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For any extension $E$ of a field $K$, let $\mathrm{Gal}(E:K)$ be the Galois group of $E$ over $K$, and for any polynomial $f$ in $K[X]$ with splitting field $L$, let $\mathrm{Gal}(f)$ stand for $\mathrm{Gal}(L:K)$.

The identity automorphism $e$ is an element of $\mathrm{Gal}(f)$, and it certainly fixes all the roots of $f$. Is $e$ the only element of $\mathrm{Gal}(f)$ that fixes all the roots of $f$?

I guess that the answer is "yes", but I'd love to see a proof (or, if my guess is wrong, of course, a counterexample).

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Yes: note the splitting field is the smallest extension (in some fixed algebraic closure) that contains all of the roots of your polynomial. So if you take $K$ and adjoin all of the roots of $f$, you must obtain $L$. So $L = K(\alpha_1,\dots,\alpha_n)$, where $\alpha_i$ are the roots of the polynomial. An element of $Gal(f)$ fixes $K$, and if it fixes each $\alpha_i$, then it must fix all of $L$, so it must be the identity.

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