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$K(x_1, x_2,\dots x_n)$ is field extension of $K$ where $x_1, x_2,\dots x_r$ is transcendental basis of $K(x_1, x_2,\dots x_n)$ over $K$.

Then, $K(x_1, x_2,\dots x_n)/K(x_1, x_2,\dots x_r)$ is algebraic extension by definition,

My question is, $K(x_1, x_2,\dots x_n)/K(x_1, x_2,\dots x_r)$ is finite extension? Atiye Macdonald uses this fact without proof.But it does not seem obvious to me. Thank you for your help.

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If $F$ is a field and $G$ is a finitely generated algebraic field extension, then $G$ is finite. This can be proved by induction with the identity $[F(\alpha,\beta):F]=[F(\alpha,\beta):F(\alpha)][F(\alpha):F]$. Now, $K(x_1,\cdots,x_n)$ is generated by $x_{r+1},\cdots, x_n$ over $K(x_1,\cdots,x_r)$.

Remark: Note that here I'm talking about a field $G$ which is finitely generated as a field extension of $F$, which is different from being finitely generated as an $F$-algebra. Namely, $F(T)$ is not a finitely generated $F$-algebra, because you have too many polynomials to take the reciprocal of. Both are different from being finitely generated as a $F$-vector space.

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  • $\begingroup$ The extension degree is exactly n-r, right ? And why we can say xr＋1,・・・,xn is a basis ? $\endgroup$
    – Poitou-Tate
    Commented Jan 12, 2021 at 10:18
  • $\begingroup$ @bellow No: the degree of the extension is $\prod\limits_{j=1}^{n-r}[K(x_1,\cdots, x_{r+j}):K(x_1,\cdots, x_{r+j-1})]$, and a basis as $K(x_1,\cdots, x_r)$-vector space may be chosen to be an appropriate family of monomials $\mathcal B_J=\left\{x_{r+1}^{\alpha_1}\cdots x_n^{\alpha_{n-r}}\,:\, (\alpha_1,\cdots, \alpha_{n-r})\in J\right\}$. $\endgroup$
    – user239203
    Commented Jan 12, 2021 at 10:22
  • $\begingroup$ For instance, $$[K(x^6,x^2,x^5):K(x^6)]=[K(x^6,x^2):K(x^6)][K(x^6,x^2,x^5):K(x^6,x^2)]=3\times 2=6$$ Mind the difference between degree $[G:K]$, i.e. the dimensione of $G$ as a $K$-vector spaces, i.e. the largest cardinality of a $K$-linearly independent subset of $G$, and transcendence degree of $G$ over $K$, which is the largest cardinality of an algebraically independent subset of $G$ over $K$. For instance, degree is multiplicative on sub-extensions, while transcendence degree is additive. $\endgroup$
    – user239203
    Commented Jan 12, 2021 at 10:29
  • $\begingroup$ In short, K[α1,・・・,αn] is obtained by adjoining a algebraic element αn＋１,・・・,αn over F to F, so the extension degree is finite from tower law, right? $\endgroup$
    – Poitou-Tate
    Commented Jan 12, 2021 at 14:34
  • $\begingroup$ Are you talking about $K[\alpha1,\cdots , \alpha_n]$, or $K(\alpha_1,\cdots, \alpha_n)$? They are two different things. For the specific case, since $x_{r+1},\cdots, x_n$ are algebraic over $K(x_1,\cdots, x_r)$, we have $K(x_1,\cdots, x_n)=K(x_1,\cdots, x_r)[x_{r+1},\cdots, x_n]$ and the degree multiplies over towers. $\endgroup$
    – user239203
    Commented Jan 12, 2021 at 14:53

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