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Let $L$ be a number field with $n:=[L:\mathbb{Q}]$

Let $0 \neq I$ be an ideal of $\mathcal{O}_L$. Then $\rvert \mathcal{O}_L/I\rvert <\infty$

Proof:

Since $\mathbb{Z}$ is a PID, and $\mathcal{O}_L$ is a free $\mathbb{Z}$-module of rank $n$, and $I$ is a submodule of $\mathcal{O}_L$, we can use the following theorem from Samuel's Algebraic theory of numbers (pg. 21)

Theorem Let $A$ be a principal ideal ring, $M$ a free $A$-module of rank $n$, and $M'$ a submodule of $M$. Then:

  1. $M'$ is free of rank $q$, $0\leq q \leq n$.
  2. If $M'\neq 0$, there exists a base $(e_1,\dots,e_n)$ of $M$ and non-zero elements $a_1,\dots,a_q\in A$ such that $(a_1e_1,\dots,a_qe_q)$ is a basis of $M'$ and such that $a_i$ divides $a_{i+1}$, $1\leq i \leq q-1$.

Then $I$ is a free $\mathbb{Z}$-module of rank $q\leq n$.

Furthermore since $\mathcal{O}_L\cong \mathbb{Z}^n$, there is a integral basis $\{\alpha_1, \ldots, \alpha_n\}$ of $\mathcal{O}_L$. Then since $I\neq0$, we can take some $a\in I$ such that $\{a\alpha_1, \ldots, a\alpha_n\}\subseteq I$. Since $\{\alpha_1, \ldots, \alpha_n\}$ is an integral basis of $\mathcal{O}_L$, it is $\mathbb{Z}$-linearly independent and hence $\{a\alpha_1, \ldots, a\alpha_n\}$ is $\mathbb{Z}$-linearly independent.Thus $I$ has rank $q\geq n$.

Putting the two together we get that $I$ has rank $n$.

Using the second part of the theorem from Samuel there is an integral basis $\{\beta_1, \ldots, \beta_n\}$ of $\mathcal{O}_L$ and integers $d_1,\ldots, d_n$ with $d_i\rvert d_{i+1}$ for all $1\leq i \leq n-1$ such that $\{d_1\beta_1, \ldots, d_n\beta_n\}$ is an integral basis for $I$.

Then define

$$\Phi : \mathcal{O}_L \rightarrow \mathbb{Z}/d_1\mathbb{Z} \times \ldots \times \mathbb{Z}/d_n\mathbb{Z}$$ $$\sum_{1\leq i \leq n} c_i \beta_i \mapsto (c_1 \ \text{mod} \ d_1, \ldots , c_n \ \text{mod} \ d_n)$$

Then this map is surjective since for $y$ in the image of $\Phi$ since we can write $y = ( y_1 \ \text{mod} \ d_1, \ldots , y_n \ \text{mod} \ d_n)$ for some integers $y_1, \ldots , y_n$. Then the the element$$\sum_{1\leq i \leq n} y_i \beta_i= x\in \mathcal{O}_L$$ is exactly the element such that $\Phi(x)=y$.

The ideal $I$ is in the kernel of this map since for any $x\in I$, $x=\sum_{1\leq i \leq n} x_i d_i\beta_i$ for integers $x_1, \ldots , x_n$. Then $\Phi(x)=(x_1 d_1 \ \text{mod} \ d_1 , \ldots , x_n d_n \ \text{mod} \ d_n)= (0 \ \text{mod} \ d_1, \ldots , 0 \ \text{mod} \ d_n)$

Hence $I\subseteq Ker \Phi$. Then if $x \in Ker\Phi$, we can write $$\Phi(x)=\Phi\left(\sum_{1\leq i \leq n} c_i \beta_i\right) = (c_1 \ \text{mod} \ d_1, \ldots , c_n \ \text{mod} \ d_n)=(0 \ \text{mod} \ d_1, \ldots , 0 \ \text{mod} \ d_n) $$ for integers $c_i$.

But this can only happen if $d_i \rvert c_i$ for all $1\leq i \leq n$. Then $c_i=k_i d_i$ for some integer $k_i$ for all $1\leq i \leq n$. Thus $x=\sum_{1\leq i \leq n} c_i \beta_i=\sum_{1\leq i \leq n} k_i d_i \beta_i \in I$. Thus $Ker \Phi \subseteq I$ and thus $Ker\Phi = I$.

Now I want to show that $\Phi$ is a ring homomorphism. We can easily obtain that $\Phi(a+b)=\Phi(a)+\Phi(b)$. The thing I'm stuck on is showing that $\Phi(ab)=\Phi(a)\Phi(b)$ and $\Phi(1_{\mathcal{O}_L})=(1 \ \text{mod} \ d_1, \ldots , 1 \ \text{mod} \ d_n)$.

If I could show these two things I can use the first Isomorphism theorem and state that

$$\mathcal{O}_L/I \cong \mathbb{Z}/d_1\mathbb{Z} \times \ldots \times \mathbb{Z}/d_n\mathbb{Z}$$

and hence $\rvert \mathcal{O}_L/I \rvert = \rvert \mathbb{Z}/d_1\mathbb{Z}\times \ldots \times \mathbb{Z}/d_n\mathbb{Z} \rvert = d_1d_2\ldots d_n<\infty$

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  • $\begingroup$ If anyone can point me in the direction of how to prove those two statements it would be greatly appreciated. $\endgroup$ – Arren May 10 '18 at 3:24
  • $\begingroup$ Do you know factorization of ideals into products of prime ideals? $\endgroup$ – mathematics2x2life May 10 '18 at 3:29
  • $\begingroup$ Yes, but a later part of the exercise is to prove that $\mathbb{O}_L$ is Dedekind so I don't think I can use the fact ideals factorize into products of prime ideals. $\endgroup$ – Arren May 10 '18 at 3:45
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    $\begingroup$ Do you really need to show that $\Phi$ is a ring homomorphism to finish your proof? It is already a homomorphism of abelian groups, so the First Isomorphism Theorem for groups applies. You're just trying to prove something about cardinality after all. $\endgroup$ – André 3000 May 10 '18 at 4:20
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    $\begingroup$ That works... Sigh. I was told as a "hint" to show it's a ring homomorphism and I looked at it too narrow mindedly. So I would say it's a group homomorphism with the group under the operation of addition of elements in $\mathcal{O}_L$ and the group under coordinate wise addition of elements in the Cartesian product. Hence first isomorphism theorem applies and the result follows. Thanks. $\endgroup$ – Arren May 10 '18 at 5:08

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