $K/F$ being Galois implies that $p(x)$ is separable.

Is this reasoning correct? Also, are there any easy counterexamples when $K/F$ is not Galois? (If an example would require some exotic construction, then please don't waste your time writing it up...)

Thanks so much!

  • $\begingroup$ Irreducible implies separable. If there exists a stem field distinct from the splitting field, then at least 2 automorphisms of the Galois group do coincide on this intermediate stem field because it is the fixed field of a subgroup of the Galois group of the polynomial so fixes the root that generates the stem field. $\endgroup$
    – NevD
    May 6, 2017 at 0:00

1 Answer 1


The reasoning is correct, but we can make a stronger observation. The proposition you mentioned can be strengthened as follows: If $\sigma \in \operatorname{Aut}(K/F)$, $f \in F[x]$, and $\alpha \in K$, then $\sigma(\alpha)$ has the same multiplicity as a root of $f$ as does $\alpha$.


For any $p \in K[x]$, let $\sigma p$ be the polynomial obtained by applying $\sigma$ to the coefficients of $p$. This gives an automorphism of the ring $K[x]$.

Now, if $\alpha$ has multiplicity $n$, then $f(x) = (x - \alpha)^n g(x)$ for some $g \in K[x]$. Then, $f(x) = (\sigma f)(x) = (x - \sigma(\alpha))^n (\sigma g)(x)$. Thus $\sigma(\alpha)$ has multiplicity at least $n$, and applying the same reasoning with $\sigma^{-1}$ we see that the multiplicities are equal. QED

Therefore, in your proposed scenario, $\sigma_1(\alpha)$ cannot be a double root unless $\alpha$ itself is.

  • $\begingroup$ Thanks for this answer. Two questions about the line $f(x) = (\sigma f)(x) = (x- \sigma(\alpha))^n(\sigma g)(x)$. (1) How do you get the first equality? (2) Why doesn't this by itself imply that $\sigma(\alpha)$ has multiplicity $n$?, ie without considering $\sigma^{-1}$ $\endgroup$
    – Sam Y.
    May 6, 2017 at 0:43
  • $\begingroup$ @SamY. (1) $f = \sigma f$ because the coefficients of $f$ lie in $F$, which is fixed by $\sigma$. (2) If it happened that $\sigma(\alpha)$ were a root of $\sigma g$, then the multiplicity would be greater than $n$. It's a bit of a stylistic issue how to rule that out. You could instead say that $g(\alpha) \ne 0$ implies $(\sigma g)(\sigma(\alpha)) \ne 0$, because otherwise applying $\sigma^{-1}$ to the equation $(\sigma g)(\sigma(\alpha)) = 0$ would yield $g(\alpha) = 0$. However you do it, the invertibility of $\sigma$ must be used somehow. $\endgroup$ May 6, 2017 at 1:43
  • $\begingroup$ Thanks! I didn't realize $f$ was in $F$- this makes question (1) obvious now... $\endgroup$
    – Sam Y.
    May 6, 2017 at 2:09

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