3

I am learning about schemes reading Ravi Vakil's notes. On page 136, he says "For example, consider the scheme $\mathbb{A}_k^2 = Spec \ k[x,y]$, where $k$ is a field of characteristic not $2$. Then $(x^2 + y^2)/x(y^2-x^5)$ is a function away from the $y$-axis and the curve $y^2 - x^5$. Its value at $(2,4)$ (by which we mean $[(x-2,y-4]$) is $(2^2 + 4^2) / (2(4^2-2^5))$, as $$ (x^2 + y^2) / (x(y^2-x^5)) \equiv (2^2 + 4^2) / (2(4^2-2^5)) $$ in the residue field".

I understand that they are equal in the residue field, because I did the calculations to verify it. But I really don't have an intuition or idea why this happens (as why evaluating the point "is" considering it in the residue field). I would greatly appreciate any explanation on this matter. Thanks!

user170778
  • 527

1 Answers1

1

The residue field of a ring $A$ at a prime ideal $p$ is $A_p/pA_p$, which is a field because $pA_p$ is (the only) maximal ideal of the localization of $A$ at $p$.

In this case, the residue field is $k[x,y]/(x-2,y-4)$. If you want, you can easily see this thanks to the fact that if $I\subset p$ is an ideal of $A$, calling $\bar{A}=A/I$, $\bar{p}=p/I$, then $A_p/IA_p \simeq \bar{A}_{\bar{p}}$ : just choose $I=p$, remembering that in this case $p$ is maximal. If you are wondering on why this happens, look here : Quotient ring of a localization of a ring.

Now, take any polynomial $f(x,y)$. Calling $[f(x,y)]$ its equivalence class in $k[x,y]/(x-2,y-4)$, you have $[f(x,y)]=f([x],[y])$. But in this residue field, $[x]=[2],[y]=[4]$, so $[f(x,y)]=f(2,4)$. Naturally, this works with rational functions as well.

Community
  • 1
Andrea
  • 691