$F(t)$ as an $F[t]$-algebra and the Weak Nullstellensatz

Question

Sorry if this question has already been answered somewhere, but it's quite hard to find if so, because of the use of the word 'algebra' in the question...

1. In the lead up to a proof of the Weak Nullstellensatz ($E/F$ field extension, $E$ finitely-generated as an $F$-algebra, then $E/F$ finite) there is an alternative proof-sketch given as a sidenote:

   $F(t)$ is not finitely generated as an $F[t]$-algebra (where $t$ is an independent indeterminate), and so if $F(x)$ is a finitely-generated $F[x]$-algebra, then $F[x]\not\cong F[t]$, and so $x$ is algebraic.

   The thing that I am struggling to understand is how $F(t)$ is not finitely generated as an $F[t]$-algebra: isn't $F(t)$ simply $F[t][t^{-1}]$?

2. Assuming that 1. is understood, how can we use this in an inductive proof of the Weak Nullstellensatz? That is, how can we show that $F(x_1)$ is a finitely-generated $F[x_1]$-algebra? (i.e. we have just used the induction hypothesis to show that $[E=F[x_1,\ldots,x_m]:F(x_1)]$ is finite, and now need to show the same for $[F(x_1):F]$)

3. In the main proof given, which works by induction on the size of the generating set, there is a part which says that

   $E=F(x_1)[x_2,\ldots,x_m]$ is finitely generated as an $F(x_1)$-module.

   I understand that the induction hypothesis is used to tell us that $[E:F(x_1)]$ is finite, but don't see how we can use this to show that it is finitely generated as an $F(x_1)$-module.

   Edit: Obviously $(E:F(x_1))$ being finite is just saying that it is finitely generated as an $F(x_1)$-module (here module = vector space).

It's very likely that I've just missed something very obvious, but I'm still struggling to get to grips with commutative algebra, so any help would be much appreciated! If a longer excerpt of the whole proof is needed then let me know. It seems that this is a pretty well-known proof, so I assumed that people might know the gist of it anyway, but I could be wrong!

Answer 1

Here is a collection of hints from my comments to OP:

• For (1), think about the analogous question: "is $\mathbf{Q}$ finited generated as a $\mathbf{Z}$-algebra (!)?"

A more explicit answer to (1): Recall that $F[t]$ is a principal ideal domain (much like $\mathbf{Z}$). We will show that $F[t]$ has infinitely many primes.

If $K$ is a finite set of primes in $F[t]$, then, by unique factorisation in $F[t]$, the element $1+\prod_{\pi \in K} \pi \in F[t]$ must have prime in its factorisation distinct from those in $K$. Thus, no finite set of primes in $F[t]$ is exhaustive!

Now any set of generators for $F(t)$ as $F[t]$-algebra must contain all these primes by the unique factorisation in $F[t]$. This finishes the proof.

• For (2), it suffices to note that $x_1$ is algebraic over $F$, for then, we have that $[F(x_1): F]$ is finite (and equals the degree of the minimal polynomial of $x_1$ over $F$). But you explain in that in the blockquote in (1).