Let $E$ be a field extension of a field $F$.

Noether's Normalization Lemma states (in this case) that if $E = F[y_1,...,y_n]$, that is E is a finitely generated $F$ - algebra, then there exists an algebraically independent (over F) set $\{x_1,...,x_k\}$ in $E$, such that $E$ is integral over $F[x_1,...,x_k]$.

The weak Nullstellensatz tells us that in this situation, $E/F$ is a finite extension.

Here is the the source of my confusion: If $\{x_1,...,x_k\}$ is algebraically independent, then is it not contradictory that the dimension of the field extension is finite? For example, take $x_1$, and assume the dimension is finite. Since the extension is finite, if we take powers of $x_1$, eventually we get a linearly dependent set of vectors over F. But then we have some polynomial in $x_1$ that is zero. But this contradicts the algebraic independence of the set of $x_i$.

I'm sure I have made a simple incorrect assumption somewhere, I would appreciate any help finding it!

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    $\begingroup$ I suppose this could be solved if k = 0. In fact maybe that is always the case when you apply NNL to a field extension. That would actually match up with other sources I have read on the issue. If someone could confirm this it would still be appreciated. I know it looks like I've just posted a question without thinking about it much but this has genuinely been bothering me for like the last 40 minutes of work. $\endgroup$ – Daven Apr 6 '18 at 11:37
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    $\begingroup$ Which statement of the weak Nullstellensatz do you have? $\endgroup$ – Bernard Apr 6 '18 at 11:52
  • $\begingroup$ Let F ⊆ E be two fields such that E is finitely generated as an algebra over F. Then E/F is a finite extension $\endgroup$ – Daven Apr 6 '18 at 12:02
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    $\begingroup$ Chapter 10 of this might be of interest. As you said, we need $k=0$. Gathman even uses NNL to prove the NSS (as I think is quite standard). $\endgroup$ – 57Jimmy Apr 6 '18 at 12:25

In general, the Noether Normalization Lemma states that for a commutative $F$-algebra $A$ over a field $F$, there exists a $k \geq 0$ and algebraically independent elements $x_1, \dots, x_k \in A$ such that $A$ is a finitely generated as a $F[x_1,\dots,x_k]$-module.

When $A$ is a domain, $k$ is the transcendence degree of (the field of fractions of) $A$ over $F$.

So, indeed, in your case $A = E$ is an algebraic field extension of $F$, so it has transcendence degree $0$ over $F$.


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