So I think it’s probably best to have a rotating study plan schedule that allows me to do certain topics on certain days. So far, I’m thinking of having a rotation that looks something like:
Differential Geometry -> Algebra -> Clifford Stuff -> Algebraic Topology (optional),
and since yesterday was (unofficially) differential geometry day, I’m going to spend today doing algebra.
First order of business: Eisenbud and Harris. And, since I’ve been meaning to write down some of the solutions to exercises I’ve passed, I guess I’ll do that here.
I-1. Find when is:
Proof. Recall that the elements of the set are the prime ideals of and that the 0 ideal is included (I’ll not write 0 hereon except when absolutely necessary). Therefore, for part 1, is the set of all where is prime. For (2), note that is a field, that every field is a PID, and that in a PID, maximal ideals and prime ideals are equivalent; moreover, note that in , for all integers relatively prime to , and because all integers are relatively prime to the prime number 3, .
On the other hand, isn’t a field – or even an integral domain – and so the possible elements of are of the form for (0 is already mentioned above, and due to 1,5 being relatively prime to 6). Note that isn’t prime since for . Simple computation verifies that are prime in , however, so . Because , part 4 is equivalent to part 2.
For part 5, being a field implies that is a PID, so it suffices to find all maximal ideals there, and because is a field (for a PID) precisely when is -irreducible, it follows that
Of course, is algebraically closed, whereby it follows that the collection of such ideal-generating polynomials is precisely the linear ones, namely those of the form , . Finally, note that consists of elements of the form for . Ignoring units, this ring therefore consists precisely of the linear polynomials , , and because of part 5, it follows that .
I-2. What is the value of the “function” 15 at the point ? At the point ?
Proof. The notation here is as follows. Elements are considered as functions on the space whose valuations are computed as: If , the quotient field of the integral domain consists of values defined to be the image of via the canonical maps . So then, because the quotient field of is itself (since is a field), the value of the “function” 15 in is simply . Similarly, is again a field, whereby the value of 15 there is given by .
I-3. (a) Consider the ring of polynomials , and let be a polynomial. Show that if is a number, then is a prime of , and there is a natural identification of with such that the value of at the point is the number . (b) Show that this holds in the general case as well.
Proof. As demonstrated above, the fact that is a PID and that is a field for all shows that is prime. Moreover, because the image of in is defined to be the image of in the quotient field of – that is, itself, since is maximal – the identification given is derived equivalently to the derivation of the fact that by way of the correspondence .
In the general case, is the coordinate ring of an affine variety over an algebraically closed field and is the maximal ideal corresponding to a point in the usual sense, whereby the conclusion is that and is the value of at in the usual sense. This result follows from a construction identical to that in case (a) due to the fact that being maximal implies that is a field that’s therefore equal to . Hence, because , corresponds to and corresponds to .