Splitting of primes in a Galois extension Let $L$ be the splitting field of $x^3+2x+1$. I want to know how 59 splits in $L$. I calculated the discriminant of $\mathbb{Z}[\alpha]$ to be $-59$, (where $\alpha$ is a root of the polynomial), which is squarefree therefore $\mathcal{O}_K = \mathbb{Z}$ (because $d(\mathbb{Z}[\alpha]) = (\mathcal{O}_K:\mathbb{Z}[\alpha])^2d_K$). Since 59 divides the discriminant, it must be ramified. But I don't know how to get anything more than that, like what's the ramification index and how many primes does it split into?
 A: Let’s call $f(x)=x^3+2x+1\in\Bbb Z[x]$, and $\alpha$ the chosen root. Let’s look at the nonnormal extension $\Bbb Q(\alpha)\supset \Bbb Q$.
Working in $\Bbb F_{59}$, we have $f(x)\equiv(x-31)(x-14)^2$, which means that over $\Bbb Z_{59}$, there’s a linear factor congruent to $x-31$ and a quadratic factor $g(x)\in\Bbb Z_{59}[x]$. In fact, this quadratic factor is $59$-Eisenstein, so that the factorization of $(59)$ in $\Bbb Q(\alpha)$ is $\mathfrak p_1\mathfrak p_2^2$.
Can you take it from there?
A: In a finite extension  of number fields $L/K$, with rings of integers $S/R$, there is a criterion attributed to Dedekind which gives you systematically and explicitly the decomposition in $S$ of a prime ideal $P$ of $R$ under a mild condition. More precisely, suppose that $L=K(\alpha)$, with $\alpha \in R$ (this is always possible), and that the rational prime $p$ under $P$ does not divide the index $[S:R[\alpha)]$. Let $f \in R[X]$ be the monic irreducible polynomial of $\alpha$ over $R$, $\bar f$ its image in $(R/P)[X]$, and $\bar f= {\bar f_1}^{e_1}...{\bar f_r}^{e_r}$ the decomposition of $\bar f$ into irreducible factors in $(R/P)[X]$. Then the prime decomposition of $P.S$ in $S$ is ${Q_1}^{e_1}...{Q_r}^{e_r}$, where $Q_i=P.S + f_i(\alpha).S$, and moreover, the inertia index $f(Q_i/P)$ is the degree of $f_i$. See e.g. Marcus' "Number Fields", chap. 3, thm. 27.
Here $K=\mathbf Q$, and you have shown that $S=\mathbf Z[\alpha]$ because the discriminant is $-59$ [there are misprints in your post], so the criterion applies to any rational prime $p$, and it remains only to compute the decomposition of $ \bar f=X^3+2X+1$ mod $p$. But here the choice of $p=59$ is surely meant to bring some simplification. For example, since disc($\bar f)=0$, the definition of the discriminant implies that $\bar f$ has a multiple root, say $\beta$. This cannot be a triple root, because  $\beta^3+2\beta+1 =3\beta^2 +2= 6\beta=0$ has no solution in $\mathbf F_{59}$. So $\beta$ must be a double root, a solution of  $\beta^3+2\beta+1 =3\beta^2 +2=0$, and a simple elimination shows that $\beta=-\frac34 =14$, which is the double root computed by @Lubin (and it is the only one possible). The third (simple) root, 31, can be deduced from the discriminant. Finally $(59)=P_1 . {P_2}^2$, with $P_1 = (59, \alpha -31)$ and $P_2 = (59,\alpha -14)$. 
