My colleague and me were confronted with a question concerning the introductory chapter of a text on algebraic geometry. The scenario of a plane curve C=Z(f)Ak2C=Z(f)\subseteq \mathbb A_{\Bbbk}^2 over a field k\Bbbk is considered, for a nonconstant polynomial fk[x,y]f\in\Bbbk[x,y]. It was stated that the function field K:=Quot(k[x,y]/f)\mathbb K:=\mathrm{Quot}(\Bbbk[x,y]/\langle f\rangle) is obviously of transcendence degree 11 over k\Bbbk because f(x,y)=0f(x,y)=0 is a new relation between xx and yy. The prerequisites to this text are basic undergraduate knowledge of algebra and topology. The question was about the obviousness of the above statement. We came up with a proof that only really requires some linear algebra. I found it rather cute. Do you want to see it?



I am writing this blag post from the second annual meeting of the DFG priority programme SPP1489 (algorithmic and experimental methods in algebra, geometry and number theory). Apart from having a lot of fun, I am catching up on the recent developments in open-source computer algebra software. Do you want to know more?