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Edit: I see my mistake now (murphy), w(X + Y) = min(w(X), w(Y)) only when the valuations are not the same.

I have the following algebraic problem, which I encountred after thinking about blowups.

My problem is as follows:

let $k$ be a field, and let $k(X, Y)$ be the field or rational functions over $k$ in $X,Y$. letting $S_1 = S_1(X,Y) = X + Y$ and $S_2 = S_2(X,Y) = X Y$ be the symmetric polynomials in $X,Y$. we can form the subfield $k(S_1, S_2) \subset k(X,Y)$. Now, let $R = k(\frac{S_2}{S_1})[S_1]_\mathfrak{p} \subset k(S_1,S_2)$ where $ \mathfrak{p} = (S_1)$ is the ideal generated by $S_1$. (i.e. $R$ is the localization of $k(\frac{S_2}{S_1})[S_1]$ at $\mathfrak{p}$) and let $A$ be its integral clousre inside $k(X,Y)$. (elements of $A$ are elements of $k(X,Y)$ which are integral over $R$) we have the following image: $\require{AMScd}$ \begin{CD} A @>>> k(X,Y)\\ @AAA @AAA\\ R @>>> k(S_1, S_2) \end{CD}

Now, I have a good reason to believe that:

  1. R is a DVR, with uniformizator $S_1$ (and also $S_2$)
  2. A is a DVR - as the integral closure of a DVR inside a finite (even galois) field extension.(with valuation $w$ that extends the valuation of $R$

$X, Y$ are roots of the polynomial $(T-X)(T-Y) = T^2 - S_1T + S_2$ which is integral over $R$, hence $X,Y$ are in $A$. since $S_1$ and $S_2$ differ by a unit, they share the same valuation under $w$ and we get the following set of equalities:

$min(w(X), W(Y)) = w(S_1) = w(S_2) = w(X) + w(Y)$ which implies either that $w(X) = w(Y) = 0$ or that one of them is $-\infty$ both are impossible.

What am I missing?

I encountred the problem after thinking about the following picture: let $C = A_k^1$. we have a morphism $C^2 \rightarrow C^{(2)}$ where $C^{(2)}$ is the symmetric product.

blowing up the symmetric product at the point $P=(0,0)$ we get the following diagram:

$\require{AMScd}$ \begin{CD} Z @>>> Bl_{(0,0)}(C^{(2)})\\ @VVV @VVV\\ P @>>> C^{(2)} \end{CD}

the local ring at the generic point of $Z$ is $R$, which lives inside the function field of $C^{(2)}$. (this is also why I believe it is indeed a DVR, because $Z$ is of codim 1) and A is the integral closure inside the function field of $C^{2}$

The above is obviously a contradiction, and after thinking for a while, I can't find my mistake, and would love to get some help :)

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1 Answer 1

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$w(S_2)=w(S_1) = 1$

Both $X,Y$ are roots of $T^2-S_1 T+u S_1$ with $u=S_2/S_1$, this polynomial is Eisenstein at $(S_1)$ so $w(X)=w(Y)=1/2$

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