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I am doing Marcus problem 21 (b) of chapter 3. The setup for this problem is given in problem 20:

Setup: Let $L/K$ be a finite extension of algebraic number fields. Write $R = \mathcal{O}_K$ and $S = \mathcal{O}_L$. Fix a prime $P$ of $R$ and let $Q_1,\ldots,Q_r$ be the distinct primes of $S$ that live above $P$. For each $Q_i$ fix a subset $B_i \subset S$ corresponding to a basis for $S/Q_i$ over $R/P$. For each $i=1,\ldots,r$ and each $j = 1,\ldots,e_i$ fix an element $$\alpha_{ij} \in (Q_i^{j-1} - Q_i^j) \cap \left( \cap_{h \neq i} Q_h^{e_h} \right). $$ Such an element exists because firstly unique factorisation guarantees that for each $i,j$ we can choose an $x_{ij} \in Q_i^{j-1} - Q_i^j$. We then know that the following system of equations has a solution by the Chinese Remainder Theorem: $$ \begin{eqnarray*} x &\equiv& \alpha_{ij} \pmod{Q_i^j} \\ x &\equiv& 0 \pmod{Q_h^{e_h}} \hspace{2mm} (\text{as $h$ runs through all indices not equal to $i$}) \end{eqnarray*}$$ We can now set $x_{ij}$ to be equal to a solution of this system. Now consider the $n = \sum_{i=1}^r e_if_i$ elements $\alpha_{ij}\beta_{il}$, where $\beta_{il} \in B_i$, $1 \leq i\leq r$,$1 \leq j \leq e_i$ and $1 \leq l\leq f_i$. I have proven that these $n$ elements are independent mod $P$, that is to say that their images in $S/PS$ are linearly independent over $R/P$.

Problem 21(b): Call the $n$ elements $\alpha_{ij}\beta_i$ as constructed above $\alpha_1,\ldots,\alpha_n$ with now $K = \Bbb{Q}$. Except we replace $Q_h^{e_h}$ by $Q_h^N$ where $N$ is some large number to be decided later. How do I show that $$p^k | \textrm{disc}(\alpha_1,\ldots,\alpha_n)$$ where $k = \sum_{i=1}^r (e_i - 1)f_i$ ? I have been stuck on this problem for some days now and I can't seem to make any progress on it. What confuses me is what does he mean by $N$ is some large number to be decided later?

Any help is appreciated, but please do not post complete solutions. Thanks.

Edit:* User Sanchez has asked for a definition of the $\beta_{il}$ which I will now define: Recall that for each prime ideal $Q_i$ lying over a prime $p \in \Bbb{Z}$, we have that $S/Q_i$ is an $f_i$ dimensional vector space over $\Bbb{Z}/p\Bbb{Z}$. Let $B_i$ be a subset of $S$ consisting of $f_i$ elements such that the image of each element of $B_i$ under the canonical homomorphism $\pi : S \to S/Q_i$ is a basis element of $S/Q_i$.

Then our $\beta_{i1},\ldots, \beta_{if_i}$ are defined to be the $f_i$ elements in $B_i$.

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The statement will be easier to follow if you suffix the $\beta$'s. If I understand, there are $f_i$ elements in $B_i$, say $\beta_{ik}$ for $1\le k \le f_i$ and then the $n$ elements $\alpha_1,\dots,\alpha_n$ are $\alpha_{ij}\beta_{ik}$ for $1\le i\le r$, $1\le j\le e_i$ and $1\le k \le f_i$. In addition I think the $x \equiv \alpha_{ij} \pmod{Q_i^{e_i}}$ should read $x \equiv 1 \pmod{Q_i^{e_i}}$. If it is so, then the "large number $N$" means possibly: $N \ge e_i$ for every $i$ as you are then free to chose $\alpha_{ij}$ with the stronger constraint $\alpha_{ij} \equiv 0 \pmod{Q_h^N}$. –  Esteban Crespi Jan 4 '13 at 23:27
@EstebanCrespi Thanks for your reply. Yes it is correct that there are $f_i$ elements in $B_i$, $f_i$ being the dimension of $S/Q_i$ over $R/P$. Oh sorry, but looking at my congruences above the condition that $x \equiv \alpha_{ij} \pmod{Q_i^{e_i}}$ should be replaced with $x \equiv \alpha_{ij} \pmod{Q_i^j}$. –  fpqc Jan 5 '13 at 1:30
I think Esteban Crespi is right. Marcus tells us to consider the $n$ elements $\alpha_{ij} \beta$ where $\beta \in B_i, 1 \leq i \leq r, 1 \leq j \leq e_i$. For each $i$ we need to consider the $f_i$ elements from $B_i$. Using only one fixed $b_i \in B_i$ we would only get $ \sum e_i \neq n$ elements in total. (I have the book in front of me) –  Hans Giebenrath Jan 5 '13 at 19:17
@HansGiebenrath I agree with you. I see where my mistake lies before. Assuming this, do you know how to proceed with the problem? Thanks. –  fpqc Jan 6 '13 at 3:41
Dear BenjaLim, I don't know how to solve this problem. But I would try to argue as in the proof of Theorem 24 taking more care of the valuations of the involved elements. But I don't know if this gets you closer to a solution or if you already have considered this. Unfortunately I have not enough time to work it out myself. –  Hans Giebenrath Jan 6 '13 at 8:36
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2 Answers

up vote 2 down vote accepted

Let us work with a fixed Galois closure $M$ of $L/K$, because then discriminant can be calculated via elements of $Gal(M/K)$. The key would be to calculate the valuations of the discriminant in terms of $R_{it}$, a generic prime lying over $p$. The answer will show why a large $N$ is important, although I will not find the optimal value of $N$.

Some terminologies

$\sigma_1,\cdots,\sigma_n$ are the $K$-embeddings of $L$ into $M$. They can be extended to elements of $Gal(M/K)$, up to an element of $Gal(M/L)$. We will denote the extension also by $\sigma_i$ should there be no confusion.

$R_{it}$ are primes of $M$ lying above $Q_i$ in $L$, where $1 \leq i \leq r$. For convenience I will not list the range of $t$.

$e_i,f_i$ are the ramification index/residue field degree for $Q_i/p$.

$e,f$ are the ramification index/residue field degree for $R_{it}/p$. The values are the same for any $i,t$ since $M/K$ is Galois. It then follows that $e/e_i, f/f_i$ is the ramification index/residue field degree for $R_{it}/Q_i$.

$A = (\sigma_i(\alpha_j))$. The discriminant $disc_{L/K}(\alpha_1,\cdots,\alpha_n) = (det A)^2$.

$B = (v_{R_{11}} (\sigma_i(\alpha_j)))$, where $v_{R_{11}}$ is the valuation for $R_{11}$ in $M$.

By definition, we have that $v_{Q_i} (\alpha_{ij}) = j-1$, so for any $t$, $$v_{R_{it}} (\alpha_{ij}) = \frac{e}{e_i} (j-1)$$ and for $k \neq i$, $v_{Q_k} (\alpha_{ij}) \ge N$ so for any $t$, $$v_{R_{kt}} (\alpha_{ij}) \ge \frac{e}{e_i}N \ge N \hspace{5mm}(*)$$

This gives $$v_{R_{it}}(\alpha_{ij} \beta_{il}) = \frac{e}{e_i}(j-1)$$ and for $k \neq i$, $$v_{R_{kt}} (\alpha_{ij} \beta_{il}) \ge \frac{e}{e_i}N \ge N \hspace{5mm} (**)$$

Here RHS of double star just follows by RHS of (*), and the precise value is not important as long as we know it’s large.

Now look at the matrix $A = (\sigma_i (\alpha_j))$ - we try to look at the valuations of the discriminant. Without loss of generality let's work with $R_{11}$. Arrange the embeddings and the $\alpha$ as follows. For the embeddings, it can be extended to an element of $Gal(M/K)$, and any such extension is off by an element of $Gal(M/L)$. List first those embeddings whose extension sends $R_{1t}$ to $R_{11}$ for some $t$ (note that this property is independent of the extension you choose), and the first element being the inclusion of $L$ into $M$. Then we list those that sends $R_{2t}$ to $R_{11}$ for some $t$ and so on. For the elements, list $\alpha_{1j}\beta_{1l}$ first, in the order you want. Then those of the form $\alpha_{2j}\beta_{2l}$, and so forth.

We wish to calculate $v_{R_{11}} (det A)$. Consider the matrix $B$ where each element is replaced by its value under $v_{Q_1}$, for this is good enough if all we care is to get a lower bound for $v_{R_{11}}(det A)$.

Look at the first diagonal block, i.e. the upper left block, of which on the embedding side it contains those whose extension sends $R_{1t}$ to $R_{11}$ for some $t$, and on the element side things are of the form $\alpha_{1j}\beta_{1l}$. Using the display equations above, it’s clear that modulo some rearrangement, the first row of the block (remember that the first automorphism listed is the inclusion) is $$\begin{bmatrix}\frac{e}{e_1}(1-1) & \cdots & \frac{e}{e_1}(1-1) & \cdots & \frac{e}{e_1}(e_1-1)\end{bmatrix} = \begin{bmatrix}0 & \cdots & 0 & \cdots &\frac{e}{e_1}(e_1 - 1)\end{bmatrix}$$ where each number appears $f_1$ times. The other rows are exactly the same by looking at the display equations way above.

Now look at the $(1,2)$-block, the one immediately to the right of the first diagonal block. On the embedding side it contains those whose extension that sends $R_{2t}$ to $R_{11}$ for some $t$, and on the element side it has $\alpha_{1j}\beta_{1l}$. Let $\sigma(R_{2t}) = R_{11}$. (For simplicity, we will denote an extension of an embedding $\sigma$ by $\sigma$ too) Then $$v_{R_{11}} (\sigma(x)) = v_{\sigma^{-1}(R_{11})} (x) = v_{R_{2t}}(x)$$ for any $x$. In particular it means that for a typical element in this block, we are looking at $v_{R_{2t}} (\alpha_{1j}\beta_{1l})$, where we know is large by (**).

Try to convince yourself that each other diagonal/off-diagonal block behaves similarly as the first diagonal/$(1,2)$-block. This means that if you only focus on valuations, AND IF THE LOWER BOUND YOU HAVE FROM (**) IS LARGE ENOUGH, then $B$ looks something like (for simplicity I will use $L$ to mean a generic large number)

$$\begin{bmatrix} 0 & \cdots & 0 & \cdots & \frac{e}{e_1}(e_1 -1) & L & \cdots & L & \cdots & L & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & L & \cdots & L & \cdots & L & \cdots\\ 0 & \cdots & 0 & \cdots & \frac{e}{e_1}(e_1 -1) & L & \cdots & L & \cdots & L & \cdots \\ L & \cdots & L & \cdots & L & * & * & * & \cdots & L & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots & \cdots & \cdots \\ \end{bmatrix}$$

So you can see that in the first batch of columns of $A$, you can pull out a factor of $$ R_{11}^{\frac{e}{e_1}f_1 ( 1 + \cdots + e_1-1)} = R_{11}^{ef_1(e_1-1)/2}$$ Doing the same for the next batch of columns and so forth, we see that in the original matrix $A$, one can pull out a factor of $$R_{11}^{e\sum_{i=1}^r f_i(e_i-1)/2}$$

Now $$v_{R_{11}} (disc (\alpha_1,\cdots,\alpha_n)) = 2 v_{R_{11}} (A) \ge e \sum_{i=1}^r f_i(e_i-1)$$ Therefore $$v_p (disc (\alpha_1,\cdots,\alpha_n)) = \frac{1}{e}v_{R_{11}} (disc (\alpha_1,\cdots,\alpha_n)) \ge \sum_i f_i (e_i -1)$$

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I have not read fully your answer yet, but I am so thankful for it! –  fpqc Jan 8 '13 at 5:43
What do you mean by the valuation of an element? –  fpqc Jan 8 '13 at 5:49
@BenjaLim, the maximum power of a prime dividing an element. For example in $\mathbb{Z}$, $v_2(4) = 2$, $v_2(8) = 3$ and so on. –  Sanchez Jan 8 '13 at 5:52
so in the case above the valuation of $\alpha$ is say the maximum power of a prime dividing the ideal generated by that element yes? In that case then the valuation is also the maximum power of the prime involved that contains the element (since in a Dedekind domain, to divide is to contain). –  fpqc Jan 8 '13 at 5:53
@BenjaLim, remember that determinant is linear in columns, so if the whole column is divisible by $Q_1$, we know that the determinant is divisible by $Q_1$. If there is another column (say in the next batch) so that the whole column is divisible by $Q_1^2$, then together with the previous column that contributes $Q_1$ to the determinant, we know that the determinant is at least divisible by $Q_1^3$. Does it make sense? –  Sanchez Jan 8 '13 at 7:10
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This is not really an answer to my question above but more of a long comment of an approach on how to solve it. Let $d = \text{disc}(\alpha_1,\ldots,\alpha_n)$. Consider the principal ideal generated by $d$ in $L$, denoted by $(d)$. $L$ is now a finite extension of $\Bbb{Q}$ of degree $n$. To show that $p^{\sum(e_i-1)f_i}$ divides $d$, it is enough to show that for every $i$, we have

$$(d) \subseteq Q_i^{(e_i-1)n}.$$

We know that in a Dedekind domain to contain is to divide so that if for every $i$ we have such a containment above, then

$$\prod_{i=1}^r Q_i^{(e_i - 1)n} | (d).$$

Taking ideal norms, (noting that the norm of $Q_i$ is $p^{f_i}$ ) we get that

$$p^{n\sum f_i(e_i-1)} | d^n$$

and so $p^{\sum f_i(e_i-1)} | d$. I then guess that in the problem above we need to set $N$ to be equal to say $n \times \max_{1 \leq i\leq r} (e_i-1)$ but even proving that we have the containment above for all $i$ is not proving so easy.

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