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I am trying to get an intuitive idea for the genus of an algebraic curve -- this seems like kind of a common question, but I am having trouble building intuition from past answers. I'm looking specifically at this curve: $$ (x^2 + y^2 - 1)(x^2 + y^2 - 4) = 0 $$ The curve itself is just two concentric circles of radius $1$ and $2$, as shown below. My vague undertstanding of topological genus as "number of holes" tells me that the genus should be $2$, since there are two disconnected components with one "hole" each. But the genus-degree formula seems to be telling me something different; the polynomial has degree $4$, and as far as I can tell has no singular points, so it seems to me like its genus should be $(4-1)(4-2)/2 = 3$. Where have I gone wrong? (I would expect that I have the wrong intuition for algebraic genus, or that there are singular points in projective coordinates that I am not considering.)

the curve in question

Graph generated by Wolfram|Alpha.

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  • $\begingroup$ I think I'm looking for the geometric genus. $\endgroup$ – Andrew Tindall Dec 24 '17 at 22:34
  • $\begingroup$ Looking at the real point is not always a good idea for fully understand the topology of the complex points of the curve. Also a first step would be to look at Harnack's inequality. $\endgroup$ – Nicolas Hemelsoet Dec 24 '17 at 22:58
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    $\begingroup$ Genus formula is for projective curves. Your curve is not irreducible, so if you think of it as a plane curve, it is the union of two conics and thus singular. So its arithmetic genus calculated by genus formula is three. Geometric genus is not meaningful for reducible curves. $\endgroup$ – Mohan Dec 24 '17 at 23:35

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