Problem:
Let $f(X)$ be an irreducible polynomial over $F$ of degree n, and let $E$ be a field extenfion of F with $[E : F]=m$. If $gcd(m,n)=1$, show that $f$ is irreducible over $E$.
Answer:
Let $f_1$ be an irreducible factor of $f$ in $E[X]$, and let $(L,a)$ be a stem field for $f_1$ over $E$. Then m divides $[L : F]$ because $E \subset L$. But $f(a)=0$, and so $(F[a],a)$ is a stem field for $f$ over $F$, which implies that $[F[a] : F]=n$. Now n divides $[L : F]$ because $F[a] \subset L$. We deduce that $[L : F]=mn$ and $[L:E]=n$. But $[L : E]=deg(f_1)$ and so $f_1=f$
My questions:
1.what does it mean for example $(L,a)$ to be a stem field for $f_1$ over $E$ (the part "over" is not clear however I would appreciate full definition because I am not sure if I fully understand it from my textbook.
2."let $(L,a)$ be a stem field for $f_1$ over $E$", why is it always possible? I mean $a$ has to be such that $f_1(a)=0$, why such $a$ exist.
3.Why $f_1=f$, we only know $deg(f_1)=deg(f)$
(1) I have never seen the term 'stem field' before, but from the usage, a stem field $(L,a)$ for the irreducible polynomial $f_1\in E[X]$ is a field extension $L/E$ in which $f_1$ has $a$ as a root, and $a$ does not lie in any proper intermediate field of $L/E$ (equivalently $L/E$ is generated by $a$). It is 'over $E$' since we are starting with $f_1\in E[X]$ and are only considering field extensions of $E$.
As an example, $f=X^3-2\in\mathbb Q[X]$ has a root $\sqrt[3]2\in\mathbb R$, but $\mathbb R$ contains many other elements, even ones transcendental over $\mathbb Q$. The subfield $L:=\mathbb Q(\sqrt[3]2)=\{a+b\sqrt[3]2+c\sqrt[3]4\mid a,b,c\in\mathbb Q\}$ of $\mathbb R$ is the smallest subfield containing $\sqrt[3]2$, so is generated by $\sqrt[3]2$, and $(L,\sqrt[3]2)$ is a stem field for $f$ over $\mathbb Q$.
As another example, if we adjoin $i=\sqrt{-1}$, then we have the subfield $M:=\mathbb Q(\sqrt[3]2,i)=L(i)$, the smallest subfield of $\mathbb C$ containing both $\sqrt[3]2$ and $i$. Its elements are those of the form $x+iy$ with $x,y\in L=\mathbb Q(\sqrt[3]2)$. Now obviously $f$ has a root in $M$, but $(M,\sqrt[3]2)$ is not a stem field since the root lies in the proper subfield $L\subset M$. On the other hand, if we consider $f$ as a polynomial in $\mathbb Q(i)[X]$, then it is still irreducible, and now $(M,\sqrt[3]2)$ is a stem field for $f$ over $\mathbb Q(i)$, since there are no proper subfields between $\mathbb Q(i)$ and $M$, so $M$ is generated by $\sqrt[3]2$ over $\mathbb Q(i)$.
I hope this helps explain how this notion is necessarily relative to the base field.
(2) If $f\in E[X]$ is irreducible, then the quotient ring $L=E[X]/(f)$ is a field, and the image $\alpha$ of $X$ in $L$ is a root of $f$ (by construction). If we further identify $E$ with its image in $L$, then $(L,\alpha)$ is a stem field for $f$ over $E$.
(3) $f_1$ is an irreducible factor of $f$ over $E$, so $f=f_1f_2\in E[X]$. If now $\deg f_1=\deg f$, then necessarily $\deg f_2=0$, so $f_2$ is a unit, and $f$ is irreducible over $E$. Often, when working over a field, one restricts to monic polynomials, having leading coefficient 1 (so something like $X^3-2$, but not $2X^3-4$). In this case we can further deduce that $f_2=1$ and $f=f_1$.