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Let $K$ be the splitting field of some irreducible polynomial $f(x)$ in $\mathbb{Z}[x]$, and let $B$ be the integral closure of $\mathbb{Z}$ in $K$. Suppose that, for some prime $p$: $f(x) \equiv (x-\bar{\alpha})^2f_1(x)\ldots f_k(x) \mod p$, where the $f_i(x)$ are irreducible, and not equal to $(x-\bar{\alpha})$.

My understanding is that this factorization of $f(x) \mod p$ implies that the inertia group $I_{\beta}$ of a prime ideal $\beta$ of $B$ lying over $(p)$ contains a transposition, and hence so too does the galois group $G=$Gal$(K/\mathbb{Q})$.

Is this right? Can anyone explain why in terms which might be comprehensible to a novice algebraic number theorist? I understand what an inertia group is, but cannot see why a double root mod some prime would necessarily imply that it contains a transposition.

Finally, does this generalize to other situations? I am particularly interested in the possibility of showing that the galois group of a bivariate polynomial $f(x,y)$ over $\mathbb{Q}(y)$ contains a transposition, by specialising $y$ to some rational value.

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    @Adam. By containing a transposition, one is usually referring to the fact that a (specific) representation of $Gal(K/\mathbb{Q})$ into the symmetric group on some set contains in its image a transposition. Is this what you mean, or do you mean $Gal(K/\mathbb{Q})$ contains an element of order 2?2011-09-06

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This is false. Let $d$ be a squarefree integer and $p$ be a prime not dividing $2d.$ Set $f(X) = X^2 - dp^2.$ Then $B$ is unramified at $p,$ so the intertia group of any prime above $p$ is trivial and hence does not contain a transposition. On the otherhand, $f(X) \cong X^2 \mod p,$ a contradiction.