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Let $L/K$ be a field extension such that $L$ is a splitting field of $f\in K[X]$, i.e. $f=\prod_{i=1}^{k} (X-u_i)^{n_i}$ for some $u_i\in L$. If we denote the coefficients of $g:=\prod_{i=1}^{k} (X-u_i)$ with $v_0,\ldots,v_k$ is then

1) $L/K(v_0,\ldots,v_k)$ a Galois extension ?

2) Does $\text{Gal}(L/K(v_0,\ldots,v_k))=\text{Gal}(L/K) $ hold ?

I also have to show that $L$ is a splitting field of $g$, but this seems trivial, since $L$ contains the $u_i$'s, so if $f$ split, also $g$ has to split - or am I missing something here ?

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Why is $L/K$ Galois? –  user38268 Dec 9 '12 at 10:48
    
It is trivial that $g$ splits in $L$, but I guess you also must show it doesn't split over any smaller field. –  Gerry Myerson Dec 9 '12 at 11:31
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1. No. Separability may fail, for example if $K$ has char $p$, $u_1, u_2$ are purely inseparable over $K$, then you still have the inseparability issue for $L/K(v_0,\cdots,v_k)$. –  user27126 Dec 9 '12 at 11:34
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2. yes. An automorphism of $L$ fixing $K$ will permute the $u_i$, and since $v_0,\cdots,v_k$ are symmetric polynomials in terms of $u_i$, they would be fixed under any such permutation. –  user27126 Dec 9 '12 at 11:34
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@user47574, do you know about inseparable extensions? What is your definition of Galois extensions? –  user27126 Dec 9 '12 at 21:02

1 Answer 1

Let $E=K(v_1,\dots,v_k)$. The extension $L/E$ is normal, this follows because $L/K$ is normal. Now $L$ is the splitting field of $g$ over $E$. By definition $g$ has no repeated roots so is separable. So the extension is also separable and thereby Galois.

For the second part I assume that $\mathrm{Gal}(L/K)=\mathrm{Aut}(L/K)$. One inclusion is clear so let $\varphi \in \mathrm{Gal}(L/K)$ and we wish to show that $\varphi$ fixes $E$. This follows from observing that $\varphi$ acts on the roots $u_1,\dots,u_k$ and $E$ is generated by the symmetric polynomials in $u_1,\dots,u_k$ so the generators of $E$ are fixed by $\varphi$. Thereby $\varphi \in \mathrm{Gal}(L/E)$.

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