Math 480, Spring 2013, Assignment 14
From cartan.math.umb.edu
Carefully define the following terms, then give one example and one non-example of each:[edit]
- Maximal ideal.
- Point.
- Prime ideal.
- Irreducible variety.
- Irreducible component (of some reducible variety).
- Associated prime (of an ideal).
For each operation on varieties, describe the corresponding operation on ideals, then describe an algorithm that computes this operation (modulo the problem of computing radicals):[edit]
- Union.
- Intersection.
- Zariski closure of the difference.
Solve the following problems:[edit]
- Let \(I, J,\) and \(K\) be ideals. Prove that \(\left(I + J\right)K = IK + JK.\) (Hint: each side is, by definition, the smallest ideal containing certain elements. Which elements?)
- Working in \(\mathbb{C}[x]\), compute the associated primes of \(\left\langle x^3-3x^2+2x\right\rangle\).
- Working in \(\mathbb{C}[x,y,u,v]\), compute the associated primes of \(\left\langle xu, xv, yu, yv\right\rangle\). Do you think the variety of this ideal has any singular points?
Questions:[edit]
- I was wondering about a couple of things. 1.) Would you be willing to briefly recap the notion of the Associated Primes and how they are calculated? I remember you saying there was an algorithm, and the textbook did as well, but for some reason I do not have it in my notes, and 2.) Will you be posting answers for this and last homework set so that we can make sure we have the right answers when we study for the final? --Robert.Moray (talk) 19:43, 15 May 2013 (EDT)
- 1) The associated primes of an ideal are the ideals corresponding to the irreducible components of the corresponding variety. There is an algorithm to compute them, but we haven't talked about it in this class (and even if we had I wouldn't recommend it for these problems). The thing to do in these problems is to understand the corresponding variety by elementary methods, then see whether it's reducible (i.e. whether you can write it as the union of two smaller varieties). If so, then repeat the same process on the smaller pieces, and so on until you've written the variety as the union of irreducible varieties. Then finds the ideals corresponding to the irreducible pieces. 2) If you or anyone else posts solutions on the wiki, then I will check them to make sure they're right. --Steven.Jackson (talk) 07:40, 16 May 2013 (EDT)
- For problem 3: The variety \(\mathbb{V}(<xu,xv,yu,yv>)\) defines a system of equations\[xu=0\]. \(xv=0\), \(yu=0\). and \(yv=0\). To satisfy these four conditions simultaneously, we examine the possible cases. If \(x=0\), then we can assign any value to u and v to satisfy the first two equations, but then we must set \(y=0\) to satisfy the last two equations. Thus \(x=y=0\). If \(x\neq0\), then u and v must both equal 0 in order to satisfy the four equations, in which case we can assign any value to y and still satisfy the system of equations. The same cases will arise if we do this with y, so if \(u=0\), we are free to set x to be anything in order to satisfy the first equation, but then we must set v to 0 to solve the second equation. Continuing in this way, and testing all possible cases, we find that:
- \[(0,0,*,*)\] and \((*,*,0,0)\) form a basis for the variety. Therefore, I claim that we can express \(\mathbb{V}(<xu,xv,yu,yv>)\) as the union of \(\mathbb{V}(<xy>)\) and \(\mathbb{V}(<uv>)\). However, I further claim that neither variety is irreducible, because neither ideal that gives rise to its variety is prime. The former is not the prime because if we let \(f=x\) and \(g=y\), \(fg \in <xy>\), but neither x or y is a multiple of xy, and therefore by the definition of prime, \(<xy>\) is not prime. The proof for the other not being prime follows from this logic. Therefore, neither are irreducible, and thus I claim that the former variety is the union of \(\mathbb{V}(x)\) and \(\mathbb{V}(y)\) and the latter is the union of \(\mathbb{V}(u)\) and \(\mathbb{V}(v)\). Therefore, this implies that:
- \[\mathbb{V}(<xu,xv,yu,yv>)\]=\(\mathbb{V}(x)\cup \mathbb{V}(y) \cup \mathbb{V}(u) \cup \mathbb{V}(v)\).
- Therefore, the set of associated primes is \(\{x,y,v,u\}\).
- I am not sure if I am missing any cases, but this is what I was able to figure out from graphing the varieities --Robert.Moray (talk) 19:18, 16 May 2013 (EDT)
- I agree with your argument up through the statement that the variety is the union of the sets \(\{(0,0,*,*)\}\) and \(\{(*,*,0,0)\}\). (Strictly speaking your argument omits some cases, but this is forgivable since there are really a lot to consider, and the conclusion you've reached at this point is correct.) However, at this point we part ways; I think the ideal corresponding to \((0,0,*,*)\) is not \(\left\langle xy\right\rangle\) but \(\left\langle x,y\right\rangle\), and this will modify the rest of your conclusions. --Steven.Jackson (talk) 22:02, 16 May 2013 (EDT)
- I see where you got this now, based on the two sets I originally proposed. Then this would imply that the ideal corresponding to the other set would be \(<u,v>\). Calculating the corresponding varieties of these ideals seems to indicate that each can contain only one point, in the former when \(x=y=0\) and in the latter when \(u=v=0\). However, since for instance the variety \(\mathbb{V}(x)\) only contains \(x=0\) and the variety \(\mathbb{V}(y)\) only contains \(y=0\), wouldn't this mean that \(\mathbb{V}(x,y)=\mathbb{V}(x) \cup \mathbb{V}(y)\)? This conclusion, however, did not rely on exploiting the relationship between primeness and irreducibility. EDIT: So I am conflicted now about whether things can be any more simplified than \(<x,y>\) and \(<u,v>\). I had come up with the alternative suggestion that since the corresponding varieties to these simplified ideals contain only the points where \(x=y=0\) and \(u=v=0\), this might mean two even more simplified ideals could be \(<x-y>\) and \(<u-v>\), but this seems to be less restrictive than the ideals that I claimed gave rise to these even more simplified ideals. How would I know when I reached the end? --Robert.Moray (talk) 10:39, 17 May 2013 (EDT)
- So the hard part is deciding whether the sets \(\mathbb{V}(\left\langle x,y\right\rangle)\) and \(\mathbb{V}(\left\langle u,v\right\rangle)\) are irreducible. Probably I haven't given you the tools to decide this with complete rigor; however, relying to some extent on intuition, you may have some feel for whether a two-dimensional plane can be written as the union two proper subvarieties (which would need to be curves). (Deciding this question rigorously is equivalent to deciding whether \(\left\langle x,y\right\rangle\) and \(\left\langle u,v\right\rangle\) are prime ideals. If you've had some abstract algebra, ask yourself whether the corresponding quotient rings are integral domains.) --Steven.Jackson (talk) 18:52, 20 May 2013 (EDT)
- So for instance the variety \(\mathbb{V}(<x,y>)\) contains only \((0,0)\in \mathbb{C}^2\). However, the origin in the two-dimensional space represents the intersection of the lines \(x=0\) and \(y=0\), so if we compressed this down to one dimension, wouldn't this just be the value of 0 on the y line and the value of 0 on the x line? If this is true, would this mean that we would have the union \(\mathbb{V}(<x>) \cup \mathbb{V}(<y>)\)? If this were true, then \(<x,y>\) is not irreducible, which would then mean that it isn't prime either. However, surely the ideals \(<x>\) and \(<y> \) are prime and irreducible. If this logic were true then it could also be applied to the ideal \(<u,v>\). Though the system suggests that the equations be solved simultaneously, thus an intersection of varieties, surely any element in an intersection qualifies as being an element in a union. --Robert.Moray (talk) 20:11, 20 May 2013 (EDT)
- I think you're getting into trouble because you're changing the ring of polynomials (and hence the dimension of the affine space, the ideal generated by a given set of polynomials, and the corresponding varieties) in the midst of the discussion. Here we should fix the polynomial ring to be \(\mathbb{C}[x,y,u,v]\) for the whole discussion; then (for example) \(\mathbb{V}(\left\langle x,y\right\rangle)\) is the two-dimensional plane in which the \(x\) and \(y\) coordinates are fixed at zero but the \(u\) and \(v\) coordinates are arbitrary. So the question is whether (intuitively speaking) this two-dimensional plane can be the union of two curves. --Steven.Jackson (talk) 20:54, 20 May 2013 (EDT)
- In that case, if we continue with \(\mathbb{C}[x,y,u,v]\), and we wanted to claim that, for instance, the variety that the ideal \(<x,y>\) gives rise to is the union of \(\mathbb{V}(x) \cup \mathbb{V}(y)\), then these varieties would consist of the points of form (0,*,*,*) and (*,0,*,*) respectively, with the *s denoting arbitrary values. This is where things could be confusing. u and v don't play any role when we only have x and y because they would always have a coefficient of 0. Now if we were to remain in four dimensional space, then the variety of the former ideal would have y being anything it wants and the variety of the latter ideal would have x being anything it wants. But in reality, the ideal that we intended these two smaller ideals to be a union of could not have x being anything it wanted, nor could it have y being anything it wanted because we have established that x and y must simultaneously be 0. Does this mean that these ideals cannot be further broken up? Would this explain why we have the ideals \(<x,y>\) and \(<u,v>\) from the original ideal because we accepted that any value of arbitrary x and y with u and v both being 0 (and vice versa) was in the variety? Because in that case it worked the breaking down was fine, but if what I have said previously about breaking up the ideals further is true, then would that mean that the problem would end with these two reduced ideals?--Robert.Moray (talk) 21:37, 20 May 2013 (EDT)
- It can't be the case that \(\mathbb{V}(\left\langle x,y\right\rangle) = \mathbb{V}(\left\langle x\right\rangle)\cup \mathbb{V}(\left\langle y\right\rangle)\), since the varieties on the right are actually larger than the variety on the left. (The correct answer to the question is that the associated primes are \(\left\langle x,y\right\rangle\) and \(\left\langle u,v\right\rangle\). Also one can show that there is a singular point at the origin, where the irreducible components meet.) --Steven.Jackson (talk) 14:07, 22 May 2013 (EDT)
- In that case, if we continue with \(\mathbb{C}[x,y,u,v]\), and we wanted to claim that, for instance, the variety that the ideal \(<x,y>\) gives rise to is the union of \(\mathbb{V}(x) \cup \mathbb{V}(y)\), then these varieties would consist of the points of form (0,*,*,*) and (*,0,*,*) respectively, with the *s denoting arbitrary values. This is where things could be confusing. u and v don't play any role when we only have x and y because they would always have a coefficient of 0. Now if we were to remain in four dimensional space, then the variety of the former ideal would have y being anything it wants and the variety of the latter ideal would have x being anything it wants. But in reality, the ideal that we intended these two smaller ideals to be a union of could not have x being anything it wanted, nor could it have y being anything it wanted because we have established that x and y must simultaneously be 0. Does this mean that these ideals cannot be further broken up? Would this explain why we have the ideals \(<x,y>\) and \(<u,v>\) from the original ideal because we accepted that any value of arbitrary x and y with u and v both being 0 (and vice versa) was in the variety? Because in that case it worked the breaking down was fine, but if what I have said previously about breaking up the ideals further is true, then would that mean that the problem would end with these two reduced ideals?--Robert.Moray (talk) 21:37, 20 May 2013 (EDT)
- I think you're getting into trouble because you're changing the ring of polynomials (and hence the dimension of the affine space, the ideal generated by a given set of polynomials, and the corresponding varieties) in the midst of the discussion. Here we should fix the polynomial ring to be \(\mathbb{C}[x,y,u,v]\) for the whole discussion; then (for example) \(\mathbb{V}(\left\langle x,y\right\rangle)\) is the two-dimensional plane in which the \(x\) and \(y\) coordinates are fixed at zero but the \(u\) and \(v\) coordinates are arbitrary. So the question is whether (intuitively speaking) this two-dimensional plane can be the union of two curves. --Steven.Jackson (talk) 20:54, 20 May 2013 (EDT)
- So for instance the variety \(\mathbb{V}(<x,y>)\) contains only \((0,0)\in \mathbb{C}^2\). However, the origin in the two-dimensional space represents the intersection of the lines \(x=0\) and \(y=0\), so if we compressed this down to one dimension, wouldn't this just be the value of 0 on the y line and the value of 0 on the x line? If this is true, would this mean that we would have the union \(\mathbb{V}(<x>) \cup \mathbb{V}(<y>)\)? If this were true, then \(<x,y>\) is not irreducible, which would then mean that it isn't prime either. However, surely the ideals \(<x>\) and \(<y> \) are prime and irreducible. If this logic were true then it could also be applied to the ideal \(<u,v>\). Though the system suggests that the equations be solved simultaneously, thus an intersection of varieties, surely any element in an intersection qualifies as being an element in a union. --Robert.Moray (talk) 20:11, 20 May 2013 (EDT)
- So the hard part is deciding whether the sets \(\mathbb{V}(\left\langle x,y\right\rangle)\) and \(\mathbb{V}(\left\langle u,v\right\rangle)\) are irreducible. Probably I haven't given you the tools to decide this with complete rigor; however, relying to some extent on intuition, you may have some feel for whether a two-dimensional plane can be written as the union two proper subvarieties (which would need to be curves). (Deciding this question rigorously is equivalent to deciding whether \(\left\langle x,y\right\rangle\) and \(\left\langle u,v\right\rangle\) are prime ideals. If you've had some abstract algebra, ask yourself whether the corresponding quotient rings are integral domains.) --Steven.Jackson (talk) 18:52, 20 May 2013 (EDT)
- I see where you got this now, based on the two sets I originally proposed. Then this would imply that the ideal corresponding to the other set would be \(<u,v>\). Calculating the corresponding varieties of these ideals seems to indicate that each can contain only one point, in the former when \(x=y=0\) and in the latter when \(u=v=0\). However, since for instance the variety \(\mathbb{V}(x)\) only contains \(x=0\) and the variety \(\mathbb{V}(y)\) only contains \(y=0\), wouldn't this mean that \(\mathbb{V}(x,y)=\mathbb{V}(x) \cup \mathbb{V}(y)\)? This conclusion, however, did not rely on exploiting the relationship between primeness and irreducibility. EDIT: So I am conflicted now about whether things can be any more simplified than \(<x,y>\) and \(<u,v>\). I had come up with the alternative suggestion that since the corresponding varieties to these simplified ideals contain only the points where \(x=y=0\) and \(u=v=0\), this might mean two even more simplified ideals could be \(<x-y>\) and \(<u-v>\), but this seems to be less restrictive than the ideals that I claimed gave rise to these even more simplified ideals. How would I know when I reached the end? --Robert.Moray (talk) 10:39, 17 May 2013 (EDT)
- For the final, will we be required to know the algorithms for computing the sum of ideals, the product of ideals or the colon ideal (I am assuming that is what is meant by the algorithm of the ideal equivalent of the variety operation modulo the problem of determining the radical)? Also, I could not find algorithms in the book for computing them; all I could find were the set definitions. Am I to derive an algorithm from these definitions? --Robert.Moray (talk) 19:00, 22 May 2013 (EDT)
- If \(I = \left\langle f_1,\dots,f_s\right\rangle\) and \(J = \left\langle g_1,\dots,g_t\right\rangle\), then \(I + J = \left\langle f_1,\dots,f_s,g_1,\dots,g_t\right\rangle\) and \(IJ = \left\langle f_1g_1, f_1g_2,\dots,f_sg_t\right\rangle\). The book contains an algorithm based on Groebner bases that computes generators for \(I\cap J\), and it's probably good to know about that one. It also contains an algorithm that computes generators for \((I:J)\), but that one is rather involved, so I won't ask you to learn it in detail, just to know that it exists. --Steven.Jackson (talk) 21:13, 22 May 2013 (EDT)
