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.
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Yes, but then how do we calculate the krull dimension? – Mohan Jan 20 '13 at 02:45
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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
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This looks incredibly slick, and certainly isn’t the way I would have argued. – Lubin Jan 19 '13 at 18:03