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It is easy to prove that for any two algebraic sets $X_1, X_2$ in $\mathbb{A}^n$ we have that $$I(X_1\cap X_2) = \sqrt{I(X_1)+I(X_2)}$$ Find an example that the radical is neccessary, i.e., an example on algebraic sets $X_1, X_2$ that $I(X_1\cap X_2) \ne I(X_1)+I(X_2)$. Can you see geometrically what it means if we have inequality here?

Many thanks in advance.

Daniel Donnelly
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9999
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    It is indeed part of the exercise 1.4.1 of Gathmann's Algebraic geometry notes, so it seems to be your homework as well. Perhaps, you should let the other know what you did try? – Ehsan M. Kermani Dec 08 '13 at 18:38
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    Hint: Play around with some examples of plane curves $X_1,X_2$. What happens when they intersect transversally, intersect non-transversally, or not intersect at all? – Martin Brandenburg Dec 08 '13 at 21:12
  • @Ehsan M. Kermani : Yes, I am trying to read Gathmann's notes by myself, not my class. His writing is quite good and I like the notes, eventhough my background is still limited. – 9999 Dec 24 '13 at 21:40

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Take $X_1 = V(y)$, $X_2 = V(y - x^2)$. Then $(y) + (y - x^2) = (x^2, y)$. However $X_1 \cap X_2 = V(x,y)$; $(x,y)$ being, of course, the radical ideal of $(x^2,y)$.

Magdiragdag
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