Factorization of $5$ in the splitting field of $x^3 + 2$

Question

I wonder if someone could help to clarify the following.

Let $\zeta$ be a primitive cube root of unity and $\alpha = \sqrt[3]{2}$. Let $K = \mathbb{Q}(\alpha)$ and $L = K(\zeta)$. Then $L$ is the splitting field of $x^3 + 2$ and thus the field extension $L/\mathbb{Q}$ is Galois, of degree $6$.

One has $10 = 2^3 + 2 = (2 + \alpha)(2 + \zeta\alpha)(2 + \zeta^2\alpha)$ and thus $5 = (\alpha^2 + 1)(\alpha^2 + \zeta)(\alpha^2 + \zeta^2)$. The ideal $(\alpha^2 + 1)\mathcal{O}_K$ is a prime ideal of $\mathcal{O}_K$ (as $\alpha^2 + 1$ is a zero of $(x - 1)^3 - 4$, from which follows that $Nm_{K/\mathbb{Q}}(\alpha^2 + 1) = 5$). Let us denote it by $\mathcal{p}$. By the statemet about the Frobenius element in section $2$ of http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf, for any prime ideal $\mathcal{q} \subset \mathcal{O}_L$ that divides $\mathcal{p}$ one must have an element $Frob_q \in Gal(L/K)$ such that for any $r \in \mathcal{O}_L$ must hold $$Frob_q(r) = r^5 \mod \mathcal{q}.$$

There are only two elements in $Gal(L/K)$, namely, the identity and the map that sends $\zeta$ to $\zeta^2$. Were $Frob_q = id$, then $\zeta - \zeta^5 = \zeta(1 - \zeta^4) = \zeta(1 - \zeta)\in \mathcal{q}$ and since $\zeta$ is a unit, one must have $1 - \zeta \in \mathcal{q}$. This, however, cannot be, as $Nm_{L/\mathcal{Q}}(\zeta - 1) = 3^3$ is coprime to $5 \in \mathcal{q}$. One thus concludes that $Frob_q(\zeta) = \zeta^2$. $Frob_q$ then is also Frobenius element of the prime ideal $5\mathbb{Z}$ with respect to the extension $L/\mathbb{Q}$. When acting on $Gal(L/\mathbb{Q})$ as a permutation it factors into three disjoint cycles of length two and one thus is led to conclude that $5\mathcal{O}_L = \mathcal{p}_1\mathcal{p}_2\mathcal{p}_3$, where each $\mathcal{O}_L/\mathcal{p}_i$ has cardinality $5^2$.

Doesn't the above disagree with the fact that $x^3 + 2$ has only one root modulo $5$?

Answer 1

If want to calculate the factorization of $5\mathcal{O}_L$ into prime ideals, by reducing a polynomial in $\mathbb{Z}[x]$ modulo $5\mathbb{Z}[x]$, then you need to make sure that $\mathcal{O}_L=\mathbb{Z}[\varepsilon]$, for some integral element $\varepsilon \in L$, and then the polynomial to be reduced must be the minimal polynomial of $\varepsilon$ (and not $x^3+2$, which is the defining polynomial of a subextension $K$ of $L$).

In this case, $\mathcal{O}_L$ has a power basis. Indeed, $\mathcal{O}_L=\mathbb{Z}[\varepsilon]$ where $\varepsilon$ is a root of $$f(x)=x^6 + 3x^5 - 5x^3 + 3x + 1.$$ You can find a proof of this fact, and a whole other bunch of interesting facts about the splitting field of $x^3+2$, in these notes by my colleague, Keith Conrad. (Note that the $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{-2})$, so the splitting field of $x^3+2$ and the splitting field of $x^3-2$ coincide.)

Now, $\mathcal{O}_L\cong \mathbb{Z}[x]/(f(x))$ and so $\mathcal{O}_L/(5)\cong \mathbb{F}_5[x]/(\tilde{f}(x))$, where $\tilde{f}(x)$ is the reduction of $f(x)$ modulo $5$. The polynomial $\tilde{f}(x)$ factors into irreducibles in $\mathbb{F}_5[x]$ as follows: $$\tilde{f}(x)\equiv (x^2 + x + 2)(x^2 + 3x + 3)(x^2 + 4x + 1)$$ and therefore, $$\mathcal{O}_L/(5)\cong \mathbb{F}_5[x]/(\tilde{f}(x))\cong \mathbb{F}_5[x]/(x^2 + x + 2) \times \mathbb{F}_5[x]/(x^2 + 3x + 3) \times \mathbb{F}_5[x]/(x^2 + 4x + 1),$$ and $5\mathcal{O}_L=\wp_1\wp_2\wp_3$ factors as the product of three distinct prime ideals of degree $2$.

Since $\mathcal{O}_K =\mathbb{Z}[\sqrt[3]{-2}]$ (this is also shown in the notes mentioned above), when you reduce $x^3+2$ modulo $5$ you are calculating the factorization of $5\mathcal{O}_K$ into prime ideals in $\mathcal{O}_K$. Since $$x^3+2\equiv (x + 3)(x^2 + 2x + 4)$$ over $\mathbb{F}_5[x]$, this implies that $5\mathcal{O}_K=\mathfrak{P}_1\mathfrak{P}_2$ factors as a product of two prime ideals, with $\mathfrak{P}_1$ of degree $1$ and $\mathfrak{P}_2$ of degree $2$. If $\mathfrak{P}_1$ was to split in $L/K$, then there would be two primes of degree $1$ dividing $5\mathcal{O}_L$, so $\mathfrak{P}_1$ must be inert in $L/K$. If $\mathfrak{P}_2$ was to stay inert in $L/K$, then there would be a prime of degree $4$ dividing $5\mathcal{O}_L$, so $\mathfrak{P}_2$ splits, i.e., up to reordering we have $$\wp_1=\mathfrak{P}_1\mathcal{O}_L,\ \text{ and }\ \wp_2\wp_3=\mathfrak{P}_2\mathcal{O}_L.$$