# Can the transcendence degree differ from the Krull dimension for the pluricanonical ring of a variety?

There is an exercise (p. 88) in Beauville's book Complex Algebraic Surfaces that claims that:

• For $X$ a smooth complex projective variety, if the Kodaira dimension (defined in this book as the maximum dimension over all $m$ of the image of $X$ under the map defined by the complete linear system $|mK_X|$) is non-negative, then it is equal to one less than the Krull dimension of the pluricanonical ring $R_K(X) = \bigoplus_m H^0(X, mK_X)$.

Without using the fact that the pluricanonical ring is finitely generated (if you like, pretend the exercise says "Iitaka dimension" and $K_X$ is an arbitrary line bundle, so in fact the ring could be non-finitely generated - in any case, this result had not been proven when Beauville wrote this book!), is this statement true/provable?

It is not too hard to see that the transcendence degree over $\mathbb{C}$ of the field of fractions of $\mathrm{Proj} \ R_K(X)$ is equal to the Kodaira dimension, but if a ring is not finitely generated over $\mathbb{C}$, it is perfectly possible that its transcendence degree is less than its dimension (e.g. the field of fractions itself has this property!).

(For what it's worth, the exercise in Beauville seems to define the Krull dimension as the transcendence degree of the field of fractions, so the exercise is fine either way).

• There are shockingly few references treating dimension theory in the non-Noetherian case. Even EGA doesn't really touch this (beyond the basic definitions). I was hoping to apply the dimension theory of local Noetherian rings (Tag 00K4) to the graded non-Noetherian setting. But even if there is a finitely generated ideal of definition, we cannot use it to compute the dimension: it seems that this method really relies on $\mathfrak m$ being finitely generated (see e.g. the discussion after Def 10.58.1). – R. van Dobben de Bruyn Dec 14 '16 at 15:33