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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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Are you assuming $f(x)$ is irreducible over $\mathbb{Q}$? Otherwise, that double factor modulo $p$ could be a reflection of a double factor in $f(x)$ and tell you nothing. Also, you are probably assuming that the $f_i$ are distinct from $x-\overline{\alpha}$... –  Arturo Magidin Sep 6 '11 at 19:10
yes, sorry...edited the question. –  Adam Sep 6 '11 at 19:18
It's true if $B=\mathbb{Z}[r]$ where $r$ is a roof of $f(x)$, because then the prime factorization of $(p)$ in $B$ is given by the factorization of $f(x)$ modulo $p$; so $(p)$ would be a product of squares of prime ideals, the inertia group would have order $2$, and hence have a transposition... –  Arturo Magidin Sep 6 '11 at 19:19
@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? –  jspecter Sep 6 '11 at 19:44

1 Answer 1

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.

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