Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $K\subseteq L$, where $L$ is a finite number field extension of $K$, we consider $p\subset R_K$ and $p'\subset R_L$ where $p'$ lies over $p$, where $R_K$ is the ring of integers of $K$ and $R_L$ defined likewise. Then valuations on $K$ and $L$ are associated with the primes of the fields, so there is a valuation associated with $p$ and $p'$.

My question is how would the way $p$ behave in $L$ (i.e. whether it is inert, split or ramified) affect the relation between $v_{p}(x)$ in $K$ and $v_{p'}(x)$ in $L$?

For example, if $L$ is a quadratic extension of $K$, I think that if:

  • $p$ is inert in $L$, then $v_p(x)=v_{p'}(x)$ (Note that this means $v_{p}(x)$ in $K$ is equal to $v_{p'}(x)$ in $L$).
  • $p$ splits in $L$, so that $pR_L=p'p''$, then $v_p(x)=v_{p'}(x)+v_{p''}(x)$.
  • $p$ ramifies in $L$, so that $pR_L=p'p'$, then $v_p(x)=2v_{p'}(x)$.

I'm would like to know how this is generalised to general finite extensions $L$ of $K$ using inertia degree and a proof or a reference to something containing a proof would be much appreciated. Thank you!

share|improve this question
1  
Are $K$ and $L$ arbitrary fields? Number fields? Local fields? What is $R_K$? –  Hans Giebenrath Jul 27 '13 at 15:32
    
@HansGiebenrath I've updated the question to correct that ambiguity. $K$ and $L$ are number fields and $R_K$ and $R_L$ are the ring of integers. –  BlackAdder Jul 28 '13 at 9:54
    
Your second conjecture about quadratic extensions doesn't work out. Consider the behavior of some prime p≡1(mod4) over Q[i]. –  Kevin Carlson Jul 28 '13 at 11:53
    
I am still puzzled by this set up. Why do you use prime elements instead of prime ideals? Why does there exist a factorization of $p$ into prime elements in $R_L$? I think we should be looking at prime ideals instead of prime elements since the involved rings of integers are not necessarily UFDs (or is this an assumption)? –  Hans Giebenrath Jul 28 '13 at 13:33
    
@HansGiebenrath You were right. I definitely meant prime ideals there, but perhaps it wasn't clear enough. I've edited the question to reflect that, hopefully that won't be anymore confusion. –  BlackAdder Jul 28 '13 at 14:08

1 Answer 1

up vote 2 down vote accepted

Let us first consider your quadratic case. Note that the valuations can be obtained from the unique prime ideal factorization, i.e., for $x$ in $K$ we have $$ xR_K = \prod_{\mathfrak p \subseteq R_K} \mathfrak p^{v_\mathfrak p(x)}, $$ where as usual $xR_K$ denotes the principal (fractional) ideal of $K$ generated by $x$ and the product runs over all nonzero prime ideals of $R_K$. Since we are only interested in the valuation at a fixed prime ideal $\mathfrak p$, let us write $$ xR_K = \mathfrak a \cdot \mathfrak p^{v_\mathfrak p(x)}, $$ where $\mathfrak a$ is an fractional ideal of $K$. In order to obtain the valuations in the bigger field $L$ we need to consider $xR_L$. Thus we need to consider $$ xR_L = (\mathfrak a R_L) \cdot (\mathfrak p^{v_\mathfrak p(x)} R_L) = \mathfrak A \cdot (\mathfrak p R_L)^{v_{\mathfrak p}(x)}.$$ Now we can consider the three cases:

  1. $\mathfrak p$ is inert, $\mathfrak pR_L = \mathfrak P$. Then plugging in we get $$ xR_L = \mathfrak A \cdot \mathfrak P^{v_\mathfrak p(x)}. $$ Note that as $\mathfrak a$ has nothing in common with $\mathfrak p$, the ideal $\mathfrak A$ has nothing in common with $\mathfrak P$. In particular $v_\mathfrak p(x)$ is the exponent of $\mathfrak P$ in the prime ideal decomposition of $xR_L$, i.e., $$ v_\mathfrak P(x) = v_\mathfrak p(x). $$
  2. $\mathfrak p$ splits, $\mathfrak pR_L = \mathfrak P_1 \mathfrak P_2$. Again, we get $$ xR_L = \mathfrak A \cdot (\mathfrak P_1 \mathfrak P_2)^{v_\mathfrak p(x)} = \mathfrak A \cdot \mathfrak P_1^{v_\mathfrak p(x)} \mathfrak P_2^{v_\mathfrak v(x)}. $$ With the same argument as in (1) we get $$ v_\mathfrak p(x) = v_{\mathfrak P_1}(x) = v_{\mathfrak P_2}(x).$$ (This implies $v_\mathfrak p(x) = (v_{\mathfrak P_1}(x) + v_{\mathfrak P_2}(x))/2$.)
  3. $\mathfrak p$ ramifies, $\mathfrak p = \mathfrak P^2$. Now something new happens. We get $$ xR_L = \mathfrak A \cdot (\mathfrak P^2)^{v_\mathfrak p(x)} = \mathfrak A \cdot \mathfrak P^{2v_\mathfrak p(x)}. $$ We conclude $$ v_\mathfrak P(x) = 2 v_\mathfrak p(x),\,\text{i.e.,}\quad v_\mathfrak p(x) = \frac{v_\mathfrak P(x)} 2.$$

I hope this shows you how the inertia degree influences extensions. Moreover you should be able to describe the situation for arbitrary number field extensions $L|K$ where a prime $\mathfrak p$ of $K$ decomposes as $$ \mathfrak p R_L = \prod_{i=1}^g \mathfrak P_i^{e_i} $$ in $L$. I think every book titled "algebraic number theory" should contain this stuff more or less explicitly.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.