I'm trying to make sense of this problem and solution.
Let $K$ be a field, and let $P(X)$ be an irreducible polynomial of degree $n$ with coefficients in $K .$ Let $Q(X)$ be a polynomial with coefficients in $K$. Show that every irreducible factor of $P(Q(X))$ has degree divisible by $n$.
Proof: Let $T(X)$ be an irreducible factor of $P(Q(X)),$ and consider the field $K(\alpha)$ where $\alpha$ is a root of $T(X) .$ Then $Q(\alpha)$ is a root of $P(X),$ so $K(Q(\alpha))$ has degree $n$ over $K .$ Since $K(Q(\alpha))$ is clearly a subfield of $K(\alpha),$ the degree of $K(\alpha)$ over $K$ (which is equal to the degree of $T(X)$) is divisible by $n$.
Question 1: I was taught that $K(\alpha)$ was an "abuse of notation", in that $\alpha := x + K[X]/\langle T(X) \rangle$ and $K(\alpha):= K[X]/\langle T(X) \rangle$. In the content above, $K(\alpha)$ almost seems to be treated as the field of fractions of the ring $K[\alpha]$, where $\alpha$ is to be treated not as $\alpha := x + K[X]/\langle T(X) \rangle$ but as an element in a larger field $L$, where $K \subset L$. Is my proof written down below correct?
Question 2: I do not understand how $\alpha$ being a root of $T(X)$ implies $Q(\alpha)$ being a root of $P(X)$. (I do not know how to justify this rigorously, in the language of $\alpha := x + K[X]/\langle T(X) \rangle$ and $K(\alpha):= K[X]/\langle T(X) \rangle$. If there is a simpler way to view things, please teach me!)
Please consider writing the answers out in detail, I get confused when they are too compact...
For question 1, I think this approach is relevant: As $K[X]/\langle T(X) \rangle$ is a field due to the irreducibility of $T(X)$ in the PID $K[X]$. $\alpha := x + K[X]/\langle T(X) \rangle$ is an element of that field. Clearly $K, \alpha \subset K[X]/\langle T(X) \rangle$, thus $K'\subset K[X]/\langle T(X) \rangle$ where $K'$ is the field of fractions of $K[\alpha]$. Elements of $K[X]/\langle T(X) \rangle$ are a $K$-linear dependence of $\alpha^n$, thus $K'\supset K[X]/\langle T(X) \rangle$.
Partial answers, e.g. only answering question 1, are very welcome!
If $T(\alpha)=0$ then $P(Q(\alpha))=0$ thus $Q(\alpha)$ is a root of $P(X)$. Here $\alpha$ is an element of a splitting field for $f(X)=P(Q(X))\in K[X]$, those splitting fields exist and are all isomorphic.
That a splitting field exists is because we can factorize $f=\prod_j f_j$ in $K_0=K$, pick a non-linear factor, let $K_1=K_0[x_1]/(f_i)$, and factorize $f=\prod_j g_j$ in $K_1$, pick a non-linear factor, let $K_2=K_1[x_2]/(g_j)$, and so on until $f$ splits completely in $K_m$.
That the splitting fields are all isomorphic is because for $b$ in a field containing $K$ and algebraic with $K$-minimal polynomial $f(X)$ then $K(b)\cong K[t]/(f(t))$, then apply again an inductive argument.
That $K(Q(\alpha))$ is a subfield of $K(\alpha)$ is obvious, why do you think $|K|\le |L|$ appears somewhere ? Yes it is the fraction field of $K[Q(\alpha)]$, but since $Q(\alpha)$ is algebraic then $K[Q(\alpha)]=K(Q(\alpha))$.