I'm trying to solve the following problem:
Let $F$ be a field and $q \in F(X)\setminus F$. Show the following:
1) Two coprime polynomials $f, g \in F[X]$ exist, so that $q=f/g$.
2) The polynomial $$a(T):=f(T)-g(T)q \in F(q)[T]$$ is the minimal polynomial of $X$ over $F(q)$. This also shows $$[F(X):F(q)]=\max[\deg(f),\deg(g)].$$
3) $F$ is algebraically closed in $F(X)$.
So I've already shown that $a(T)$ is irreducible, but I still have to show that $a(X)=0$. This doesn't seem too hard, but how am I supposed to do that without knowing $f,g$? Since $[F(X):F(q)]$ is defined as the degree of the minimal polynomial of $X$ with coefficients in $F(q)$ and $a(T)$ is the sum of two polynomials, its degree must be the $\max[\deg(f),\deg(g)]$, right?
Can someone help me out here?
Part 1 follows directly from the definition of $F(X)$ and the fact that $F[X]$ is a PID, so one can express $$ q=\frac{f_0}{g_0}=\frac{f}{g} $$ where $f=f_0/\gcd(f_0,g_0)$ and $g=g_0/\gcd(f_0,g_0)$.
If you have already proved that $a(T)$ is irreducible, then you're done: $$ a(X)=f(X)-g(X)\frac{f(X)}{g(X)}=0 $$ The degree of $a(T)$ is clearly $\max\{\deg f(T),\deg g(T)\}$.
Part 3 is nothing more than showing that, if $q\in F(X)$ is algebraic over $F$, then $q\in F$. Note that, if $q\in F(X)$ is algebraic over $F$, then $[F(q):F]$ is finite. If $q\notin F$, then also $[F(X):F(q)]$ is finite. Then…