I have encountered once a book on algebraic geometry, "An Invitation To Algebraic Geometry", which my classmate recommended much. Inside the book, I found a confusing exercise of the Chapter II, which states that maxspec$\mathbb{C} [x,y]/(x^2)$ is homeomorphic to the affine plane $\mathbb{A}^1$. But this I really had a hard time fighting with.
In fact, it can already be shown that this maximal spectrum is isomorphic to a closed subspace of $\mathbb{A}^2$; has this anything to do with the formal proof? I am now trying only to get a hint, and then proceed. Therefore any help will be appreciated.
Thanks in advance.
Why is the maximum spectrum of $C[x,y]/(x^2)$ homeomorphic to $A^1$?
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algebraic-geometry
1 Answers
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The element $x$ is nilpotent in your ring, so it belongs to every prime ideal. It follows from this very easily that there is a map from $\mathrm{Spec}\;\mathbb C[x,y]/(x^2)$ to $\mathrm{Spec}\;\mathbb C[x,y]/(x)$, mapping a prime ideal $\mathfrak p$ in $\mathbb C[x,y]/(x^2)$ to the prime ideal $\mathfrak p/(x)$ in $\mathbb C[x,y]/(x)$. This map is in fact an homeomorphism.
Finally, it should not be difficult to see that $\mathrm{Spec}\;\mathbb C[x,y]/(x)$ and $\mathrm{Spec}\;\mathbb C[y]$ are homeomorphism.
You wanted the maximal spectrum, and I did the whole spectrum... Mutatis mutandis you get the same result in the same way :)
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0Does that map come from the algebra inclusion $\mathbb C[y]\to\mathbb C[x,y]/(x^2)$? – 2011-11-15
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0@Rasmus: yes; but the natural map goes the other way. – 2011-11-15
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0@MarianoSuárez-Alvarez: Thanks for the answer. Is *Mutatis mutandis* latin? The first word looks like the ablative... – 2011-11-15
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0@awllower, see http://en.wikipedia.org/wiki/Mutatis_mutandis – 2011-11-15
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0@Mariano: Really? Should directions be reversed because $\mathrm{Spec}$ is contravariant? – 2011-11-15
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0@MarianoSuárez-Alvarez:Thanks again. Also I agree with Rasmus's comment. – 2011-11-16