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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!

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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
up vote 3 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.

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