1. ## Extension

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

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

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

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

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