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this is my first question here, I'll apologise in advance for any kind of noob mistakes because I'm aware that I might to do them. I'm solving one assignment for my studies and I can't do anything smart on one part of it. I'm a bit new to field theory since my studies were focused on other things.

Here's the assignment:

Let $E$ be intermediate field of the extension $K\subset F$ and assume that $E$ = $K(u_1,\ldots,u_r)$ where the $u_i$ 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 $E$.

Next, I'll show you my try (I'll write the case for the normed polynomial($a_n=1$):

$\Rightarrow$ We have to show that $F$ is a splitting field of $f$ over $E$, knowing that $F$ is a splitting field of $f$ over $E$. Firstly, we know that $f$ can be written as product of linear factors $a_n(x-u_1)(x-u_2)\ldots(x-u_n)$ where each $u_i$ is in $F$ and $a_n$ is in $K$. Now because of the definition of the intermediate field $E$ and the fact that $F$ is a splitting field of $f$ over $K$, statement holds. Why? Because by the definition, we know that $a_n$ will be in $E$. Now we know that we can write $f$ as a product of linear factors $a_n(x-u_1)(x-u_2)(x-u_3)\ldots(x-u_n)$ where each $u_i$ is in $F$ and $a_n$ is in $E$ now. So basically, we can conclude that $F$ is a splitting field of $f$ over $E$.

$\Leftarrow$ We have to prove converse and this is where I have problem. I sense that I should use minimality of spliting field and the fact that $K$ is generated by some roots of $f$.

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  • $\begingroup$ Because $F$ is the splitting field of $f$ over $E$, it contains all its roots. Because $f$ is also an element of $K[x]$, F is also the splitting field of $f$ over $K$. Isn't this enough? This is an honest question. $\endgroup$ – TPace May 22 '18 at 20:10
  • $\begingroup$ You can take a look on that answer: math.stackexchange.com/questions/24680/splitting-fields $\endgroup$ – Igor Sikora May 22 '18 at 21:35

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