A field extension $k\subseteq F$ is finitely generated if there exist $\alpha_a,\alpha_2,\dots,\alpha_n\in F$ such that $$F=k(\alpha_1)(\alpha_2)\dots(\alpha_n)$$ This is not the same as saying $F$ is a finite type $k$-Algebra. What if $\alpha_i$ is transcendental for some $i\in\{1,2,3\dots,n\}$?

Why is $F$ not a finite type $k$-Algebra? Isn't $F=k[\alpha_1,\alpha_2\dots,\alpha_n]$?

  • $\begingroup$ I think you meant to ask why "not every [...] is [...]" that is why there are some that are not. $\endgroup$ – quid Jan 18 '16 at 2:45

The element $\alpha^{-1}$ is contained in $k(\alpha)$ basically by definition. (Excluding the degenerate case $\alpha=0$.) As the notation $k(\alpha)$ means the smallest field containing $k$ and $\alpha$.

By contrast $k[\alpha]$ may well not contain $\alpha^{-1}$, informally as it is formed by the elements one gets plugging $\alpha$ into a polynomial with coefficients in $k$.

If $\alpha$ is algebraic the two happen to coincide but this is a result that needs to be proved. And the converse holds as well.

In case you actually want a proof that they are not equal for transcendental elements, rather than just the plausibility argument I gave above, then show that $\alpha^{-1}$ is actually not a polynomial expression in $\alpha$ by assuming the converse and deriving that $\alpha$ would be a root of a polynomial.

  • $\begingroup$ I have a further question, and it would be great if you could resolve it. The Nullstellensatz says "Let $K\subseteq F$ be a field extension, and assume that $F$ is a finite type $k$-algebra. Then $k$ is a finite (and hence algebraic) extension." Is this not obvious? If $F=k[\alpha_1,\alpha_2,\dots,\alpha_n]$, then we can just re-write $F$ as $k(\alpha_1,\alpha_2,\dots,\alpha_n)$. $\endgroup$ – fierydemon Jan 18 '16 at 2:58
  • $\begingroup$ Why is the Nullstellensatz a non-trivial statement then? $\endgroup$ – fierydemon Jan 18 '16 at 2:59
  • $\begingroup$ A finite field extension is not the same thing as a finitely generated field extensions. An estension $F$ over $k$ is called a finite extension if $[F:k]$ is finite, this is the dim of $F$ as a $k$-vector space, that is in this case it is even fin gen as a vector space not just as field. $\endgroup$ – quid Jan 18 '16 at 3:02

If $t$ is transcendental, $k(t)$ is finitely generated as a field extension of $k$, but it's not finitely generated as a $k$-algebra (if $k$ is too large). In fact if $k$ is uncountable, then the dimension of $k(t)$ as a vector space over $k$ is uncountable, whereas the dimension of a finitely generated $k$-algebra as a vector space over $K$ is at most countable (exercise). This observation can actually be used to prove special cases of the Nullstellensatz.


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