1
$\begingroup$

My question comes out of Exercise 5.16 of the book by Atiyah-Macdonald.

Let $k$ be an infinite field and let $A \neq 0$ be a finitely generated $k$-algebra. Let $x_1, \dots , x_n$ generate $A$ as a $k$-algebra. We can renumber the $x_i$ so that $x_1, \dots , x_r$ are algebraically independent over $k$ and each of $x_{r+1}, \dots , x_n$ is algebraic over $k[x_1, \dots , x_r]$.

I don't understand the sentence "We can renumber the $x_i$ so that $x_1, \cdots , x_r$ are algebraically independent over $k$ and each of $x_{r+1}, \cdots , x_n$ is algebraic over $k[x_1, \cdots , x_r]$"

[My try]

There is a maximal algebraically independent subset of generators and I renumber these elements by $x_1, \cdots , x_r$. But I can't show that each of the remainder ,say $x_{r+1}, \cdots , x_n$, is algebraic over $k[x_1, \cdots , x_r]$.

How can I suitably renumber the generators.

Thanks in advance.

EDIT

There was a typo. $x_{r+1}, \cdots x_r$ is algebraic over $k[x_1, \cdots, x_r]$ not $k[x_1, \cdots , x_n]$.

$x_{r+1}, \cdots , x_{n}$ is trivially algebraic over $k[x_1, \cdots, x_n]$. But I don't know why it is algebraic over $k[x_1, \cdots, x_r]$.

$\endgroup$
  • $\begingroup$ Well, if they weren't algebraic, then they would be transcendental... $\endgroup$ – Qiaochu Yuan Mar 14 '17 at 8:33
  • $\begingroup$ @QiaochuYuan Sorry. I had a mistake in my question. So I edited it. $\endgroup$ – Jeong Mar 15 '17 at 0:00
0
$\begingroup$

(Just an elaboration of Qiaochu Yuan's comment.) Suppose $x_{r+1}$ is not algebraic over $k[x_1,\ldots,x_r]$. Then it is not the root of any polynomial with coefficients in $k[x_1,\ldots,x_r]$. Since $x_1,\ldots,x_r$ are algebraically independent, this implies that $x_1,\ldots,x_r,x_{r+1}$ are algebraically independent, contrary to the maximality of $r$.

$\endgroup$
  • $\begingroup$ Sorry. I had a mistake in my question. So I edited it. $\endgroup$ – Jeong Mar 15 '17 at 0:01
  • $\begingroup$ @user26857 $x_{r+1}$ is algebraic over $k[x_1, \cdots,x_{r},x_{r+2}, \cdots x_n]$ not yet over $k[x_1, \cdots,x_{r}]$. Why $x_{r+1}$ has a nontrivial polynomianl independent of $x_{r+2}, \cdots, x_n$? $\endgroup$ – Jeong Mar 16 '17 at 5:03
  • $\begingroup$ The renumbering can be performed to avoid this problem. Suppose, inductively, that $x_{i_1},\ldots,x_{i_s}$ are algebraically independent. Take $x_j$ with $j\not\in \{i_1,\ldots, i_s\}$. Either $x_j$ is algebraic over $k[x_{i_1},\ldots,x_{i_s}]$, in which case ignore $x_j$ from now on, or $x_{i_1},\ldots,x_{i_s},x_j$ are algebraically independent. (As in my answer.) We end with $x_{i_1},\ldots,x_{i_r}$ algebraically independent and the remaining $x_j$ all algebraic over $k[x_{i_1},\ldots,x_{i_r}]$. (Some may be algebraic over a subfield.) Hope this helps. $\endgroup$ – Mark Wildon Mar 16 '17 at 20:54
  • $\begingroup$ Now I understand. Thank you very much! $\endgroup$ – Jeong Mar 17 '17 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.