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I am new to algebraic geometry, and am reading the book An Invitation to Algebraic Geometry. Exercise 2.6.4 in this book asks you to prove the topological space $\displaystyle\operatorname{maxSpec}\frac{\mathbb{C}[x,y]}{(x^2)}$ is homeomorphic to $\mathbb{A}^1$. Can anyone show me how to prove?

Intuitively I think $\mathbb{A}^1$ should be homeomorphic to $\displaystyle\operatorname{maxSpec }\frac{\mathbb{C}[x,y]}{(x)}$ which is maxSpec $\mathbb{C}[x]$, and is different from $\displaystyle\operatorname{maxSpec}\frac{\mathbb{C}[x,y]}{(x^2)}$.

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The maximal ideals of ${\mathbb C}[x,y]$ containing $(x^2)$ are the same ideals as the maximal ideals of ${\mathbb C}[x,y]$ containing $(x)$.

After that, as you mention in the question, use that ${\mathbb A}^1$ is (homeomorphic to) $\text{maxSpec } {\mathbb C}[x]$.

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  • $\begingroup$ Thanks! And therefore maxSpec $\mathbb{C}[x,y]/(x^2)$ and maxSpec $\mathbb{C}[x,y]/(x)$ are the exactly the same, right? $\endgroup$ – hxhxhx88 Apr 22 '14 at 23:12
  • $\begingroup$ Yes, indeed. (Well, they're canonically isomorphic, as maximal ideals of ${\mathbb C}[x,y]/(x^2)$ correspond to maximal ideals of ${\mathbb C}[x,y]$ containing $(x^2)$, which are the maximal ideals of ${\mathbb C}[x,y]$ containing $(x)$, which correspond to the maximal ideals of ${\mathbb C}[x,y]/(x)$. $\endgroup$ – Magdiragdag Apr 22 '14 at 23:15

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