# Ring of regular functions of isomorphic quasiprojective varieties

I know that if two quasi-projective varieties are isomorphic then their coordinate rings may not be isomorphic. But what can we say about their ring of global regular functions? Are they isomorphic?

Let $X\subset\mathbb P^n$ be a quasi-projective variety. Denote $$K[X]=\{f:X\rightarrow k\mid f\text{ is regular at every point of }X\}.$$ So my question is if $X\cong Y$ then is it true that $K[X]\cong K[Y]$?

My guess is they are isomorphic.

• $X\cong Y$ means we have regular maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ which are inverse of each other. I defined $K[X]$ above in the question. – user436053 Apr 16 '17 at 17:54
• I meant a definition like this (from Milne p.15) the field $k(C)$ being not so easy to understand, and $k[C]$ is the subring of functions not diverging at every point of $C$ ? – reuns Apr 16 '17 at 18:13
• See only the definitions on the above link. I think $K[X]$ and $K[Y]$ are isomorphic via the maps $f^*:K[Y]\rightarrow K[X]$ and $g^*:K[X]\rightarrow K[Y]$ where $f:X\rightarrow Y$ and $g:Y\rightarrow X$ are regular maps which are inverse of each other. – user436053 Apr 16 '17 at 18:21
Yes, it is true that if $X\cong Y$ with this isomorphism realized by $f:X\to Y$ and $g:Y\to X$ their rings of regular functions are isomorphic. Consider $h:X\to k$ a regular function on $X$. Then $g^*(h)=h(g)$ is a regular function on $Y$. It is clearly a ring map. We note that $f^*g^*(h)=h(f(g))=h$, so $f^*g^*=id$, so the rings are isomorphic.
NB: The ring of regular functions on a quasi-projective variety can be a "dumb" invariant in that it might not tell you much good information. Consider $\mathbb{P}^a\subset \mathbb{P}^N$. No matter what $a$ you choose, the ring of regular functions is always just $k$, the base ring. This sort of thing is one motivating factor for usage of the structure sheaf replacing the ring of regular functions.