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In general $\mathbf I(Y_1 \cap Y_2)=\sqrt{\mathbf I(Y_1)+\mathbf I(Y_2)}$. In remark 1.27 of these notes by Gathman, an example is given to illustrate the RHS is not radical if the varieties $Y_1,Y_2$ are in some sense tangent. I'm confused about the example he gives. First of all, he works in $\mathbb A^1_\mathbb C$, but uses ideals like $ \left\langle x_2 -x_1^2\right\rangle $. Aren't we really working in $\mathbb A^2_{\mathbb R^2}$?

If so, I'm having trouble formalizing this intuition of "tangency". If we stick with $\mathbb C[x]$ which is a PID then the generator of $\mathbf I(Y_1)+\mathbf I(Y_2)$ must have at least a double root (since it's a UFD and a principal ideal in a UFD is radical iff the prime decomposition of its generator has no repeated elements), so every element in this ideal will also have at least a double root at the same point, which sounds good. But $\mathbb R[x,y]$ is not a PID so I'm not sure what to do.

What's the geometric picture here and is it really $\mathbb A^1_\mathbb C$ we're working in?

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    $\begingroup$ I think it would be good to just give the whole example here. Anyway, you should work in $\mathbb C^2$ but of course you can only draw real pictures and hope that they're good enough. The correct notion of tangency is the Zariski (co)tangent space. Here the cotangent space is cut out by the cokernel of $(\partial f_j/\partial x_i)_{i,j}$. Here that's $\binom{-2x_1}{1}$ and $\binom{0}{1}$. At the origin these are lining up. Of course he'll get to these concepts later in the notes. $\endgroup$ – Hoot Dec 13 '15 at 13:17
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It seems he meant $\mathbb A^2_\mathbb C$. The geometric picture is the one he gives in the example. (Probably that's not a helpful comment, but I don't know of a better picture.)

The proper way to formalize the notion you want is through scheme theory. (And this is exactly why scheme theory is such a useful tool -- the language of varieties is simply not rich enough do to intersection theory properly, among other things.) Here is a brief, somewhat ad-hoc way to think about it.

The usual ideal of the point $(0,0)$ is $I=(x,y)$. The ideal Gathmann gives is $J=(x^2,y)$. As vector spaces, $k[x]/I$ is one dimensional, while $k[x]/J$ is two dimensional. This accounts for the fact that the intersection occurs "with multiplicity two." (If you wiggle the line a little, you get two points of intersection.)

In other words, we have two closed subschemes supported at the same point of $\mathbb A^2$, and we can tease them apart by looking at their structure sheaves.

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  • $\begingroup$ But if we work in $\mathbb C[x]$ how can we have two variables with which to form polynomials? $\endgroup$ – user153312 Dec 13 '15 at 16:21
  • $\begingroup$ @Exterior We work in $\mathbb C[x,y]$. The base field should always be algebraically closed. $\endgroup$ – Potato Dec 13 '15 at 20:09
  • $\begingroup$ Sorry, I somehow misread your answer. $\endgroup$ – user153312 Dec 13 '15 at 20:23

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