1. ## Extension

Theorem. Let $\displaystyle f(x)$ be an irreducible polynomial in $\displaystyle F[x]$. Then all the zeros of $\displaystyle f(x)$ in $\displaystyle \overline{F}$ have the same multiplicity.

Proof. Suppose you have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \to F(\beta)$. You extend this to an isomorphism $\displaystyle \tau: \overline{F} \to \overline{F}$. Then you form the following isomorphism $\displaystyle \tau_x: \overline{F}[x] \to \overline{F}[x]$ where $\displaystyle \tau_x$ leaves $\displaystyle f(x)$ fixed. We are assuming that $\displaystyle \alpha$ and $\displaystyle \beta$ are zeros of $\displaystyle f(x)$ in $\displaystyle \overline{F}$. So then $\displaystyle \tau_{x}((x-\alpha)^v) = (x-\beta)^v$ where $\displaystyle v$ is the multiplicity of $\displaystyle \alpha$ and $\displaystyle \beta$. Why does this imply that the multplicity of $\displaystyle \alpha \leq$ multiplicity of $\displaystyle \beta$?

2. Originally Posted by Sampras
Theorem. Let $\displaystyle f(x)$ be an irreducible polynomial in $\displaystyle F[x]$. Then all the zeros of $\displaystyle f(x)$ in $\displaystyle \overline{F}$ have the same multiplicity.

Proof. Suppose you have the conjugation isomorphism $\displaystyle \psi_{\alpha, \beta}: F(\alpha) \to F(\beta)$. You extend this to an isomorphism $\displaystyle \tau: \overline{F} \to \overline{F}$. Then you form the following isomorphism $\displaystyle \tau_x: \overline{F}[x] \to \overline{F}[x]$ where $\displaystyle \tau_x$ leaves $\displaystyle f(x)$ fixed. We are assuming that $\displaystyle \alpha$ and $\displaystyle \beta$ are zeros of $\displaystyle f(x)$ in $\displaystyle \overline{F}$. So then $\displaystyle \tau_{x}((x-\alpha)^v) = (x-\beta)^v$ where $\displaystyle v$ is the multiplicity of $\displaystyle \alpha$ and $\displaystyle \beta$. Why does this imply that the multplicity of $\displaystyle \alpha \leq$ multiplicity of $\displaystyle \beta$?
If you apply $\displaystyle \tau_x$ to a decomposition $\displaystyle f(x)=C\prod_\rho (X-\rho)^{m_\rho}$, you get another decomposition $\displaystyle f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}$, hence by unicity of the decomposition into irreducible factors in $\displaystyle \overline{F}[x]$, $\displaystyle m_{\tau(\rho)}=m_\rho$ for every root $\displaystyle \rho$, and in particular the multiplicities of $\displaystyle \alpha$ and $\displaystyle \beta$ coincide. Does that look correct?

3. Originally Posted by Laurent
If you apply $\displaystyle \tau_x$ to a decomposition $\displaystyle f(x)=C\prod_\rho (X-\rho)^{m_\rho}$, you get another decomposition $\displaystyle f(x)=\tau_x(f(x))=C\prod_\rho (X-\tau(\rho))^{m_\rho}$, hence by unicity of the decomposition into irreducible factors in $\displaystyle \overline{F}[x]$, $\displaystyle m_{\tau(\rho)}=m_\rho$ for every root $\displaystyle \rho$, and in particular the multiplicities of $\displaystyle \alpha$ and $\displaystyle \beta$ coincide. Does that look correct?
It does look correct, but how would you do it using trichotomy? E.g $\displaystyle \tau_{x}(x-\alpha)^{v} = (x-\beta)^v \implies \text{multiplicity of} \ \alpha \leq \text{multiplicity of} \ \beta$ and $\displaystyle \tau_{x}(x-\beta)^{v} = (x-\alpha)^v \implies \text{multiplicity of} \ \beta \leq \text{multiplicity of} \ \alpha$. Would you use $\displaystyle \psi_{\alpha, \beta}$ and $\displaystyle \psi_{\beta, \alpha}$?