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I know that if $K$ is a field and $f\in K[x]$, then there exists a splitting field of $f$ on $K$.

If one has two isomorphic fields $K_1$ and $K_2$ (say $\sigma$ an isomorphism) and $f\in K_1[x]$, then we may consider the splitting field $F_1$ of $f$ on $K_1$ and the splitting field $F_2$ of $\bar{f}\in K_2[x]$ (where $\bar{f}$ is the image of $f$ in the obvious isomorphism between $K_1[x]$ and $K_2[x]$). I know that $F_1$ and $F_2$ are isomorphic and that the number of autmorphisms which coincide with $\sigma$ over $K_1$ is bounded by $(F_1:K_1)$ and coincide with it if all the roots of $f$ are simple.

My question is the following one: if the number of such automorphisms coincide with $(F_1:K_1)$, does $f$ have simple roots? If not, where I can find a counterexample?

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The answer is no. For example, let $y$ be an element of $F$ and let $g(x) = (x-y)^2f(x)$, where $f(x)$ is in $F[x]$ and has all simple roots. Then the splitting field of $g$ over $F$ is the same as that of $f$.

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