= Solution
Let
$$
F(X)=X^3+25X^2-50X+40.
$$
Over $\mathbb Q_2$, the <Newton polygon> has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from <Hensel lemma>. There is one unit root because $F(1)$ is even and $F'(1)$ is odd. For the two remaining roots, put
$$
F(2Y)=4(2Y^3+25Y^2-25Y+10),
$$
whose parenthesized polynomial has a simple root $Y\equiv1\pmod2$, and
$$
F(4Z)=8(8Z^3+50Z^2-25Z+5),
$$
whose parenthesized polynomial has a simple root $Z\equiv1\pmod2$. Thus $F$ splits completely over $\mathbb Q_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one
$$
\boxed{L_q=\mathbb Q_2,\qquad \pi_q=2,\qquad k_q=\mathbb F_2.}
$$
Over $\mathbb Q_5$, $F$ is an <Eisenstein polynomial>. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root then
$$
\boxed{L_q=\mathbb Q_5(\alpha),\qquad \pi_q=\alpha,\qquad k_q=\mathbb F_5,}
$$
with $e=3$ and $f=1$.
The polynomial is irreducible over $\mathbb Q$ by the <Eisenstein criterion> at $5$. Its <polynomial discriminant> is
$$
\Delta=-4\cdot25\cdot13807,
$$
which is not a <square number>, so the <Galois group of an irreducible cubic> is
$$
\boxed{\operatorname{Gal}(E/\mathbb Q)\cong S_3.}
$$
At $2$, all three roots already lie in $\mathbb Q_2$, so the local splitting field is trivial and
$$
\boxed{D_2=I_2=1.}
$$
At $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv3\pmod5$, a nonsquare unit, so the <quadratic resolvent field of a cubic> is the unramified quadratic extension of $\mathbb Q_5$. The local splitting field therefore has degree six, with
$$
\boxed{D_5\cong S_3,\qquad I_5\cong A_3\cong C_3.}
$$
These calculations are summarized by <local factorization of X3 plus 25X2 minus 50X plus 40>.
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