2
$\begingroup$

Let $\mathcal O_K$ be a Dedekind domain and $\mathfrak p$ a prime ideal in $\mathcal O_K$. Assume that we have shown the existence of a fractional ideal $\mathfrak p^{-1}$ such that $\mathfrak{pp^{-1}} = \mathcal O_K$. I want to now show that every ideal $\mathfrak a$ in $\mathcal O_K$ can be written as a product of prime ideals:

Let $\mathfrak a$ be an arbitrary ideal in $\mathcal O_K$. It is contained in a prime/maximal ideal $\mathfrak p$. Define $\mathfrak a_1 = \mathfrak{ap^{-1}}$. Clearly $\mathcal O_K \subset \mathfrak p^{-1} \Rightarrow \mathfrak a \subset \mathfrak a_1$. Also, $\mathfrak a \subset \mathfrak p \Rightarrow \mathfrak{ap^{-1}} \subset \mathcal O_K$ and so $\mathfrak a_1$ is also an ideal in $\mathcal O_K$.

We can repeat this process with $\mathfrak a_1$ instead of $\mathfrak a$ and get an increasing chain of ideals $\mathfrak{a \subset a_1 \subset a_2 \subset\cdots}$ which has to be stationary after a finite number of terms since $\mathcal O_K$ is noetherian. This stationary ideal has to be $\mathcal O_K$ and so we can write $$\mathcal O_K = \mathfrak a_n = \mathfrak {ap_1^{-1}\cdots p_n^{-1}} \Rightarrow \mathfrak a = \mathfrak{p_1p_2\cdots p_n}.$$

Does this proof work? All the proofs I have seen of this first prove that every ideal contains a product of prime ideals and proceed from there (for instance Theorem 3.6 here). That seems unnecessary to me.

$\endgroup$
  • 1
    $\begingroup$ I'm pretty sure your argument works. This is the "slick" approach given in Cassels and Frohlich. $\endgroup$ – D_S Jul 6 '15 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.