# Splitting Fields

This is a problem from Hungerford's book:

Let $E$ be an intermediate field of extension $K\subset F$ and assume that $E=K(u_1, \cdots ,u_r)$ where $u_i$ are (some of the) roots of $f\in K[x]$. Then $F$ is a splitting field of $f$ over $K$ if and only if $F$ is a splitting field of $f$ over E.

This is my attempt.

Let $u_1, \cdots ,v_n$ be the roots of $f$. Then since $F$ is a splitting field of $f$ over $K$, $F=K(v_1, \cdots ,v_n)$ $\implies$ $F=E(v_1, \cdots ,v_n)$. So $F$ is a splitting field of $f$ over $E$ .

Conversely, suppose $v_1, \cdots ,v_n$ are the roots of $f$ in $F$. Then $F=E(v_1, \cdots ,v_n)=K(u_1, \cdots ,u_r,v_1, \cdots, v_n)$. Thus $F$ is a splitting field of $f$ over $K$.

My Question: First I want to know if the proof above is right. The second part of the problem asks to extend the above to splitting fields of arbitrary set of polynomials. This is where I'm lost. I'd like a little help on how to begin.
Thanks.

Update: I think I've got it now. I had some difficulty with the indexing. Thanks.

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There's nothing to do here other than understand and unpack the definition of "splitting field" twice. I recommend you try to do this yourself. If you continue to have difficulty, your next question should be something more elementary about the definition / concept of a splitting field. – Pete L. Clark Mar 2 '11 at 19:32
@Pete: yeah I saw trick after posting the question. Thanks.lol – Nana Mar 2 '11 at 20:02
After two months, perhaps you can start asking questions instead of just copying problems and giving orders? – Arturo Magidin Mar 2 '11 at 20:49
@Arturo: Sorry for the mistake. It wasn't my intention to give orders. It won't happen again. – Nana Mar 2 '11 at 23:05

$\left(\implies\right)$ Let $v_{1},\cdots,v_{n}$ be the roots of $f$ in $F$. Since $F$ is a splitting field of $f$ over $K$, $F=K\left(v_{1,}\cdots,v_{n}\right)$ which implies that $F=E\left(v_{1},\cdots,v_{n}\right)$. So $F$ is a splitting field of $f$ over $E$.
$\left(\Longleftarrow\right)$ Suppose $v_{1},\cdots,v_{n}$ are the roots of $f$ in $F$. Then $F=E\left(v_{1},\cdots,v_{n}\right)$. But each $u_{i}$ is a one of the $v_{j}$, so $F=E\left(v_{1},\cdots,v_{r}\right)=K\left(u_{1},\cdots,u_{r},v_{1},\cdots,v_{n}\right)$.
Thus $F$ is a splitting field of $f$ over $K$.