Study Plan, tentatively, + Algebraic Geometry Exercises

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 \text{Spec} R when R is:

  1. \mathbb{Z}
  2. \mathbb{Z}/(3)
  3. \mathbb{Z}/(6)
  4. \mathbb{Z}_{(3)}
  5. \mathbb{C}[x]
  6. \mathbb{C}[x]/(x^2).


Proof. Recall that the elements of the set \text{Spec}R are the prime ideals of R and that the 0 ideal is included (I’ll not write 0 hereon except when absolutely necessary). Therefore, for part 1, \text{Spec}R is the set of all \mathbb{Z}/(p) where p is prime. For (2), note that \mathbb{Z}/(3) 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 \mathbb{Z}/(n), (d)=R for all integers d relatively prime to n, and because all integers are relatively prime to the prime number 3, \text{Spec}R = 0.

On the other hand, \mathbb{Z}/(6) isn’t a field – or even an integral domain – and so the possible elements of \text{Spec}R are of the form (d) for d = 2,3,4 (0 is already mentioned above, and (1)=(5)=\mathbb{Z}/(6) due to 1,5 being relatively prime to 6). Note that (4) isn’t prime since 4=2\cdot 2 for 2\not\in(4). Simple computation verifies that (2),(3) are prime in \mathbb{Z}/(6), however, so \text{Spec}R=\{(2),(3)\}. Because \mathbb{Z}_{(3)}\cong\mathbb{Z}/(3), part 4 is equivalent to part 2.

For part 5, \mathbb{C} being a field implies that \mathbb{C}[x] is a PID, so it suffices to find all maximal ideals there, and because K[x]/(f(x)) is a field (for K[x] a PID) precisely when f(x) is K[x]-irreducible, it follows that

\text{Spec}R=\{(f(x))\,:\,f(x)\text{ is irreducible in }\mathbb{C}[x]\}.

Of course, \mathbb{C} is algebraically closed, whereby it follows that the collection of such ideal-generating polynomials is precisely the linear ones, namely those of the form f(x)=x+\alpha, \alpha\in\mathbb{C}. Finally, note that \mathbb{C}[x]/(x^2) consists of elements of the form ax+b for a,b\in\mathbb{C}. Ignoring units, this ring R therefore consists precisely of the linear polynomials x+\alpha, \alpha\in\mathbb{C}, and because of part 5, it follows that R=\text{Spec}R.  \square

I-2. What is the value of the “function” 15 at the point (7)\in\text{Spec }\mathbb{Z}? At the point (5)?
Proof. The notation here is as follows. Elements f\in R are considered as functions on the space \text{Spec} R whose valuations are computed as: If x=[\mathfrak{p}]\in\text{Spec}R, the quotient field \kappa(x)=\kappa(\mathfrak{p}) of the integral domain R/\mathfrak{p} consists of values f(x) defined to be the image of f via the canonical maps R\to R/\mathfrak{p}\to\kappa(x). So then, because the quotient field of R_1=\mathbb{Z}/(7) is R_1 itself (since R_1 is a field), the value of the “function” 15 in R_1 is simply 15\mod 7 = 1. Similarly, R_2=\mathbb{Z}/(7) is again a field, whereby the value of 15 there is given by 15\mod 5 = 0.  \square

I-3. (a) Consider the ring of polynomials \mathbb{C}[x], and let p(x) be a polynomial. Show that if \alpha\in\mathbb{C} is a number, then (x-\alpha) is a prime of \mathbb{C}[x], and there is a natural identification of \kappa((x-\alpha)) with \mathbb{C} such that the value of p(x) at the point (x-\alpha)\in\text{Spec }\mathbb{C}[x] is the number p(\alpha). (b) Show that this holds in the general case as well.
Proof. As demonstrated above, the fact that \mathbb{C}[x] is a PID and that \mathbb{C}[x]/(x-\alpha) is a field for all \alpha\in\mathbb{C} shows that (x-\alpha) is prime. Moreover, because the image of p(x) in \kappa is defined to be the image of p in the quotient field of \mathbb{C}[x]/(x-\alpha) – that is, \mathbb{C}[x]/(x-\alpha) itself, since (x-\alpha) is maximal – the identification given is derived equivalently to the derivation of the fact that \mathbb{C}\cong\mathbb{R}[x]/(x^2+1) by way of the correspondence a+bi\in\mathbb{C}\sim ax+b\in\mathbb{R}/(x^2+1).

In the general case, R is the coordinate ring of an affine variety V over an algebraically closed field K and \mathfrak{p} is the maximal ideal corresponding to a point x\in V in the usual sense, whereby the conclusion is that \kappa(x)=K and f(x) is the value of f at x in the usual sense. This result follows from a construction identical to that in case (a) due to the fact that \mathfrak{p} being maximal implies that V/\mathfrak{p} is a field that’s therefore equal to K. Hence, because x\in V\sim\mathfrak{p}\in\text{Spec}R, \kappa(x) corresponds to K and f(x)\in\kappa(x) corresponds to f(x)\in K.  \square

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s