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In Qing Liu, Algebraic Geometry and Arithmetic Curves, page 116, exemple 4.1.8, one has $\mathcal{O}_K$ a discrete valuation ring with uniformizing parameter $t$, $P\in\mathcal{O}_K[S]$ an Eisenstein polynomial in $t$ (then irreducible), $\mathcal{O}_L:=\mathcal{O}_K[S]/(P(S))$ (then a integral domain) and with notation $s=\overline{S}$ one has $L=\mathrm{Frac}(\mathcal{O}_L)=K[S]/(P(S))=\bigoplus_{0\leq i\leq n-1} s^i K$.

Then we want to prove that, with $\nu_L\left(\sum_{i=0}^{n-1}\alpha_i s^i\right)=\mathrm{min}\{n\nu_K(\alpha_i)+i,i=0,\ldots,n-1\}$, then $\mathcal{O}_L$ is the discrete valuation ring of $L$.

I'm ok with $\nu_L(x+y)\geq\mathrm{min}\{\nu_L(x),\nu_L(y) \}$.

I'm ok with $\nu_L(x)\geq0\iff x\in\mathcal{O}_L$.

I'm ok with surjectivity onto $\mathbb{Z}$ if I prove the next point.

But for $\nu_L\left(\sum_{i=0}^{n-1}\alpha_i s^i \sum_{j=0}^{n-1}\beta_j s^j\right)=\nu_L\left(\sum_{i=0}^{n-1}\alpha_i s^i\right)+\nu_L\left(\sum_{i=0}^{n-1}\beta_i s^i\right)$ I can't find because as $\nu_L$ is defined only on the form $\sum_{i=0}^{n-1}\alpha_i s^i$ the calcul of $\nu_L\left(\sum_{i=0}^{n-1}\alpha_i s^i \sum_{j=0}^{n-1}\beta_j s^j\right)=\nu_L\left(\sum_k\left(\sum_{i+j=k}\alpha_i+\beta_j\right)s^k\right)$ is too difficult for me (because of the $s^ k$, $k\geq n$).

Maybe with a formula for $\nu_L$ not only with the restricted form $\sum_0^{n-1}\alpha_i s^i$?

Maybe a classical trick for simpllifying the $s^k, k\geq n$?

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Probably one has to find a description of the valuation which does not depend on the basis. Use the minimal polynomial. – Martin Brandenburg Feb 2 '13 at 18:46
If I understand well what is written there, in @QiL's book there is a typo: when define $v_L$ it is written $\max$ instead of $\min$. (Actually there is another one: when define $v_L$ one uses the elements $a_i$ which are already used as being the coefficients of the Eisenstein polynomial.) – user26857 Feb 2 '13 at 21:44
@YACP, yes I have seen this two mistakes (the min instead of max comes from the official errata on his site). – Macadam Feb 3 '13 at 11:06
Okay. I wasn't aware about the existence of an official errata of the book. – user26857 Feb 3 '13 at 16:52

I think it is better to give an equivalent definition of $v_L$: for all $b\in L$, $$ v_L(b)=v_K(N_{L/K}(b)).$$
Then use the multiplicativity of the norm $N_{L/K}$.

Let us prove the equivalence with the definition given in the book. Let $b=\sum_{i\le n-1}\alpha_i s^i\in L$. Suppose that the minimum is reached at $i_0$. Then $$\alpha_is^i=(\alpha_{i_0}s^{i_0})u_is^j, \quad u_i\in \mathcal O_K $$ where $j=i-i_0$ if $i\ge i_0$ and $j=n+(i-i_0)$ if $i<i_0$. So $b=(\alpha_{i_0}s^{i_0})(1+sc)$ with $c\in\mathcal O_L$. As $v_K(N_{L/K}(s))=1$, we are reduced to prove $N_{L/K}(1+sc)$ is a unit. But this is clearly true (either by the very definition of the norm, or compute the norm modulo the maximal ideal of $\mathcal O_K$, or in the separable extension case use the product of the conjugates).

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Did you use somewhere that $P$ is an Eisenstein polynomial or it is enough to be irreducible? – user26857 Feb 4 '13 at 9:40
@YACP: irreducible is not enough. Eisenstein is used here to say $v_K(N_{L/K}(s))=1$. – user18119 Feb 4 '13 at 9:52
I understand the general framework but I have some "details" questions: In the beginning you introduce $v_L(b)$ for $b\in\mathcal{O}_L$. Why don't define on $L$? (and espacially $v_K(N_{L/K}(b))$ has a sense for $b\in L$) It is a general methode to define $v$ on a ring $A=\mathcal{O}_K$ and expand it then on $\mathrm{Frac}(A)=K$? – Macadam Feb 4 '13 at 19:45
Second question: for the writting $\alpha_is^i=(\alpha_{i_0}s^{i_0})u_is^j$, telling that $u_i\in\mathcal{O}_K$ is equivalent to $\frac{\alpha_i}{\alpha_{i_0}}\in\mathcal{O}_K$ or $\frac{\alpha_i}{\alpha_{i_0}\sum_{k\leq n-1} a_ks^k}\in\mathcal{O}_K$ but I don't see why it is true: we have $nv_K(\alpha_i)+i\geq nv_K(\alpha_{i_0})+i_0$ but this not implicate that $v_K(\alpha_i)\geq v_K(\alpha_{i_0})$, and what doing with $\sum_{k\leq n-1} a_k s^k$? – Macadam Feb 4 '13 at 19:49
Last question: for $N_{L/K}(1+sc)\in\mathcal{O}_K^*$: with the definition of norm one has to show that $x\mapsto(1+sc)x$ is a automorphism of $\mathcal{O}_L$ as $\mathcal{O}_K$-module ie $(1+sc)\in\mathcal{O}_L^*$ but I don't find the inverse. With the modulo approach one has to prove that $N_{L/K}(1+sc)\neq0$ but then? And with the conjugates... what are the conjugates? – Macadam Feb 4 '13 at 19:54

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