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This question comes from Vakil's FOAG 3.7A:

Let $A = k[x,y]$ with $S = \{[(y)],[(x,y-1)]\}\subset \Bbb{A}_k^2$,try to find generator of $I(S)$

My attempt since both $(y), (x,y-1)$ are prime therefore $I(S) = (y)\cap (x,y-1)$, therefore needs to compute the generator of intersection of this two ideals. I don't know how to do?

The second idea is if $\bar{S} = V(\frak{a})$ then $I(S) = I(\bar{S}) = \sqrt{\frak{a}}$ therefore only needs to find the ideal $\frak{a}$ and compute its radical?

It's reasonable to guess the closure of $S$ is $V((y))\cup V((x,y-1)) = V((y)(x,y-1)) = V((xy,y^2-1))$ therefore it reduces to compute the generator for the radical of $(xy,y^2-1)$

yi li
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    The notations are not defined here, so i am trying to guess their meanins. ($S$ is a set of two ideals, there is no $I(S)$ defined in my mind for these...) The first candidate for the intersection $(y)\cap(x,y-1)$ is $(xy, y(y-1))$, double inclusion... – dan_fulea Jan 02 '23 at 15:19
  • @dan_fulea $I(S)$ is set of functions vanishing on $S$ i.e. $I(S) = \cap_{_{[p]\in S}}p$ , sorry I made a typo here $S$ contains two points in the Spec – yi li Jan 02 '23 at 15:21
  • These two idea essentially the same since $\sqrt{IJ} = \sqrt{I}\cap \sqrt{J}$ – yi li Jan 02 '23 at 15:24
  • Ok I see since $IJ = I\cap J$ if $I+J = A$. In our example it's clear that $(y) + (x,y-1) = A$ as $y+1-y = 1$ therefore $(y)\cap (x,y-1) = (y)(x,y-1) = (xy,y^2 - 1)$ therefore the generator is $xy, y^2-1$ – yi li Jan 02 '23 at 15:38
  • The notation $I(S)$ is not the main problem the main problem is that generator $y^2-1$, it should be $y(y-1)$. So i would try first to show $(y)\cap(x,y-1)=(\ yx\ ,\ y(y-1)\ )$, one inclusion is trivial, the generators $xy$ and $y(y-1)$ from the R.H.S. are in the intersection, of course. Conversely. Pedestrian way... Start with $f$ in the intersection. Since $f\in (y)$ we have $f=yg$ for some $g$. Now $yg$ is of the shape $yg=xh+(y-1)k$... So $y$ divides the R.H.S. so $$0=xh(x,0)-k(x,0)\ ,$$and so on. – dan_fulea Jan 02 '23 at 15:39
  • if $(y)+(x,y-1) = A$ then the other direction holds. @dan_fulea – yi li Jan 02 '23 at 15:40
  • Yes, this is much better the going bare foot. I typed explicitly to point to the typo (of course) $y^2-\color{red}y$ instead of $y^2-1$. – dan_fulea Jan 02 '23 at 15:48
  • yeah thank you that's a typo. – yi li Jan 02 '23 at 15:48

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