If $F/K$ is normal extension and $f \in K[x]$ irreducible and $f=\prod_{i=1}^{n}g_{i}^{m_{i}}$ in $F[x]$ then $m_{i}=m_{j}$ for all $i,j$

Question

So I have that question:

Let $F/K$ be a normal extension and $f$ irreducible polynomial in $K[x]$ assume that $f=\prod_{i=1}^{n}g_{i}^{m_{i}}$ where $g_i$ is irreducible in $F[x]$ ($m_i \geq 1$) then $\deg(g_i) = \deg(g_j)$ and $m_i = m_j$

So the first part I've done after some time (and also I've found answers to that question here: $f$ is factored into many same degree irreducible polynomials)

but I couldn't figure out why $m_i = m_j$ I know it somehow related to the fact that roots of irreducible polynomial has the same multiplicity in the algebraic closure but maybe the multiplicity of $\alpha_i$ (a root of $g_i$) has multiplicity 2 in $g_i$ and $\alpha_j$ has multiplicity $1$ in $g_j$ and $m_j=2m_i$, I couldn't think of any example and maybe it can't happen at all but I'll be glad for some help to approach that (or a way to give a counter example for the statement)

Thank you all in advance

Answer 1

So one of my friends ask the prof. and that's the reason why $m_i = m_j$ (for every one who is interested in that), I needed to add that $g_i \neq g_j$ for all $i \neq j$ (but I thought it was clear):

So by the proof of the first part that $\deg(g_i) = \deg(g_j)$ we know that there is exist $\sigma \in \text{Aut}_{K}(F,F)$ s.t. $\alpha_i$ a root for $g_i$ is map to $\sigma(\alpha_i)$ which is a root of $g_j$ isomorphism from $F$ to $F$ is irreducible preserving i.e. the natural extension $$\sigma(\sum_{j=1}^{m} a_{j}x^{j}) = \sum_{j=1}^{m} \sigma(a_j)x^{j}$$ send an irreducible polynomial to irreducible polynomial, because if for example $\sigma(g_i)$ is reducible (we know that the natural extension send a polynomial in $F[x]$ to polynomial in $F[x]$ then $$\sigma(g_i) = pq$$ then we get $$g_i = \sigma^{-1}(p) \cdot \sigma^{-1}(q)$$ which implies that $g_i$ is reducible.

Now we know that $\sigma(\alpha_i)$ satisfies $$\sigma(g_i)(\sigma(\alpha_i))=0$$ which implies that $g_j | \sigma(g_i)$ because $g_j$ is unit multiplication of $m_{\sigma(\alpha_i)}$ but $\sigma(g_i)$ is irreducible then we get that $g_j = \sigma(g_i)$

The last part is to note that $$\sigma(f) = f$$ because $\sigma$ is the identity on $K$ and also $$f = \prod_{k=1}^{n}g_{k}^{m_k} = \sigma(\prod_{k=1}^{n}g_{k}^{m_k}) =\prod_{k=1}^{n}\sigma(g_{k})^{m_k}=\prod_\underset{k \neq i}{k=1}^{n}\sigma(g_{k})^{m_k} \cdot g_{j}^{m_{i}}$$ so $$f = \prod_{k=1}^{n}g_{k}^{m_k} = \prod_\underset{k \neq i}{k=1}^{n}\sigma(g_{k})^{m_k} \cdot g_{j}^{m_{i}}$$ and because $F[x]$ is a UFD we get that $m_i = m_j$ for any $i,j$ as required