Splitting of $2$ in a cubic extension

Question

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.

Answer 1

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 $(-1-\lambda)(2-\lambda)^2+1 = -\lambda^3+3\lambda^2-3 = \lambda(3\lambda-\lambda^2)-3$.

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 $\mathcal O_L$.

Answer 2

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 $\theta\in\mathcal{O}$ with minimal polynomial over $K$ being $p(X)\in\mathcal{o}[X]$, 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 $$ \overline{p}(X)=\overline{p}_1(X)^{e_1}\cdots\overline{p}_r(X)^{e_r} $$ 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 $p_i(X)\in\mathcal{o}[X]$ 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 $$ \mathfrak{p}=\mathfrak{P}^{e_1}_1\cdots\mathfrak{P}^{e_r}_r. $$

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