Calculating $[\mathcal{O}_K : \mathfrak{p}]$ using basis

Question

Given the number field $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, the ring of integers $\mathcal{O}_K = \mathbb{Z}[\gamma]$ where $\gamma = \frac{\sqrt{2} + \sqrt{6}}{2}$.

$$\langle 5 \rangle = \langle \sqrt{6} - 1 \rangle \langle \sqrt{6} + 1 \rangle $$ And I have already verified that both are unramified, and calulated their norm to both be 25. I understand that $N_{K/\mathbb{Q}}(\langle \sqrt{6} - 1 \rangle) = |\mathcal{O}_K / \langle \sqrt{6} - 1 \rangle| = |N_{K/\mathbb{Q}}(\sqrt{6} - 1) = 25$.

But I want to verify for myself that this is the extension degree of the ideal in the ring of integers.

The minimum polynomial for $\gamma$ is $x^4 - 4x^2 + 1$ so we can see $$\mathcal{O}_K \cong \mathbb{Z}[x] / \langle x^4 - 4x^2 + 1 \rangle$$ And we can also see that $-\gamma^3 + 5\gamma - 1 = \sqrt{6} - 1$. So then $$\mathcal{O}_K / \langle \sqrt{6} - 1 \rangle \cong \mathbb{Z}[x] / \langle x^4 - 4x^2 + 1 , x^3 - 5x + 1 \rangle $$

How do we see that $$|\mathcal{O}_K / \langle \sqrt{6} - 1 \rangle| = 5^2$$ and is my reasoning until this point correct?

What is the mapping $\mathbb{Z}[X] \rightarrow \mathbb{Z} / \langle 25 \rangle$ with kernel $\langle x^4 - 4x^2 + 1 , x^3 - 5x + 1 \rangle$?

Thanks

Answer 1

If $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer with minimal polynomial $f(x)$ in $\mathbf Z[x]$ and $p$ is a prime not dividing the discriminant of $f(x)$, then (i) $p$ is unramified in $K$ and (ii) when the (monic) irreducible factorization of $f(x) \bmod p$ is $$ f(x) \equiv \pi_1(x)\cdots \pi_g(x) \bmod p $$ then the prime ideal factorization of $p\mathcal O_K$ is $\mathfrak p_1\cdots\mathfrak p_g$ where ${\rm N}(\mathfrak p_i) = p^{\deg \pi_i}$. Property (i) is due to ${\rm disc}(K)$ dividing ${\rm disc}(f)$ and property (ii) is a special case of the Dedekind-Kummer theorem.

In your case, $K = \mathbf Q(\sqrt{2},\sqrt{3}) = \mathbf Q(\sqrt{2}+\sqrt{3})$, so we can use $\alpha = \sqrt{2}+\sqrt{3}$, whose minimal polynomial over $\mathbf Q$ is $x^4 - 10x^2 + 1$. Its discriminant is $147456$, which is not divisible by $5$, so the irreducible factorization $$ x^4 - 10x^2 + 1 \equiv x^4 + 1 \equiv (x^2+2)(x^2-2) \bmod 5 $$ tells us that $5\mathcal O_K = \mathfrak p\mathfrak p'$ where $\mathfrak p$ and $\mathfrak p'$ both have norm $5^2 = 25$.

Note we did not need to compute $\mathcal O_K$ explicitly to do this.

Answer 2

sage: R.<x> = PolynomialRing(ZZ, 1)
sage: I = Ideal([x^4 - 4*x^2 + 1, x^3 - 5*x + 1])
sage: I.groebner_basis()
[x^2 + 4*x + 1, 5]

Which is $\mathbb{F}_5 / \langle x^2 + 4x + 1 \rangle \cong \mathbb{F}_{5^2}$.

Thanks @franz for his comment.