most recent 30 from http://mathoverflow.net 2013-05-23T05:11:29Z http://mathoverflow.net/feeds/question/47499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47499/multiplicity-under-specialization vu viet 2010-11-27T11:03:39Z 2010-12-01T06:16:02Z

<p>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 $r ?$ </p>

http://mathoverflow.net/questions/47499/multiplicity-under-specialization/47873#47873 vu viet 2010-12-01T06:16:02Z 2010-12-01T06:16:02Z

<p>After a few days thinking about this question, I found out the answer. I write it down here, it might help someone get trouble like me.</p> <p>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 $Q$. </p> <p>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</p> <p>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 $F(Y)$.</p> <p>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.</p>