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Let $V$ be any irreducible variety over $\mathbb{R}$, prove that if $dim_{\mathbb{R}}V(\mathbb{R})= dim(V)$, then $V(\mathbb{R})$ is dense in $V$.

Any hints to make a start?

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    Dear CC, How do you conclude that $\dim_{\mathbb R}V(\mathbb R) = 1$. What if $V$ were the affine plane? Regards,2012-10-19
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    Why is $\dim_{\Bbb R}V(\Bbb R)=1$? Isn't $\Bbb A^2(\Bbb R)$ two-dimensional?2012-10-19
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    Oh yes.. I think I misunderstood the definition of dimension of an algebraic variety.2012-10-19
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    What do you mean by $\dim_{\mathbb{R}} V(\mathbb{R})$? $V(\mathbb{R})$ isn't necessarily, say, a manifold, so it's not clear (to me) what notion of dimension is meant here.2012-10-19
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    @Matt: it's not clear to me that each component will have the same dimension, though. Do we take a supremum over all components?2012-10-19
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    Dear CC, How about some context? What do you know? Regards,2012-10-19
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    @QiaochuYuan: Dear Qiaochu, That's I would do (at least, the exercise is true with that interpretation!). Regards,2012-10-19
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    @QiaochuYuan In fact, we don't have such definition in our lecture note (We always have HW assignments like this that contain unknown notations..). However, I guess since $V$ is determined by a prime ideal $P$, does $dim_R(V(R))$ means the kul dimension of $P$ in the ring $\mathbb{R}[x_1,..,x_n]$?2012-10-19
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    @MattE I'm currently seraching notations from the book "Algebraic Curves, Algebraic Manifolds and Schemes" by Danilov and Shokurov.2012-10-19
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    @CC_Azusa: I assumed that that was what $\dim(V)$ meant.2012-10-19
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    Dear CC, No, $\dim_{\mathbb R}V(\mathbb R)$ means the topological dimension of the real points. Regards,2012-10-19
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    @MattE Thanks Matt, that makes more sense2012-10-19

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