In $R^2$, we have those solutions of $x^2+y^2-1=0$ describe a closed unit disk and $y^2 = x^3 − x + 1$ describes an unclosed elliptic curve. When we consider corresponding algebras e.g. ${R^2}/ \langle {x^2} + {y^2} - 1 \rangle $ and ${R^2}/ \langle {y^2} - {x^3} + x - 1 \rangle $, which things in their algebras describe the closed property? It seems that homological algebras may involve, for example, as we know, dimension or multiplicity was defined.

  • $\begingroup$ I think what you mean is not "closed", but "compact". However, you're asking about the compactness of the set of real solutions under the topology inherited from $\mathbb{R}^n$. This type of compactness is not a purely algebraic question, so I don't see why you should be able to detect it at the level of commutative algebra (as you mention in your comment to Asvin's question). You could try to make it algebraic by asking about compactness in the Zariski topology, but then the both of your spaces are quasi-compact, though neither are Hausdorff so they are not compact (depending on your def) $\endgroup$
    – Will Chen
    Sep 5, 2020 at 4:28
  • $\begingroup$ You can see the shape of the elliptic curve in "en.wikipedia.org/wiki/…." and it is not closed. I do not talk about the topology of algebraic varieties here because I only care about the properties of its coordinate rings. $\endgroup$ Sep 5, 2020 at 4:48
  • $\begingroup$ What do you mean by closed? $\endgroup$
    – Will Chen
    Sep 5, 2020 at 8:15
  • $\begingroup$ Closed like a Jordan curve. $\endgroup$ Sep 5, 2020 at 17:08
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    $\begingroup$ The OP's question may not have been precisely stated but I think it's reasonable. We know that many geometric and topological properties of the complex points of a variety are detectable via algebra (for an extreme example, the Betti numbers via counting points over finite fields); the OP asks a similar question but for the real points. $\endgroup$ Sep 5, 2020 at 19:02

1 Answer 1


One way of asking this is: Does the hyperplane at infinity (in the projective plane) intersect the curve at a real point? That is, if we look at the homogeneous terms of the highest degree, does that polynomial have a real, non trivial solution? In the first case, $x^2+y^2$ has no real solution except $(0,0)$ while in the second case $x^3$ has $(0,1)$ has a solution.

  • $\begingroup$ But you still are using algebraic geometry, in the case, projective algebraic geometry. We have a duality of commutative algebra and algebraic geometry, so I want to see a corresponding between them. $\endgroup$ Sep 5, 2020 at 2:29
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    $\begingroup$ Everything in here can be done at the level of rings: homogenize, set the variable you introduced to zero and look for solutions. $\endgroup$
    – user347489
    Sep 5, 2020 at 10:33

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