$f\in K[t]$ is irreducible if and only if $G$ $(\Gamma(L:K))$ acts transitively on the roots of $f$. Let $K$ be a field, $f\in K[t]$ separable over $K$, $L$ the splitting field of $f$ over $K$, and $G=\Gamma(L:K)$. I need to prove that $f$ is irreducible if and only if $G$ acts transitively on the roots of $f$.   I was able to show one direction, that $f$ irreducible implies $G$ acts transitively on the roots. I'm stuck on the other direction of implication; supposing $G$ acts transitively on the roots of $f$. Any guidance or a point in the right direction would be much appreciated!
$f$ is irreducible $\iff$ $G$ act transitively on the roots Let $F$ be a field and $f\in F[X]$ a separable polynomial. Let $K_f$ be the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$ its Galois group. Show that $f$ is irreducible $\iff$ $G_f$ acts transitively on the roots on $f$. My work $\implies :$ Let $f\in F[x]$ irreducible, $\alpha \in K_f$ a root and $\alpha _1,...,\alpha _n$ the other roots of $f$ (so $K_f=F(\alpha ,\alpha _1,...,\alpha _n)$). What I have to show (I think) is that for all $i$ there is a $\sigma \in G_f$ s.t. $\sigma \alpha =\alpha _i.$ Q1) Is it right ? Q2) Is $f$ the minimal polynomial of $\alpha $ ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since $\alpha $ and $\alpha _1$ have the same minimal polynomial, there is an isomorphism $\sigma :\alpha \longmapsto \alpha _1$. We can observe that $\sigma : F(\alpha )\longrightarrow K_f$. Now $F(\alpha ,\alpha _1)/F(\alpha )$ is algebraic, then we can extend $\sigma $ to a morphism $F(\alpha ,\alpha _1)\longrightarrow K_f$. Now, we can continue like that and extend $\sigma $ as $\psi: K_f\longrightarrow K_f$ s.t. $\psi(\alpha )=\alpha _1$. Q3) The fact that $\psi(\alpha )=\alpha _1$ comes from the fact that $\psi|_{F(\alpha )}=\sigma $, right ? The bijectivity is obvious since it's a field homomorphism and that $\sigma (K_f)=K_f$. Q4) How to conclude that $G_f$ act transitively on the roots ? I don't have particular problem for the converse.
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An  registered a domain name that acted as a kill-switch to the ransomware. I presume the non-existent domain was used by the hackers on their private intranet so they didn't infect themselves.  Next time they will be smarter so don't rely on this current kill-switch. Installing the Microsoft patch, which prevents exploiting a vulnerability in the SMBv1 protocol, is the best method.  On May 14, 2017 Red Hat Linux said they are not affected by "Wanna Cry" ransomware. This might mislead Ubuntu users along with Red Hat, CentOS, ArchLinux and Fedora users. Red Hat supports wine which answers below confirm can be effected. In essence Ubuntu and other Linux distro users googling this issue might be mislead by the Red Hat Linux Support answer .  May 15, 2017 Update. Over the last 48 hours Microsoft released patches called KB4012598 for  to protect against "Wanna Cry" ransomware. These Windows versions are no longer on automatic updates. Although I applied security update MS17-010 on my Windows 8.1 platform yesterday, my old Vista Laptop still needs patch KB4012598 downloaded and manually applied.       Moderator note: This question is not off topic - it asks about whether or not any Linux users need to do any steps for protecting against the risk.      It is perfectly on topic here, because it's relevant to Linux (which Ubuntu is), and it's also relevant for Ubuntu users running Wine or similar compatibility layers, or even VMs on their Ubuntu Linux machines.