Edit: I should point out that I'm working over an algebraically closed field $k$.

Let $X_1,X_2\subset\mathbb{A}^n$ be affine algebraic sets. Show that $I(X_1\cap X_2)=\sqrt{I(X_1)+I(X_2)}$. Show by example that taking the radical here is necessary. Can you see geometrically what it means if $I(X_1\cap X_2)\neq I(X_1)+I(X_2)$?

I showed this using that for ideals $\mathfrak{a}_1,\mathfrak{a}_2$ in a ring, $\sqrt{\mathfrak{a}_1+\mathfrak{a}_2}=\sqrt{\sqrt{\mathfrak{a}_1}+\sqrt{\mathfrak{a}_2}}$. As for an example, I took $X_1=V(x)$, $X_2=V(x+y^2)$ in $\mathbb{A}^2$. Then $$I(X_1)+I(X_2)=\langle x\rangle+\langle x+y^2\rangle=\langle x,y^2\rangle,$$ which is not radical. But I don't know exactly how to interpret that geometrically.

The intersection of those two varieties is just the origin. I would have supposed that it has something to do with the origin occurring "more than once" (read from the ideal $\langle x,y^2\rangle$), but I'm not sure why that is the case. Is it because the variety $X_1$ is a tangent to $X_2$ at the origin, and thus it counts as a "double" intersection? Then I don't understand it, because the point $0$ occurs only once in $X_1$, so why twice in the intersection? And what is the difference - geometrically - between $\langle x,y^2\rangle$ and $\langle x^2,y\rangle$?

Thank you very much for helping me understand these things!


1 Answer 1


First let me congratulate you: your intuition is remarkably correct and your example is excellent.

In the land of scheme theory things are astonishingly simple:
Given two subschemes $X_1= V(I_1), X_2=V(I_2)\subset \mathbb A^n _k$, their intersection $X_1\cap X_2=V(I_1+ I_2)$ is defined by the sum of their ideals, period .
In your example $I_1=(x), \: I_2=(x+y^2) $ and so the intersection $X_1\cap X_2$ is given by the ideal $I_1+ I_2=(x, y^2)$.
This is carefully to be distinguished from the subscheme $$(X_1\cap X_2)_{red}=V(\sqrt{I_1+ I_2})= V(x,y)\subsetneq X_1\cap X_2 $$ which is smaller in the sense that it has the same underlying set (the origin $(0,0)$), but less functions living on it: only the elements of $k$, whereas on the genuine intersection $X_1\cap X_2$ the set of regular functions is the non reduced ring $k[y]/(y^2)$.
The non reducedness is the translation of the tangency of $X_1$ and $X_2$ at $(0,0)$.

In the more barren land of classical varieties you are constrained to reduced ideals and their corresponding varieties.
Unfortunately this implies that even if you want to intersect two reduced varieties, the intersection will not be reduced, as in the example examined above, and you will be forced to reduce that intersection by taking the root $\sqrt{I_1+ I_2}$ of the genuine ideal $I_1+ I_2$ of the intersection.

I'll add a few words about your last question, the difference between $(x,y^2)$ and $(y,x^2)$, which I forgot to address.
The first ideal can be seen as that of the intersection $S=V\cap L=V(x,y^2)$ of the parabola $P=V(x-y^2)$ and the vertical line $L=V(x)$.
From the classical varieties point of view this is just the origin $O=(0,0)$.
But scheme-theoretically you add the information that the ring of regular functions is $k[x,y]/(x, y^2)=k[y]/( y^2)$, so that a function on $S$ is of the form $q+r\bar y$.
In other words, the intersection $S$ is a little bigger than just$O$, in that if you restrict a polynomial $F(x,y)=a+bx+cy+dx^2+...$ to $S$, you will get $a+c\bar y$ :you can compute $c=\frac {\partial F}{\partial y}(0,0)$ for a function on $S$!
A similar analysis applies to $T=V(y,x^2)$ on which you can compute $\frac {\partial F}{\partial x}(0,0)$

  • 2
    $\begingroup$ Hello @Georges, first of all, thanks for all the great answers you already wrote in my topics! I'm not so sure I understand most of the scheme stuff yet, but I'm happy my intuition was okay there :) What about the difference between $\langle x,y^2\rangle$ and $\langle x^2,y\rangle$. They should both be the origin "doubled" in some way. Is it maybe some kind of indication of what direction the curves are intersection at $0$? $\endgroup$ Feb 19, 2012 at 15:51
  • $\begingroup$ Dear @Rand: sorry, I forgot about that part of the question. I have added a few words in an edit on it. $\endgroup$ Feb 19, 2012 at 16:51
  • 3
    $\begingroup$ Edit: @Georges: No need to be sorry, it was a very insightful answer anyway, now even more. Thank you! $\endgroup$ Feb 19, 2012 at 16:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .