# $K(X,Y)$ is not finitely generated $K$ algebra.

I want to show that $K(X,Y)$ is not finitely generated $K$ algebra($K$ is a field) where $X ,Y$ are algebraically independent over $K$.

I proved it using Noether Normalisation as following. If it was finitely generated $K$ algebra then there was $X_1, \ldots,X_m$ algebraically independent over $K$ such that $K(X,Y)$ were integral over $K[X_1,\ldots,X_m].$ Then $m=0$ since $dim(K(X,Y))=0.$ Thus $K(X,Y)$ becomes algebraic over $K,$ which is not true since $X$ is not algebraic over $K.$

But I want to prove it in the sprit of the problem that $\mathbb{Q}$ is not finitely generated $\mathbb{Z}$ algebra as there are infinitely many non associate primes in $\mathbb{Z}.$ If there is any solution involving the number of primes then it will be helpful for me. Though I know that there are infinitely many primes in the polynomial ring $K[X,Y].$

Thank you.

• My gut says that all the different possible $\frac1{Y+f(X)}$ for $f\in K[X]$ are algebraically independent. However, my gut has been wrong before, and I have no idea whether this is easy to prove even if it were true. – Arthur Dec 11 '17 at 17:16
• Can you not literally copy the proof for $\mathbb{Q}$ word-for-word, just replacing $\mathbb{Q}$ by $K(X,Y)$ and $\mathbb{Z}$ by $K[X,Y]$? – Eric Wofsey Dec 11 '17 at 18:05
• @Eric Wofsey: How do I show that there are infinitely many non associate primes in $K[X,Y].$ It is clear that there is infinitely many primes in $K[X,Y].$ – user371231 Dec 12 '17 at 1:31
• Your proof also works for $K(X)$, right? – Juan Carlos Ortiz Apr 22 at 21:04
• Yes, it works for any function field involving finitely many variables. – user371231 Apr 24 at 7:51