most recent 30 from http://mathoverflow.net 2013-06-19T09:07:47Z http://mathoverflow.net/feeds/question/77980 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77980/reducibility-or-not-of-algebraic-curves Anirbit 2011-10-13T01:19:52Z 2011-10-13T04:10:20Z

<p><em>[ 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? ]</em> </p> <p>Here by a "curve" I shall tend to think of algebraic curves in $\mathbb{CP}^2$</p> <ul> <li>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? </li> </ul> <p>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? </p> <ul> <li>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. </li> </ul> <p>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. </p> <ul> <li><p>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..)</p></li> <li><p>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..) </p></li> <li><p>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? </p></li> </ul>

http://mathoverflow.net/questions/77980/reducibility-or-not-of-algebraic-curves/77993#77993 roy smith 2011-10-13T04:01:29Z 2011-10-13T04:10:20Z

<p>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. </p> <p>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.</p> <p>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.</p> <p>hence this is a stackexchange level question.</p>