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Example. Let $K$ be a number field with $O_K=\mathbb{Z}[\alpha]$, where $\alpha$ is a root of $f(X)=X^3+X+1$. Let us check the prime factorizations of $(2),(3) \in \mathbb{Z}$.

$X^3+X+1 $ is irreducible $(\mod 2 )$ therefore $2\mathbb{Z}[\alpha]=(2)=\mathfrak{p}$, it inerts. $X^3+X+1 \equiv (X-1)(X^2 + X + 2 )(\mod 3 ) $ so it is reducible. Thus, $3\mathbb{Z}[\alpha]=(3,\alpha -1)(3, \alpha^2+\alpha+ 2)=\mathfrak{q_1}\mathfrak{q_2}$.

Now, I think that $[K:\mathbb{Q}]=3$ so that $Norm(\mathfrak{p})=2^3=8$. Also, $Norm(\mathfrak{q_1}\mathfrak{q_2}) = 3^3$. So, I have few questions:

  1. Can we say that $2\mathbb{Z}[\alpha] = (2,\alpha^3 + \alpha +1)=(2)?$ $Norm((3,\alpha -1)(3, \alpha^2+\alpha+ 2))=Norm(\mathfrak{q_1}\mathfrak{q_2})=27.$
  2. How do we decide whether $Norm(q_i) =3$ or $9$, for $i=1,2$?

  3. What can be said about norm of an ideal with two generators, e.g. $Norm(\alpha,\beta)$ in general? I know that in $O_K$, ideals has at most $2$ generators.

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  • $\begingroup$ It is not mentioned in the lecture notes that I am following. $\endgroup$ – Ninja Dec 20 '17 at 20:35
  • $\begingroup$ Of course, I edited the question. $\endgroup$ – Ninja Dec 20 '17 at 20:52
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Here's a partial answer to your question that applies to the cases in the example. Suppose $\mathcal O_K$ is generated by a single element $\alpha$ with minimal polynomial $f(X)$. Then for every rational prime $p \in \mathbf Z$, we can factor $p \mathcal O_K$ as a product of prime ideals. Each prime ideal $\mathfrak p \mid p$ has the form

$$\mathfrak p = (p,g(\alpha))$$

where $g$ is one of the irreducible factors of $f \bmod p$.

In this case $\mathfrak N(\mathfrak p) = p^{\deg g}$. We can prove this as follows:

\begin{align} \mathfrak N(\mathfrak p) &= \left\lvert \mathcal O_K / \mathfrak p \right\rvert \tag{1}\\ \mathcal O_K/\mathfrak p &= \frac{\mathbf Z[\alpha]}{(p, g(\alpha))} \tag{2}\\ &\cong \frac{\mathbf Z[X]/(f(X))}{(p,g(X),f(X))/(f(X))} \tag{3} \\ &\cong \frac{\mathbf Z[X]}{(p,g(X),f(X))} \tag{4} \\ &\cong \frac{\mathbf Z/p\mathbf Z[X]}{(g(X),f(X))} \tag{5} \\ &\cong \frac{\mathbf Z/p\mathbf Z[X]}{(g(X))} \tag{6} \\ &= \operatorname{GF}(p^{\deg g}) \tag{7} \end{align}

Lines 1, 2 are by definition. Line 3 is since $\mathbf Z[\alpha] \cong \mathbf Z[X]/(f(X))$. Lines 4 and 5 are the third isomorphism theorem. Line 6 is since $g(X) \mid f(X) \pmod p$.

Finally, since $g$ is irreducible, $\mathbf Z/p\mathbf Z[X]/(g(X))$ is a field. Moreover it is a field extension of $\mathbf Z/p\mathbf Z$ of degree $\deg g$. Thus it is a finite field with $p^{\deg g}$ elements. It follows that $\mathfrak N(\mathfrak p) = p^{\deg g}$.

So to answer your first two questions (where $f(X) = X^3 + X + 1$):

  1. Yes, $2 \mathbf Z[\alpha]$ is a prime ideal so it factors as $(2, \alpha^3 + \alpha + 1) = (2)$ (since $\alpha^3 + \alpha + 1 = 0$).

  2. $\mathfrak N((3, \alpha - 1)) = 3$ and $\mathfrak N((3, \alpha^2 + \alpha + 2)) = 3^2$.

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