Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Yes, but then how do we calculate the krull dimension? – Mohan Jan 20 '13 at 2:45
up vote 9 down vote accepted

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

share|cite|improve this answer
This looks incredibly slick, and certainly isn’t the way I would have argued. – Lubin Jan 19 '13 at 18:03
Thank you for this beautiful answer. – Mohan Jan 19 '13 at 18:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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