# multiplicity under specialization

Let $C$ be a projective non-singular curve defined over a field $K$ with the characteristic zero. Let $y,z$ be non-constant rational functions defined $K$ such that $y$ is defined at all poles and zeros of $z$ and gives an injective mapping of this set into the algebraic closure of $K.$ Let $(y_1,y_2,...,y_m)$ be all conjugates of $y$ over $K(z)$ and $r$ be the largest of the multiplicities of the zeros of $z.$ If $(y'_1,...,y'_m)$ is a specialization of $(y_1,y_2,...,y_m)$ over $z \rightarrow 0.$ \ Is this true that the multiplicity of any $y'_k$ in $(y'_1,...,y'_m)$ is less than $r ?$

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Let $\mathcal{o}$ be valuation ring generated by maximal ideal $(z)$ in $K(z)$ and $v$ be the absolute value corresponding to it. Let $K(z)_v$ be the completeness of $K(z)$ with respect to $v$. Let $K(w)$ be the function field of $C$, then it is easy to see that if $\mathcal{o}'$ is an valuation ring extending $\mathcal{o}$ to $K(z)$ then $\mathcal{o}'$ is the valuation ring of a zero point $Q$ of $z$ and its ramification index is the multiplicity of $Q$ in the zero divisor of $z$. Hence, the ramification index of $\mathcal{o}’ \cap K(z,y)$ is less than the multiplicity of $Q$.
Because $y$ defines at zero points of $z$, we see that $y$ is integral over $\mathcal{o}$. Thenfore, the absolute value corresponding to $\mathcal{o}’ \cup K(z,y)$ in $K(y,z)$ is an unramified extension of $v$. Hence, its ramification index equals its local degree. This imlies the local degree is less than the multiplicity of $Q We know that each extension of$\mathcal{o}$to$K(y,z)$is induced by an embedding$\sigma: K(y,z)\rightarrow \overline{K(z)_v}$(the algebraic closure of$K(z)_v$). Let$F(Y)$be the irreducible polynomial of$y$over$K(z)$and$F(Y)=F_1(Y)\cdots F_k(Y)$be its irreducible factorization in$K(z)_v$(the completeness of$K(z)$with respect to$v$). Note that$F_i\not = F_j$by$K$having zero characteristic. Then,$(y_1,\ldots,y_m)$are zeros of$F(Y)$. Consider an extension$\overline{\mathcal{o}}$of$\mathcal{o}$to$\overline{K(z)_v}$. We claim that$y_i,y_j$are mapped to the same image by the canonical place$\phi$of$\bar{\mathcal{o}}$iff they are zero points of the same$F_s(Y)$. Indeed, suppose$\sigma_i, \sigma_j: K(z,y) \rightarrow \overline{K(z)_v}$are embeddings which map$y$to$y_i,y_j$, respectively. Then,$\phi \circ \sigma_i =\phi \circ \sigma_j$as places of$K(y,z)$. Hence, they induce the same absolute value in$K(y,z)$. This follows our claim. Now, we can easily imply the answer of the question, as the multiplicity of$y’_i$in$(y’_1,\ldots,y’_m)$equals degree of some$F_s\$ which equal the local degree.