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Let $E|F$ be a finite Galois extension and $f(x) \in F[x]$ an irreducible polynomial. Prove that each of the irreducible factors of $f(x)$ in $E[x]$ have the same degree.

An idea: Let $\phi \in Gal(E|F)$ and $f(x)=g_1(x) \ldots g_m(x)$, where $g_i(x) \in E[x]$ is irreducible. Since $\phi$ fixes F, $f(x)=\phi f(x)=\phi g_1(x) \ldots \phi g_m(x)$. Therefore, since factorization is unique $\phi$ permutes the $g_i$'s. Note that $\phi$ preserves the polynomial's degree. So, if we can show that $Gal(E|F)$ acts transitively on $\{g_1(x), \ldots ,g_m(x)\}$, then we are done.

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    $\begingroup$ Jyrki: Why are the roots of $f(x)$ distinct? The extension is separable, but that only means minimal polynomials of elements in E are separable, right? $\endgroup$ – Marco Flores Oct 15 '14 at 6:10
  • $\begingroup$ A good question, Marco. I completely missed the boat this time. I was thinking about the case of a tower of Galois extensions $L/E/F$ with $L$ being the splitting field of $f(x)$ and $L/F$, $E/F$ both Galois. $\endgroup$ – Jyrki Lahtonen Oct 15 '14 at 7:18
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Let $g_1(x)$ be one of the irreducible factors in $E[x]$. Let $H\le G$ be the subgroup of automorphisms that fixes all the coefficients of $g_1$, and let $D=\{\tau_1,\tau_2,\ldots,\tau_k\}$ be a set of representatives of left cosets of $H$.

Consider the polynomial $$ g(x)=\prod_{i=1}^k(\tau_i g_1)(x). $$ Fix an element $\phi\in G$. We have that $\phi D$ is another set of reprensentatives of left cosets of $H$. In other words there is a permutation $\alpha\in S_k$ such that $$ \phi\tau_i=\tau_{\alpha(i)}h_i $$ for all $i=1,2,\ldots,k,$ and some elements $h_i\in H$. Therefore $$ \begin{aligned} (\phi g)(x)&=\prod_{i=1}^k(\phi\tau_i g_1)(x)\\ &=\prod_{i=1}^k(\tau_{\alpha(i)}h_i g_1)(x)\\ &=\prod_{i=1}^k(\tau_{\alpha(i)}g_1)(x)\\ &=g(x). \end{aligned} $$ As $\phi$ was arbitrary this means that $g(x)$ is fixed under all of $G$. Therefore $g(x)\in F[x]$.

As you observed, all the polynomials $\tau_i g_1$ are factors of $f$. As they are distinct, so is their product $g(x)$. But $f(x)$ was assumed to be irreducible, so we must have $g(x)=f(x)$. The claim follows from this the way you outlined.

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Consider the action of $\text{Gal}(E/F)$ on the factors $g_1,\ldots, g_m$ and assume it is not transitive, say no $\sigma$ sends $g_1$ to $g_2$.

Then we have $\{\sigma(g_1) : \sigma\in G\}\cap \{\sigma(g_2) : \sigma\in G\}=\varnothing$, i.e. the orbits $G\cdot g_1, G\cdot g_2$ are disjoint. But then we can find a rational function $q(x)\in E(x)$ so that $g_1|\sigma(q)$ and $g_2|\sigma(q)^{-1}$, for each $\sigma\in G$ thanks to the CRT where division means it divides the numerator and the $^{-1}$ means reciprocal.

However,

$$f\bigg|\prod_\sigma \sigma(q)$$

because $G$ permutes all the roots of $f$ transitively. At the same time, we have

$$f\bigg |\prod_\sigma \sigma(q)^{-1}$$

for the same reason--a blatant impossibility since $q(x)$ can be taken to be reduced.

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