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The curve $C \colon x^3 + x^2 = y^2$ is a singular affine variety with a node at zero. How would one show that as an real affine variety $C \subseteq \mathbb{A}_\mathbb{R}^2 = \mathbb{R}^2$ it is no topological manifold?

More generally: How can someone tell and show which singular affine varieties over the reals given by equations are topological manifolds (when equipped with the subspace topology)?

I tried to find a property of a node which makes it impossible to have euclidian neighbourhoods (like having a closed neighbourhood which is the finite union of closed non-neighbourhoods), but this seems way too complicated – how could one even show that this curve has this property at the node an euclidian space hasn't?

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    You might find this relevant: http://math.stackexchange.com/questions/205489/is-it-possible-to-compute-if-an-algebraic-variety-is-a-differential-manifold2012-10-18
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    @Andrew Yes, I do. But for examples curves can have cusps which make them not differentiable. But they are still topological manifolds.2012-10-19

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