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[ I am a bit clueless about why this question is getting downvotes!? I put it up with a genuine serious interest and I don't seem to be making any egregious error either - apart from those unsure sentences which I have made with a "?" in the bracket. Please explain if something is terribly wrong with this question! Is this question too elementary for this forum? ]

Here by a "curve" I shall tend to think of algebraic curves in $\mathbb{CP}^2$

  • Is reducibility or not of a curve a question of whether the defining equation factorizes (necessarily into linear factors?) or is something more demanded from the factors? If the curve is thought of as a monic (in $y$) element in $\mathbb{C}[x][y]$ (which it can always be) has even $1$ root isn't that sufficient to say that it is reducible?

The degree of an algebraic curve will be the highest degree homogeneous component in it and hence if it has a triple point that would imply that the third degree term is the only term. Hence further if this is an ordinary triple point that would mean that this only term (of third degree) has 3 distinct roots and hence the curve is reducible. Is the argument right?

  • I would like to understand the other related such statements that I face like - a fourth degree curve with 4 singular (whether or not ordinary? whether or not distinct?) is also reducible, that if a fourth degree irreducible curve has 3 distinct singular points then they are necessarily double.

As the framing itself suggests, I am not sure of the statements and would like to know what is the precise statement that is correct and why.

  • Thought of as the monic polynomial (as in the first bullet point) if the curve has $0$ discriminant then it will have repeated roots. Is that then equivalent to saying that irreducibility implies that the discriminant is not identically $0$? (..or is some further work required?..)(I guess the converse is not true - a non-zero discriminant curve can still be reducible?--I guess so..)

  • What is the meaning of an "ordinary singular" point on a curve? I am aware of the notion of an "ordinary k-tuple" point. (...is it true that for $k>1$ such a point has to be singular?..seems so..)

  • What is the general way to connect reducibility or not of a curve with the fact as to whether or not it has singular points or how many of them does it have?

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    $\begingroup$ The question is too elementary. It also suggests that you are not clear enough on the basic concepts to ask this type of question here. $$ $$ If you think of curves in $\mathbb{CP}^2$, why do you write about it as being in $\mathbb{C}[x][y]$? And why talk about highest degree homogeneous component? Also, reducible curves must have singular points, since their irreducible components must intersect. $\endgroup$ Oct 13, 2011 at 3:30
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    $\begingroup$ As others have mentioned, your question is more appropriate for math.stackexchange.com or other sites listed in the FAQ. Before asking more algebraic geometry questions here, you should probably read a basic treatment of projective geometry and especially consider how to make polynomials homogeneous by adding a variable. $\endgroup$
    – S. Carnahan
    Oct 13, 2011 at 4:37
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    $\begingroup$ @S.Carnahan I had asked an almost similar question on math.stackexchange and it lay there unanswered. Hence I closed the question on that forum and posted a modified version here. So is MathOverflow now going to be out of bounds for beginners in any subject? That would be a very severe change of policies - atleast over the last year or more I never had this experience of my questions being closed and I don't think I ever asked any research question - always graduate level stuff. I wonder if this is a way MathOverflow is trying to restrict the level to even higher than beginning graduate level. $\endgroup$
    – Anirbit
    Oct 13, 2011 at 6:35
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    $\begingroup$ @Anirbit: you completely misunderstand my comment, which has nothing to do with organization. If you want to talk about a curve in $\mathbb{CP}^2$, you should think of a homogeneous polynomial in $\mathbb{C}[x,y,z]$. If you dehomogenize it, then you are talking about an affine curve in $\mathbb{C}^2$. Of course, going back and forth between the two points of view can provide helpful insight, but first I would recommend to understand each of them really well separately, because it's really easy to get tripped up going back and forth from projective to affine. $\endgroup$ Oct 13, 2011 at 14:42
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    $\begingroup$ Since S. Carnahan made a similar remark, it seems the consensus is that the wording of your question suggests that you are insufficiently aware of these distinctions between affine and projective. It may not really be the case, but that's what the question makes you sound like. And since this distinction is elementary undergraduate-level, you will find little interest for it in MO users. Like S. Carnahan, I would suggest carefully working on more elementary considerations before tackling your question again. Walker's book is a good place to start, I have not read Griffith's book. $\endgroup$ Oct 13, 2011 at 14:50

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read walker's algebraic curves, the first few chapters, for a nice discussion of this. there you will find I believe something like e.g. that a curve of degree d with more than (1/2)(d-1)(d-2) singular points is reducible, in the sense that its equation is a product of two equations, and a curve with more than (1/2)d(d-1) singular points is non reduced, in the sense that its equation is not square free.

the reason for these facts is visible topologically if you know that a complex curve of degree d is topologically a specialization of a surface of genus g = (1/2)(d-1(d-2), and the maximum finite number of singularities occurs for a union of d lines.

these results are proved by the strong bezout theorem. e.g. if a cubic curve has two singularities, then the line through them meets the curve with multiplicity 2x2 = 4, hence lies inside the curve.

hence this is a stackexchange level question.

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  • $\begingroup$ @Roy Thanks for the references. By "specialization" do you mean normalization? Your comment about the topological motivations reminds me of the genus formula but I can't connect them. In the genus formula one has that an irreducible algebraic curve with degree d with only singular double points ( say m) of them is normalized by a Riemann surface of genus (d-1)(d-2)/2 - m. But I can't see how this can be turned around to use it to decide on irreducibility or not given the degree and nature/number of singularities. $\endgroup$
    – Anirbit
    Oct 13, 2011 at 7:04
  • $\begingroup$ @Roy BTW I posted thus question here only after it remained unanswered on mathstackexchange $\endgroup$
    – Anirbit
    Oct 13, 2011 at 7:05
  • $\begingroup$ I agree that it's too bad that there isn't a better website for basic early-graduate-level questions, but there are books, courses, TAs, and other students to draw on. This graduate student avoids posting any question unless he thinks it will be of some interest to a nonzero number of research mathematicians. By the way, William Fulton has made his lovely little book on algebraic curves available for free online: math.lsa.umich.edu/~wfulton/CurveBook.pdf $\endgroup$ Oct 13, 2011 at 16:38
  • $\begingroup$ i should apologize for sending you to stack exchange and yet i myself do not regularly peruse stack exchange for questions to answer. i will try to improve in that regard. $\endgroup$
    – roy smith
    Oct 14, 2011 at 5:27
  • $\begingroup$ @Andrew I think your argument is a bit misplaced.You are having in mind only students fortunate to be in big enough places where there are all these resources?Is MathOverflow going to be out of bounds for others?Even just a few months ago I had no access to any TA (and still don't) and there weren't any students around me doing anything like this. Even more than that I think MathOverflow can be a great resource for people like me who are not even in any mathematics department. But then if the cut-off for the questions is going to be raised so high then one is shutting out many graduate student $\endgroup$
    – Anirbit
    Oct 15, 2011 at 0:57

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