Show for an irreducible polynomial $f(x) \in F[x]$ of degree $n$, $n$ divides $[E:F]$ where $E/F$ is the splitting field of $f(x)$ Let $F$ be a field. I want to show that for an irreducible polynomial $f(x)$ in $F[x]$ of degree $n$, and for a splitting field $E/F$ of $f(x)$, we have $n \mid [E:F]$. 
Can anyone provide any hints for me to think about this?
Thank you.
 A: We assume familiarity with the result if $F$ is a field, $L$ a finite extension of $F$ containing subfield $K$ such that $F \subseteq K$, then $$[L:F]=[L:K][K:F]$$
If this is not familiar, it is not to difficult to show using the concept of finding a basis for $L$ over $F$.
Now consider the irreducible polynomial $f\in F[X]$, with $\deg(f)=n$ and $E$ the splitting field of $f$ over $F$. By definition, $E$ is the field containing all roots of the polynomial $f$. We wish to construct a field $F(x)$, with $x$ a root of $f$, such that $F\subseteq F(x) \subseteq E$ and $[F(x):F]=\deg(f)$.
Let $\pi:F[X]\rightarrow F[X]/(f)$ be the canonical projection of the coordinate ring onto the quotient ring (i.e. $\forall g\in F[X], g\mapsto g+(f)$). We consider the the following: $L$ contains a root of $f$ and all elements of $F$, $L=F[X]/(f)$ is a field, $[L:F]=\deg(f)$. 
Clearly the quotient ring is indeed a ring. We make the distinction now that $\pi(X)=X+(f)=x$ so that for $g = \sum_{k=0}^n a_k X^k \in F[X]$ we have $\pi(g)=\pi(\sum_{k=0}^n a_kX^k)=\sum_{k=0}^n a_kx^k$ where we have now dropped the notation $(f)$ for convenience. Since $\pi(f)=f(\pi(X))=f(x)=0$, we have $x\in L$ is a root of $f$. 
Also, for every $g\in F[X]$ we have $f\mid g$ or $(f,g)=1$ since $f$ is an irreducible polynomial. In the former case, $g\mapsto 0$. Thus, every nonzero element $g$ of $L$ is realizable as the image of a relatively prime polynomial to $f$. By Bézout's identity there are polynomials $s,t\in F[X]$ such that $1=fs+gt$. Since $x$ is a root of $f$, evaluating the polynomials in this expression at $x$ yields $1=g(x)t(x)$. Thus every nonzero element $g(\pi(X))=g(x)\in L$ has an inverse, showing $L$ is a field.
Finally, we show $[L:F]=\deg(f)$. To do this we show explicitly $\{1,x,x^2,...,x^{\deg(f)-1}\}$ is a basis for $L$. Linear independence is clear since if there existed scalars $a_0,...,a_{\deg(f)-1}$ such that $\sum_{k=0}^{\deg(f)-1} a_kx^k=0$, then $p(X)=\sum_{k=0}^{\deg(f)-1} a_kX^k$ has lesser degree than $f$ and vanishes at $x$. Since $f$ is an irreducible polynomial $f\mid p$ which forces $p(X)=0$ and all the scalars are $0$.
Consider $g\in F[X]$. By Euclidean division we can find $q,r\in F[X]$ such that $g=fq+r$ with $0\leq\deg(r)<\deg(f)$. Then $g(x)=r(x)$ which can be realized via the basis suggested showing that we were not wrong to claim $\{1,x,...,x^{\deg(f)-1}\}$ is a basis for $L$.
Our assertion is then made using the result we assumed familiarity with since, for $E$ a splitting field of $f$, a polynomial of degree $n$, we have $$[E:F]=[E:L][L:F]=[E:L]n$$
in other words $n|[E:F]$.
