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Let $L$ be the splitting field of $X^3-3X+1=0$. How does the prime $2$ split in $L$? I have figured out that either $2=mathfrak{P}$ or $2=mathfrak{P}mathfrak{Q}mathfrak{R}$. I guess it is the second form, but I do not know how to find the prime divisors.

The discriminant of $X^3-3X+1$ is $81 = 9^2$, so the splitting field is $L = Bbb Q[X]/(X^3-3X+1)$. Let $t in L$ with $t^3-3t+1=0$. Let $alpha, beta in L$ be the remaining two roots.

Then $X^3-3X+1 = (X-t)(X^2+tX+(t^2-3)) = (X-t)(X-alpha)(X-beta)$, so:
$$(X-alpha)(X-beta) = X^2 + tX + (t^2-3)$$

From Vieta, we have $t+alpha+beta = 0$, so $alpha+beta = -t$.

Its discriminant is $81 = 9^2$, so:

$$begin{array}{rcl}
(t-alpha)(t-beta)(alpha-beta) &=& 9 \
(t^2 + tt + (t^2-3))(alpha-beta) &=& 9 \
(3t^2-3)(alpha-beta) &=& 9 \
(t^2-1)(alpha-beta) &=& 3 \
end{array}$$

To find the inverse of $t^2-1$, we calculate the characteristic polynomial of the matrix of multiplication by $t^2-1$, with basis ${1, t, t^2}$:

• $1 times (t^2-1) = -1 + 0t + t^2$


• $t times (t^2-1) = t^3 - t = 3t - 1 - t = 2t - 1 = -1 + 2t + 0t^2$


• $t^2 times (t^2-1) = 2t^2 - t = 0 + (-1)t + 2t^2$

So the matrix is:

$$begin{bmatrix} -1 & -1 & 0 \ 0 & 2 & -1 \ 1 & 0 & 2 end{bmatrix}$$

whose characteristic polynomial is $(-1-lambda)(2-lambda)^2+1 = -lambda^3+3lambda^2-3 = lambda(3lambda-lambda^2)-3$.

Therefore:

$$begin{array}{rcl}
(t^2-1)(3(t^2-1)-(t^2-1)^2)-4 &=& 0 \
dfrac3{t^2-1} &=& 3(t^2-1)-(t^2-1)^2 \
&=& 2t^2+t-4
end{array}$$

So:
$$alpha - beta = frac3{t^2-1} = 2t^2+t-4$$

And we conclude:

• $alpha = dfrac12 [(-t) + (2t^2+t-4)] = t^2-2$


• $beta = (t^2-2)-(2t^2+t-4) = -t^2-t+2$

Now to calculate the relevant discriminant, we take the squared determinant of:

$$begin{bmatrix}
1 & t & t^2 \
1 & t^2-2 & -t^2-t+4 \
1 & -t^2-t+2 & t+2
end{bmatrix}$$

whose determinant is $-9$, so the discriminant is $81$.

So potential additional integers have the form $dfrac13(a+bt+ct^2)$ with $a,b,c in {0,1,2}$.

So I wrote a program to check:

(21:43) gp > f=(a,b,c)->charpoly([a,-c,-b;b,a+3*c,3*b-c;c,b,a+3*c])

%30 = (a,b,c)->charpoly([a,-c,-b;b,a+3*c,3*b-c;c,b,a+3*c])

(21:43) gp > for(i=0,2,for(j=0,2,for(k=0,2,print(f(i/3,j/3,k/3)))))

x^3

x^3 - 2*x^2 + x - 1/27

x^3 - 4*x^2 + 4*x - 8/27

x^3 - 1/3*x + 1/27

x^3 - 2*x^2 + x - 1/9

x^3 - 4*x^2 + 13/3*x - 19/27

x^3 - 4/3*x + 8/27

x^3 - 2*x^2 + 1/3*x + 1/27

x^3 - 4*x^2 + 4*x - 8/9

x^3 - x^2 + 1/3*x - 1/27

x^3 - 3*x^2 + 8/3*x - 17/27

x^3 - 5*x^2 + 7*x - 19/9

x^3 - x^2 + 1/9

x^3 - 3*x^2 + 8/3*x - 19/27

x^3 - 5*x^2 + 22/3*x - 71/27

x^3 - x^2 - x + 19/27

x^3 - 3*x^2 + 2*x - 1/3

x^3 - 5*x^2 + 7*x - 73/27

x^3 - 2*x^2 + 4/3*x - 8/27

x^3 - 4*x^2 + 5*x - 17/9

x^3 - 6*x^2 + 32/3*x - 136/27

x^3 - 2*x^2 + x - 1/27

x^3 - 4*x^2 + 5*x - 53/27

x^3 - 6*x^2 + 11*x - 17/3

x^3 - 2*x^2 + 8/9

x^3 - 4*x^2 + 13/3*x - 37/27

x^3 - 6*x^2 + 32/3*x - 152/27

As one can see, the only integer polynomial is $x^3$ which corresponds to $0$. So we conclude that $mathcal O_L = Bbb Z[t] = Bbb Z[X]/(X^3-3X+1)$.

And finally to see the behaviour of $(2)$:

$$begin{array}{rcl}
mathcal O_L / 2 mathcal O_L
&=& Bbb Z[X]/(X^3-3X+1,2) \
&=& Bbb F_2[X]/(X^3-3X+1)
end{array}$$

which is a field, as $X^3-3X+1 in Bbb F_2[X]$ is irreducible, as the comments have pointed out.

Therefore $(2)$ is a prime in $mathcal O_L$.

Use this proposition from p. 47 of Algebraic Number Theory by Neukirch: Let $L/K$ be a separable extension of number fields and let $mathcal{o}$ and $mathcal{O}$ be the rings of integers of $K$ and $L$, respectively. Assume that $L=K(theta)$ for some $thetainmathcal{O}$ with minimal polynomial over $K$ being $p(X)inmathcal{o}[X]$, then we have:

(8.3) Proposition. Let $mathfrak{p}$ be a prime ideal of $mathcal{o}$ which is relatively prime to the conductor $mathfrak{F}$ of $mathcal{o}[theta]$, and let
$$
overline{p}(X)=overline{p}_1(X)^{e_1}cdotsoverline{p}_r(X)^{e_r}
$$
be the factorization of the polynomial $overline{p}(X)=p(X)pmod{mathfrak{p}}$ into irreducibles $overline{p}_i(X)=p_i(X)pmod{mathfrak{p}}$ over the residue class field $mathcal{o}/mathfrak{p}$, with all $p_i(X)inmathcal{o}[X]$ monic. Then
$$
mathfrak{P}_i=mathfrak{p}mathcal{O}+p_i(theta)mathcal{O},quad i=1,ldots,r,
$$
are the different prime ideals of $mathcal{O}$ above $mathfrak{p}$. The inertia degree $f_i$ of $mathfrak{
P}_i$ is the degree of $overline{p}_i(X)$, and one has
$$
mathfrak{p}=mathfrak{P}^{e_1}_1cdotsmathfrak{P}^{e_r}_r.
$$

Now we apply this to our problem.

In our problem $K=mathbb{Q}$ and $L=mathbb{Q}(theta)$, where $theta$ is a root of $p(X)=X^3-3X+1$. $mathcal{o}=mathbb{Z}$ and $mathcal{O}=mathbb{Z}[theta]$. The minimal polynomial of $thetainmathcal{O}$ over $mathbb{Q}$ is $p(X)inmathbb{Z}[X]$. Let $mathfrak{p}=2mathbb{Z}$. Since the conductor of $mathbb{Z}[theta]$ is $mathbb{Z}[theta]$ itself, $2mathbb{Z}$ is relatively prime to it. Now $overline{p}(X)=p(X)pmod{2}=X^3-3X+1pmod{2}$. $overline{p}(X)$ is irreducible over $mathbb{Z}/2mathbb{Z}$ because it is of degree $3$ and has no roots in $mathbb{Z}/2mathbb{Z}$. Whence the factorization of $overline{p}(X)$ into irreducibles over $mathbb{Z}/2mathbb{Z}$ is just itself, namely
$$
overline{p}(X)=X^3-3X+1pmod{2}.
$$
Notice that $X^3-3X+1inmathbb{Z}[X]$ is monic. Then
$$
mathfrak{P}=mathfrak{p}mathcal{O}+p(theta)mathcal{O}=2mathbb{Z}mathbb{Z}[theta]+0mathbb{Z}[theta]=2mathbb{Z}[theta]
$$
is the only prime ideal of $mathbb{Z}[theta]$ above $2mathbb{Z}$. The inertia degree of $2mathbb{Z}[theta]$ is the degree of $X^3-3X+1pmod{2}$, i.e. $3$, and one has
$$
2mathbb{Z}=2mathbb{Z}[theta].
$$
In particular, $2mathbb{Z}$ is inert in $mathbb{Q}(theta)$, i.e. $2$ is inert in $mathbb{Q}(theta)$.