This is a question from Janusz 'Algebraic Number Theory'. Let $R$ be a DVR with maximal ideal $\mathfrak p=\pi R$. Let $K$ be the quotient field of $R$ and $\mid\cdot\mid_{\mathfrak p}$ the non-archimedean absolute value of $K$ associated to $\mathfrak p$. Let $L$ be a finite separable field extension of $K$ and $R'$ the integral closure of $R$ in $L$. We have $\pi R'=\mathfrak P_1^{e_1}...\mathfrak P_g^{e_g}$ with $\mathfrak P_i$ distinct primes in $R'$.

Let $R'_{\mathfrak P_i}$ be the localization of $R'$ with maximal ideal $\tau R'_{\mathfrak P_i}=\mathfrak P_iR'_{\mathfrak P_i}$.

Why is it true that $\pi/1=u\tau^{e_i}$ for some unit $u$ of $R'_{\mathfrak P_i}$ as claimed in the book?

What I have so far: $\pi\in \pi R'$, so $\pi$ is a sum of products of the form $x_1...x_g$ where $x_j\in \mathfrak P_j^{e_j}$ for all $j$. We have $x_j/1\in R'^{\times}_{\mathfrak P_i}$ for all $j\neq i$ and $x_i/1\in \tau^{e_i}R'_{\mathfrak P_i}$. Therefore $x_1...x_g/1$ belongs to $v\tau^{e_i}R'_{\mathfrak P_i}$ for some unit $v$ of $R'_{\mathfrak P_i}$. Am i on the right track?

Many thanks for your help!

  • $\begingroup$ Norms of things of absolute value $1$ are units in $R$. $\endgroup$ – Adam Hughes Jul 18 '14 at 17:33

If $\tau$ generates the maximal ideal in the DVR $R^\prime_{\mathfrak{P}_i}$, then every element in the field of fractions of $R^\prime_{\mathfrak{P}_i}$ (which is the same as the field of fractions of $R^\prime$, i.e. $L$) can be written as $\tau^n u$ with $n\in\mathbf{Z}$ and $u\in (R^\prime_{\mathfrak{P}_i})^\times$. In particular this is true of $\pi\in R\subseteq R^\prime\subseteq R^\prime_{\mathfrak{P}_i}$. As for the precise exponent, we have

$$\pi R_{\mathfrak{P}_i}^\prime=\prod_{j=1}^g\mathfrak{P}_j^{e_j} R^\prime_{\mathfrak{P}_i}=\prod_{j=1}^g (\mathfrak{P}_jR_{\mathfrak{P}_i}^\prime)^{e_j}=(\mathfrak{P}_iR_{\mathfrak{P}_i}^\prime)^{e_i}=\tau^{e_i}R_{\mathfrak{P}_i}^\prime$$

because if $i\neq j$, $\mathfrak{P}_j R_{\mathfrak{P}_i}^\prime=R_{\mathfrak{P}_i}^\prime$ as some element of $\mathfrak{P}_j$ must be a unit in $R_{\mathfrak{P}_i}^\prime$. So $\pi=\tau^{e_i}u$ with $u\in(R_{\mathfrak{P}_i}^\prime)^\times$.

The argument you suggest in "What I have so far" probably isn't the way to go. Note that $x_i\in\mathfrak{P}_i$ does not necessarily implly $x_i\notin\mathfrak{P}_j$, so its image in $R_{\mathfrak{P}_j}^\prime$ need not be a unit.

  • $\begingroup$ Many thanks for your answer- both clear and elegant. It shows that I was definitely on the wrong track! Regards $\endgroup$ – pete m Jul 18 '14 at 18:04

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