Another Sunday, or Awaiting Week 4

3 weeks.

I’ve officially survived the first three weeks of my second year of grad school (twice, actually). Again, I know keeping count of the days is a terrible thing to do to myself, particularly when there’s been a very small amount of good to come from my weeks thus far, but at this point, I’m sort of using that countdown as some sort of badge of accomplishment. Or something.

The coming weeks are going to be very very stressful and busy and stressful. Besides my usual load of stuff (I’m enrolled in 6 classes, I have a reading class in algebraic geometry starting up on Tuesday, I’m TAing for 1 lecture and 7 labs, and I’m trying to pick advisors / plan presentations I’ll need to give some timem soon), I also thought it was a good idea (remind me why?!) to make a poster to present at FSU’s upcoming Math Fun Day. That particular endeavor shouldn’t be especially difficult, but it requires time and time, ladies and gentlemen, is precisely what I have zero of.

Daunting is the adjective that comes to mind.

Also daunting is / was / has been the thought of continuing my goal to do all the problems in Hatcher. As you may recall, I spent the first half of summer slaving to acquire the information needed for the Chapter 0 exercises, only to have my plan for Chapter 1 totality derailed by that little piece of awesome that was my Wolfram internship. Long story short: The obsessive-compulsive part of me wants to not move forward until I hash out a Chapter 1 plan, but the This will benefit me in the class I’m taking now which, subsequently, hinges on my ability to understand Chapters 2 and 3 of Hatcher part wants to press forward.

I’m pleased to announce that the second guy won out.

In particular, my Hatcher Solutions page is showing signs of progress. It didn’t take as long as I’d predicted it to take to build that framework, and due to a random, unforeseen bout of sleeplessness at 3am this morning, I had precisely the opportunity needed to seize the moment. Right now, all those are empty pages, but I’m pleased to report that I seem to have accumulated approximately six solutions; if everything goes as planned, I’ll be taking time to update by including those as soon as possible.

In the meantime, I’m going to continue to hash out what to do about this paper. And what to do about the professors I’m emailing regarding potential advisor-hood. And what to do about the fact that I severely cut my weekend work time by spending yesterday ballin’ out of control in celebration of my wife’s birth. And what to do about….

Au revoir, internet. I bid thee well.

Oh, I just remembered: I have my first exam of the semester Friday. It’s on field theory. I’m less than pleased. :\

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The long, hard road to updates

So since coming back around here last week, I’ve been working on an update.

Of course, I’d be lying if I said I’d been working non-stop on an update, but I have, in fact, been working on one. I’d say I’m a solid 75% finished with it now, even though it’s (a) not going as quickly as I’d expected and (b) probably not going to be written the way I’d anticipated. Oh well; such is life, I guess.

I’m down to my last seven days of Wolfram employment, and to say I’m a sad robot is understatement of the year. I’m hoping the hustle and bustle of a new school term with new responsibilities and opportunities and excitements will curb that somewhat, but at this point, I’m not 100% convinced.

Among new things that have happened in the last week:

  • FSU made office assignments for the new year. Apparently I’m staying put. As much as I’d have liked a new office (you know, since my CEILING COLLAPSED AND DESTROYED ALMOST EVERYTHING I HAD THERE(!!!)), I don’t like the hassle that comes with doing something new. I’ll already be doing enough new things; figuring out a new office situation isn’t something I want to add to that list.
  • Fall schedules have been entered. I’m officially taking the third semesters of abstract algbera (field theory + categories, I think) and topology (advanced algebraic topology), as well as a course on complex manifolds (taught from an algebraic geometry perspective, I’d guess) and Riemannian manifolds. I’m excited. Sincerely.
  • I’ve gotten my change of residency file 95% compiled. Monday will be the day to finish it off and submit it.
  • I’ve finally started training for my new gig with Pearson. I’m less than thrilled with the progress so far. We’ll see.

Otherwise, things have been kinda the same: I’ve been doing a lot of Wolfram stuff, I’ve been taking more time away to hang with my family, and I’ve been somehow managing to not think about the nervousness I’ll invariably feel when new TA responsibilities, etc., pick up.

Things are good, I’d say, even though I’ve got a lot of things I need to start doing otherwise. I need to start reading the books potential advisers have suggested; I need to start doing more independent research; I need to start getting back in school mode.

For the first time in a really really long time, I’m enjoying being in not-school mode. I wonder if this is that changing tide I always heard so much about.

Anyway, expect a new content entry soon enough. And maybe some to follow that one. We’ll see.

Peace.

Verifying Easy Properties, or Nowhere, Going Nowhere

Whenever I decide to learn something – and especially when it’s learning for learning’s sake – I make sure to be meticulous with things. In particular, whenever I see propositions stated without proof, I break out the old pen and paper and start verifying.

The purpose of this post is to examine a few of the properties on page 661 of Dummit and Foote. Some of the background notation needed is discussed in this previous entry.

Claim. The following properties of the map \mathcal{I} are very easy exercises. Let A and B be subsets of \mathbb{A}^n.
   (7) \mathcal{I}(A\cup B)=\mathcal{I}(A)\cap\mathcal{I}(B).
   (9) If A is any subset of \mathbb{A}^n, then A\subseteq\mathcal{Z}(\mathcal{I}(A)), and if I is any ideal, then I\subseteq\mathcal{I}(\mathcal{Z}(I)).
   (10) If V=\mathcal{Z}(I) is an affine algebraic set then V=\mathcal{Z}(\mathcal{I}(V)), and if I=\mathcal{I}(A) then \mathcal{I}(\mathcal{Z}(I))=I, i.e. \mathcal{Z}(\mathcal{I}(\mathcal{Z}(I)))=\mathcal{Z}(I) and \mathcal{I}(\mathcal{Z}(\mathcal{I}(A)))=\mathcal{I}(A).

Proof. (7) Note that A\cup B= (A\setminus B)\sqcup (B\setminus A)\sqcup (A\cap B). In particular, a polynomial f vanishes on A\cup B if and only if it vanishes on each disjoint component of (A\setminus B)\sqcup (B\setminus A)\sqcup (A\cap B) separately, which happens if and only if it vanishes on the entirety of A and on the entirety of B.

(9) Suppose A\subset\mathbb{A}^n. Clearly, any polynomial f which vanishes on A is in \mathcal{I}(A), and because f\in\mathcal{I}(A) if and only if f\equiv 0 on A, A is certainly contained in the zero set of f. Therefore, A\subseteq\mathcal{Z}(\mathcal{I}(A)). Note that the inclusion doesn’t necessarily reverse since f(x_1,\ldots,x_n) may vanish for some element (x_1,\ldots,x_n)\not\in A.

Next, suppose that I is an ideal of the ring k[x_1,\ldots,x_n] and that f\in I. By definition, the locus \mathcal{Z}(I)\subset\mathbb{A}^n is the collection of all points (a_1,\ldots,a_n)\in\mathbb{A}^n for which f(a_1,\ldots,a_n)=0 for all f\in \mathcal{I}. Then, certainly, for all a=(a_1,\ldots,a_n)\in \mathcal{Z}(I), f(a)=0, i.e., f is an element in the ideal \mathcal{I}(\mathcal{Z}(I)) that vanishes on \mathcal{Z}(I). Hence, f\in I implies that I\subseteq \mathcal{I}(\mathcal{Z}(I)).

(10) First, suppose that I is an ideal and that V=\mathcal{Z}(I) is an affine algebraic set. It suffices to show that V=\mathcal{Z}(\mathcal{I}(V)) by way of two-sided inclusion. To that end, let a=(a_1,\ldots,a_n)\in V. Then a is in the zero-set of the ideal I, whereby it follows that f(a)=0 for all f\in I. But for any f satisfying f(a)=0 for arbitrary a\in V, f\in\mathcal{I}(V) and a\in\mathcal{Z}(\mathcal{I}(V)) by definition. Hence, V\subset\mathcal{Z}(\mathcal{I}(V)).

Conversely, if a\in\mathcal{Z}(\mathcal{I}(V)), then f(a)=0 for all f\in\mathcal{I}(V). But all such functions f disappear for all values v\in V=\mathcal{Z}(I) by definition of \mathcal{I}(V). This means that the polynomials f\in\mathcal{I}(V) for which f(a)=0 are precisely the functions which satisfy f(v)=0 for all v\in{Z}(I). Hence, a itself must be an element of \mathcal{Z}(I)=V, whereby the equality is proved.

The other expression is proved similarly and is omitted for brevity. Therefore, as claimed,

\mathcal{Z}(\mathcal{I}(\mathcal{Z}(I)))=\mathcal{Z}(I) and \mathcal{I}(\mathcal{Z}(\mathcal{I}(A)))=\mathcal{I}(A),

from which it follows that the maps \mathcal{Z} and \mathcal{I} are inverses of one another under the construction given here.    \square

Dummit and Foote Example

I found one particular example – namely Example 2 on page 660 of Dummit and Foote – to be a good exercise in the sense that it tied together a lot of ideas from earlier parts of the book. In order to share, though, I need to give a little background.

Throughout, let k denote a field. The affine n-space \mathbb{A}^n over k is the set of all n-tuples (k_1,k_2,\ldots,k_n) where k_i\in k for all i. For a general subset A\subset \mathbb{A}^n, the ideal \mathcal{I}(A) is called the ideal of functions vanishing at A and is defined to be

\mathcal{I}(A)=\{f\in k[x_1,\ldots,x_n]\,:\,f(a_1,\ldots,a_n)=0\text{ for all }(a_1,\ldots,a_n)\in A\}.

It’s easily verified that \mathcal{I}(A) is indeed an ideal and that it is, by definition, the unique largest ideal of functions which are identically zero on all of A. With that in mind, consider the aforementioned example:

Page 660, Example 2. Over any field k, the ideal of functions vanishing at (a_1,\ldots,a_n)\in\mathbb{A}^n is a maximal ideal since it is the kernel of a surjective (ring) homomorphism from k[x_1,\ldots,x_n] to the field k given by evaluation at (a_1,\ldots,a_n). It follows that

\mathcal{I}((a_1,\ldots,a_n))=(x_1-a_1,x_2-a_2,\ldots,x_n-a_n).

Proof. As mentioned above, \mathcal{I}((a_1,\ldots,a_n)) is certainly an ideal. To show that it’s a maximal ideal in k[x_1,\ldots,x_n], first define \varphi:k[x_1,\ldots,x_n]\to k by the action that sends f=f(x_1,x_2,\ldots,x_n) to the constant f(a_1,a_2,\ldots,a_n)\in k. Verifying that this is a ring homomorphism is trivial, and given an element x, x=f(x,0,\ldots,0) where f(x_1,\ldots,x_n)=x_1 is a polynomial in k[x_1,\ldots,x_n]. Moreover, the kernel of \varphi consists precisely of those elements in f\in k[x_1,\ldots,x_n] for which f(a_1,\ldots,a_n)=0, whereby it follows that \ker(\varphi)=\mathcal{I}(A) where A=\{(a_1,\ldots,a_n)\}\subset\mathbb{A}^n.

Hence, \varphi is a surjective ring homomorphism with kernel \mathcal{I}(A). In particular, by the first (ring) isomorphism theorem, k[x_1,\ldots,x_n]/\mathcal{I}(A)\cong\text{Im}(\varphi) where \text{Im}(\varphi)=k by surjectivity of \varphi. Clearly, then, k[x_1,\ldots,x_n]/\mathcal{I}(A) is a field, something that happens if and only if the ideal \mathcal{I}(A) is maximal. Therefore, the first claim is proved.

For the second claim, note that the ideal (x_1-a_1,\ldots,x_n-a_n) is an ideal that vanishes precisely on A. Clearly, then, \mathcal{I}(A)\supseteq (x_1-a_1,\ldots,x_n-a_n). On the other hand, an element f\in\mathcal{I}(A) is an element that vanishes at the point (a_1,\ldots,a_n), whereby it follows that

\displaystyle f(x_1,\ldots,x_n)=g(x_1,\ldots,x_n)\prod_{i=1}^n(x_i-a_i)^{\alpha_i}

for some positive powers \alpha_i. Then, clearly, f\in (x_1-a_1,\ldots,x_n-a_n) and so \mathcal{I}(A)\subseteq (x_1-a_1,\ldots,x_n-a_n). This concludes the proof.   \square