1
$\begingroup$

Suppose we have K,a field extension of L.Now consider a polynomial f in L[x].Let s be a element of Aut(K|L).Now we know if r is a root of f then s(r)is also a root of f.Now assume this situation that K contains exactly one root of f.So if r is the root does that mean s(r)=r?

My confusion is 'r' might not come from L and s(r) is clearly a root,so what happens when K contains exactly one root of f?

  • 0
    What is so strange about $s\left(r\right) = r$? – 2017-02-06
  • 0
    What you describe implies that if $r$ does not belong to $L$, it is nonetheless fixed by the automorphism $s\in Aut(K|L)$, adopting your notation. This is not terribly unusual, although if $r$ were fixed by *every* $s\in Aut(K|L)$, that would mean the extension $L\subset K$ is *not Galois*. – 2017-02-06
  • 0
    What we are doing basically ,we are looking at a particular type of extension namely which contains exactly one root of f.Now f is arbitrary.But why that information affecting the global behaviour of the group Aut(K|L) neamely it has a fixed point at r? – 2017-02-06
  • 0
    I would not say that $f$ is arbitrary, but rather that $s \in Aut(K|L)$ is arbitrary. Your argument applies to *any* automorphism $s$ to show $s(r)=r$, and therefore the field extension is not Galois if $r\not\in L$. [By definition, a field extension $K:L$ is *Galois* when the subfield of $K$ that is fixed by $Aut(K|L)$ is exactly $L$.] – 2017-02-06

0 Answers 0