Math 480, Spring 2013, Assignment 9

Carefully define the following terms, then give one example and one non-example of each:

1. Projection map.
2. Zariski closure.
3. Polynomial map.
4. Rational map.
5. Graph of a map.

Carefully state the following theorems (you need not prove them):

1. Geometric extension theorem.
2. Closure theorem.
3. Polynomial implicitization theorem (Theorem 3.3.1).
4. Rational implicitization theorem (Theorem 3.3.2).

Do the following problems:

1. Let \(S\) be the image of the polynomial map given by \((x, y, z) = (uv, uv^2, u^2)\). Find the ideal of the Zariski closure of \(S\). Then (working over \(\mathbb{C}\)) find all the points of the Zariski closure of \(S\) which do not lie in \(S\) itself.

Questions:

1. I am a little bit confused by the notation used for the problem. So I assume we have a polynomial map here and it is taking a subvariety to another subvariety, and it looks like there are three variables in the preimage and two variables in the image. The problem is that, for whatever reason, I am having a problem translating the concept from the class notes to the homework problem. For instance, with the twisted cubic, we introduced (x,y,z) to make a variety, but we took one variable to more variables, not three variables to less variables. In the twisted cubic, we introduced three variables which led to a variety whose terms we could use Buchberger's algorithm on to obtain a lexicographic G.B with respect to the three variables only. Is this the same principle that would be applied here? If so, should I assume I should find a way of translating between (x,y,z) and (u,v)? --Robert.Moray (talk) 16:03, 15 April 2013 (EDT)

The map is meant to take (all of) \(k^2\) into \(k^3\). The input variables are \(u\) and \(v\) while the output variables are \(x, y,\) and \(z\). Hope that helps. -Steven.Jackson (talk) 16:19, 15 April 2013 (EDT)

1. Please remind me again, will the test include material through this assignment?--Matthew.Lehman (talk) 13:41, 17 April 2013 (EDT)

Yes. -Steven.Jackson (talk) 15:48, 17 April 2013 (EDT)

1. On this homework problem I have been doing Buchberger's algorithm and keep generating terms in the grobner basis. I do not know if the algorithm is supposed to keep iterating like this. Did anyone get anything differently or am I generating the ideal incorrectly? I have an ideal \(I=<uv-x,uv^2-y,u^2-z> \) to start and am using a lexicographic order where u > v > x > y > z. --Robert.Moray (talk) 18:38, 17 April 2013 (EDT)

I am working on it now, and seem near an end. Currently at 8 equations in the basis, but haven't added a new one the past 9 S-pairs --Paul.Shook (talk) 18:53, 17 April 2013 (EDT)

Sorry for the tedium. It's hard to produce problems that illustrate something without requiring a lot of calculation. Really one should do each type of problem by hand once in one's life, then use machines. -Steven.Jackson (talk) 20:38, 17 April 2013 (EDT)

So Robert's start is correct. I am getting a reduced Groebner basis with eight elements\[\{u^2-z, uv-x, ux-vz, uy-x^2, v^2z-x^2, vx-y, vyz-x^3, x^4-y^2z\}\]. Of these, only the last does not involve any input variables; hence the Zariski closure of \(S\) is the variety defined by \(x^4-y^2z=0\). The business about finding the points of the closure not in \(S\) itself is probably less important; it involves a lot of case-work but the reasoning is elementary. Anyway, the answer is that all points of the form \((x,y,z)=(0,t,0)\) with \(t\neq0\) are in the closure but not in \(S\) itself. -Steven.Jackson (talk) 21:09, 17 April 2013 (EDT)

I may have gotten mixed up in my math, as I got some of these elements and not others. Actually, I ended up with three elements that were not in terms of u and v, although I switched my notation because the letters look too similar. Had I got to the answer you had, I would likely have come to the same conclusion by declaring the variety of the elimination ideal to be the zariski closure of S and then noting that since \(x^3-y^2z=0\), this implies that \(x^3=y^2z\), and thus by setting x and y to zero, we can see that any arbitrary point of y will satisfy the equality. I probably got lost somewhere in my polynomial LD. --Robert.Moray (talk) 21:40, 17 April 2013 (EDT)