# How to prove a splitting field $\Sigma$ for $f$ over $K$ can be one for $L$ in the following case?

Q.If $\Sigma$ is the splitting field for $f$ over $K$ and $K\subseteq L \subseteq \Sigma$, show that $\Sigma$ is the splitting field for $f$ over $L$.

I give up, I have tried for 2 hours and no I don't get even how to solve this. I have seen he exact same question(not mine) in a past post but it did not provide any insight, what is more, it had flawed arguments as you can see.

(Splitting fields for $f$ over field $K$) A subfield $\Sigma$ of $\mathbb{C}$ is a splitting field for the nonzero polynomial $f$ over the subfield $K$ of $\mathbb{C}$ if

1.$K \subseteq \Sigma$

2.$f$ splits over $\Sigma$

3.If $K \subseteq \Sigma' \subseteq \Sigma$ and if $f$ splits over $\Sigma'$ then $\Sigma = \Sigma'$

Now, I know that $\Sigma$ is a splitting field of some $f$ over $K$. That is one information.

Also, $k \subseteq L \subseteq \Sigma$. Information two.

But with this scarce information, how can I draw the conclusion that $\Sigma$ is a splitting field for $f$ over $L$?

I first thought conditions $3$ in the definition is the trick. It seems like $L= \Sigma'$. But I quickly discarded this because it's "IF $f$ splits over $\Sigma"$ then $\Sigma = \Sigma'$" which does not help me in actually proving it splits over $\Sigma'$ or, in our case $L$.

So I tried to go by definition. Namely, show that

1.$L \subseteq \Sigma$

2.$f$ splits over $\Sigma$

3.If $L \subseteq \Sigma' \subseteq \Sigma$ and if $f$ splits over $\Sigma'$ then $\Sigma = \Sigma'$

Well, conditions $1$ and $2$ are satisfied. So all good up to there. But again, conditions $3$ is a pain.

I have tried to argue as follows but am very NOT confident with it.

Suppose that $\Sigma'$ such that $L \subseteq \Sigma' \subseteq \Sigma$ exists. And, if $f$ does not split over $\Sigma'$, we are done.(...Right?)

If $f$ does split over $\Sigma'$, then $\Sigma'= \Sigma$ because $\Sigma$ is the smallest field extension of $K$ such that $f$ splits and $K \subseteq L \subseteq \Sigma' \subseteq \Sigma$. If $\Sigma ' \neq \Sigma$ then $\Sigma'$ would be the smallest field extension of $K$ that $f$ splits and thus a contradiction.

Thus all $3$ conditions have been satisfied and hence shown.

I need more practice with this area of abstract math, but can someone look at this problem and point me in the right direction at least if I am wrong up there? Please help

So $\;\Sigma/K\;$ is the splitting field of $\;f(x)\in K[x]\;$ , and this means in simple words that

(1) $\;\Sigma\;$ contains all the roots of $\;f\;$ , or what is the same : $\;f\;$ splits as a product of linear factors in

$\;\Sigma[x]\;$, and

(2) $\;\Sigma\;$ is the minimal (w.r.t set inclusion) field extension of $\;K\;$ that fulfills (1).

Now, if $\;K\subset L\subset \Sigma\;$ , then clearly also $\;f(x)\in L[x]\;$ and all the roots of this polynomial are in

$\;\Sigma\;$ . If there was a field $\;L\le M\lneqq\Sigma\;$ containing all the roots of $\;f\;$ , then this field would also be

an extension of $\;K\;$ since $\;K\subset L\subset M\;$ and we'd get a contradiction to the minimality of $\;\Sigma\;$ as

the minimal field extension of $\;K\;$ where $\;f\;$ splits in linear factors.

• Thanks for helping out. One question, what does $\;L\le M\rlap{\;\,/}<\Sigma\;$ mean in field theory? I know it means subgroup in groups but is that supposed to mean "a subfield of"? And, so assuming such an $M$ exists for $f(x) \in L[x]$ leads to a contradiction, and thus $\Sigma$ must be the splitting field over $L$ for $f(x)$... yes? – John Trail Feb 29 '16 at 20:25
• @JohnTrail Sorry, that was a typo. I already fixed. It should have been $\;L\le M\lneqq\Sigma\;$, meaning: proper subfield. – DonAntonio Feb 29 '16 at 20:53