Edit: In the time since I've posted this, the questions seem to have resolved themselves. I would still appreciate feedback on whether the resolutions themselves are correct or not and whether there might be more effective solutions.
This is gonna be a long one. I'm working through Jacobson's proof of Luroth's theorem in "Basic Algebra II" (pg. $522$) and there are several points where I want to check my reasoning. I will now recount the proof in my own words (this is not a copy-paste from Jacobson), with indication where I want to check myself.
Let $F$ be our base field and $t$ transcendental over $F$. Our goal is to show that if $F\subset K\subset F(t)$ is an intermediate field, then $K=F(u)$ for some $u$ transcendental over $F$. Let $v\in K\setminus F$, then by the previous lemma $t$ is algebraic over $F(v)$ and hence over $K$. Let $f(x)=x^n+k_1x^{n-1}+\cdots+k_n,k_i\in K,$ be the minimal polynomial of $t$ over $K$. Since $t$ is transcendental over $K$, some $u:=k_j\notin F$. The goal is to show $K=F(u)$ ($u$ is transcendental over $F$ by the previous lemma). Write $u=g(t)h(t)^{-1}$ for $g(x),h(x)\in F[x],(g,h)=1,$ and $m=\max(\deg g,\deg h)$. Then by the previous lemma, $|F(t):F(u)|=m$ and since $F(t)\supset K\supset F(u)$, we have $m\geq n$ with equality if and only if $K=F(u)$. We will show that $m=n$. Now $t$ is a root of $g(x)-uh(x)\in K[x]$, so we have $$g(x)-uh(x)=q(x)f(x)\tag{1}$$ for some $q(x)\in K[x]$. Since the coefficients $k_i$ of $f$ are rational functions of $t$, there exists a non-zero $c_0(t)\in F[t]$ of least degree such that $c_i(t):=c_0(t)k_i\in F[t],1\leq i\leq n$. Then $$f(t,x):=c_0(t)f(x)=c_0(t)x^n+c_1(t)x^{n-1}+\cdots+c_n(t)\in F[t,x]=F[t][x]$$ is primitive (otherwise $c_0$ wouldn't be the common denominator of least degree). The $x$-degree of $f$ is $n$ (clearly) and since $k_j=g(t)h(t)^{-1}$ with $g,h$ coprime, the $t$-degree of $f$ is $\geq m$.
It is enough to show that the $t$-degree of some $c_i$ is $\geq m$, since in that case the $t$-degree of $f$ must also be $\geq m$ (there's no chance of different $t$-terms cancelling out because the $c_i$ have properly differing $F[x]$ coefficients). We have that $c_j=c_0\cdot\frac{g}{h}$ and since $(g,h)=1$, $h$ must divide $c_0$. If $\deg g=m$, then since $h$ divides $c_0$, we have $\deg c_j=\deg(g\cdot\frac{c_0}{h})\geq\deg(g)=m$, so the $t$-degree of $f$ is $\geq m$. If $\deg g<\deg h=m$, then the degree of $c_0$ must be $\geq m$ (since $h|c_0$) and $c_0$ is the leading coefficient of $f$, so the $t$-degree of $f$ is again $\geq m$. Is this correct?
We substitute the expression for $u$ into $(1)$ and get $$g(x)h(t)-g(t)h(x)=q(x)f(x)=q(x)\frac{1}{c_0(t)}c_0(t)f(x)=q'(x)f(t,x),$$ where $q'\in F(t)[x]$ and the rest is in $F[t,x]$. This is a divisibility problem in $F(t)[x]$ and since $f$ is primitive, it reduces to divisibility in $F[t][x]$, i.e. there is a $q(t,x)\in F[t,x]$ such that $$g(x)h(t)-g(t)h(x)=q(t,x)f(t,x)\tag{2}.$$ The $t$-degree of the RHS is $\geq m$ (since the $t$-degree of $f$ is $\geq m$), while the $t$-degree of the LHS is $\leq m$, therefore the $t$-degree is $m$ and $q(t,x)=q(x)\in F[x]$. We have $$g(x)h(t)-g(t)h(x)=q(x)f(t,x)\tag{3}.$$ The RHS is primitive as a polynomial in $x$ and so is the LHS.
We know $f(t,x)$ is primitive in $F[t][x]$, so $(c_0,\dots,c_1)=(1)$ in $F[t]$. If $q(x)=b_kx^k+\cdots+b_0,b_i\in F,$ then the content ideal of $q(x)f(t,x)$ is $I=(b_0c_n,b_1c_n+b_0c_{n-1},\dots)$. Clearly $c_n\in I$, and $c_{n-1}\in I$, and proceeding inductively $c_i\in I$, so $I=(1)$ and $q(x)f(t,x)$ is primitive in $F[t][x]$. The primitivity of the LHS follows. Is this correct?
By symmetry the LHS of $(3)$ is primitive as a polynomial in $t$ also. Hence $q(x)=q\in F$. Then $f(t,x)$ has the same $x$-degree and $t$-degree, so $m=n$.
The definition of the LHS is symmetric in $t$ and $x$ (up to a minus sign), so if it is primitive in one variable, it must also be primitive in the other. Since the LHS is primitive in $t$, $q(x)$ must be a unit of $F[x]$ (and hence a constant), otherwise the RHS would have non-trivial content. Again by symmetry of definition, the $t$- and $x$- degrees of the LHS coincide, whence the same holds for $f$ and $m=n$. Correct?
