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.
Please help me. Thank you
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.