5
$\begingroup$

Let $K$ be field. How do I proof that transcendence degree of $K[X_1,X_2,\ldots,X_n]$ is $n$? The set $\{X_1,X_2,\ldots,X_n\}$ is algebraically independent over $K$. So, I have to show that every subset of size greater than $n$ is algebraically dependent.

$\endgroup$
  • $\begingroup$ Yes, but then how do we calculate the krull dimension? $\endgroup$ – Mohan Jan 20 '13 at 2:45
17
$\begingroup$

suppose $P_0, \ldots, P_n$ are $n+1$ polynomials, of degree less than $d$. Then by multiplying the $P_i$ among themselves up to $k$ times, you can build at least about $k^{n+1}/(n+1)!$ polynomials of the form $\prod P_i^{\alpha_i}$ of degree less than $dk$.

But the dimension of the vector space of polynomials of degree less than $dk$ in $K[X_1,\ldots X_n]$ is only about $(dk)^n/n!$.

So if you pick $k$ large enough you get more things of the form $\prod P_i^{\alpha_i}$ than there are dimensions in $K_{dk}[X_1,\ldots,X_n]$, which means that there is a combination of the $\prod P_i^{\alpha_i}$ with coefficient in $K$ that gives $0$, which means that you have a polynomial in the $P_i$ that gives $0$, hence they are algebraically linked

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This looks incredibly slick, and certainly isn’t the way I would have argued. $\endgroup$ – Lubin Jan 19 '13 at 18:03

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.