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Definition: Let $K/F$ be a field extension and let $p(x)\in F[x]$, we say that $K$ is splitting field of $p$ over $F$ if $p$ splits in $K$ and $K$ is generated by $p$'s roots; i.e. if $a_{0},...,a_{n}\in K$ are the roots of $p$ then $K=F(a_{0},...a_{n})$.

What I am trying to understand is this: in my lecture notes it is written that if $K/E$,$E/F$ are field extensions then $K$ is splitting field of $p$ over $F$ iff $K$ is splitting field of $p$ over $E$.

If I assume $K$ is splitting field of $p$ over $F$ then $$\begin{align*}K=F(a_{0,}...,a_{n})\subset E(a_{0,}...,a_{n})\subset K &\implies F(a_{0,}...,a_{n})=E(a_{0,}...,a_{n})\\ &\implies K=E(a_{0,}...,a_{n}). \end{align*}$$

Can someone please help with the other direction ? help is appreciated!

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It isn't true. Take $F = \Bbb Q, E = \Bbb Q(\sqrt{2}), K = \Bbb Q(\sqrt{2}, \sqrt{3})$. Then $K$ is the splitting field of $p(x) = x^2-3$ over $E$, but $\Bbb Q(\sqrt{3}) \neq K$ is the splitting field of $p$ over $\Bbb Q$. –  Brandon Carter Jul 7 '12 at 20:26
    
It is true that the splitting field of $p$ over $F$ is a subextension of $K$ if $K$ is the splitting field of $p$ over $E$. –  Brandon Carter Jul 7 '12 at 20:27
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@Brandon: I guess we both thought of the same minimal counterexample, though you beat me to it by a minute :) I invite you to post your comments as an answer. –  Zev Chonoles Jul 7 '12 at 20:35
    
@Zev: Indeed! Your answer is more detailed and I don't need the reputation, so I'm happy leaving mine here as comments. –  Brandon Carter Jul 7 '12 at 21:12

1 Answer 1

up vote 5 down vote accepted

This is false. Let $F=\mathbb{Q}$, let $E=\mathbb{Q}(\sqrt{2})$, let $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, and let $p=x^2-3\in F[x]$. Then the splitting field for $p$ over $E$ is $K$, but the splitting field for $p$ over $F$ is $\mathbb{Q}(\sqrt{3})\subsetneq K$.

Let's say that all fields under discussion live in an algebraically closed field $L$. Letting $M$ be the unique splitting field for $p$ over $F$ inside $L$, then the splitting field for $p$ over $E$ inside $L$ is equal to $M$ if and only if $ME=M$, which is the case if and only if $E\subseteq M$.

In other words, you'll get the same splitting field for $p$ over $F$ and over $E$ if and only if $E$ were already isomorphic to a subfield of the splitting field for $p$ over $F$. When $E$ does not have that property, there is "extra stuff" in $E$ (for example, $\sqrt{2}\in E$ in the example) that will need to also be contained in the splitting field for $p$ over $E$.

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I don't understand, $K,E$ are both not splitting field of $p$ over $F$ –  Belgi Jul 7 '12 at 20:43
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@Belgi: $K$ is the splitting field of $p$ over $E$, but not over $F$. –  Arturo Magidin Jul 7 '12 at 20:44

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