The inertia field is ramified?

Question

I can't seem to figure out how this example is possible. When I put the following into SageMaths:

K.<a> = NumberField(x^6 - 3*x^5 + 5*x^4 - 5*x^3 + 5*x^2 - 3*x + 1)
G = K.galois_group()
H1=G.inertia_group(K.primes_above(23)[0])
H1.fixed_field()

I get:

(Number Field in a0 with defining polynomial x^3 - x - 1 with a0 = a^4 - 2*a^3 + 3*a^2 - 2*a + 2,
 Ring morphism:
   From: Number Field in a0 with defining polynomial x^3 - x - 1 with a0 = a^4 - 2*a^3 + 3*a^2 - 2*a + 2
   To:   Number Field in a with defining polynomial x^6 - 3*x^5 + 5*x^4 - 5*x^3 + 5*x^2 - 3*x + 1
   Defn: a0 |--> a^4 - 2*a^3 + 3*a^2 - 2*a + 2)

Now the fixed field of the inertia group should be unramified, yet for some reason:

M.<b>=NumberField(x^3 - x - 1)
M.discriminant()

gives:

-23

I would expect that the inertia group is equal to the Galois group since only one prime ramifies, and so the inertia field would be an unramified extension of $\mathbb{Q}$ unless we had that the inertia field was equal to $\mathbb{Q}$, in which case the inertia group is equal to the Galois group. So, where does this code go wrong?

Answer 1

$K/\Bbb{Q}$ is Galois (it is the splitting field of $x^3-x-1$)

$$23O_K=\mathfrak{P}_1^2\mathfrak{P}_2^2\mathfrak{P}_3^2$$

You are looking at the fixed subfield $L=K^{I(\mathfrak{P}_1)}$ where $I(\mathfrak{P}_1)=\{ \sigma\in Gal(K/\Bbb{Q}),\forall a\in O_K,\sigma(a)-a\in \mathfrak{P}_1\}$ is the inertia of $\mathfrak{P}_1$.

You get $L=\Bbb{Q}[x]/(x^3-x-1)$ and $$23O_L=\mathfrak{p}_1\mathfrak{p}^2$$ where $\mathfrak{p}_1=\mathfrak{P}_1\cap O_L$ and $\mathfrak{p}=\mathfrak{P}_2\cap O_L=\mathfrak{P}_3\cap O_L$.

Only $\mathfrak{p}_1$ is unramified, not all prime above $23$.