It's a great book. I love the style. Not many words, but always enough to work out what they're saying.
I think when I'm handed a perfect, complete proof, it tends to wash over me much more easily.
It's a bit slower going though. I've spent a good number of hours trying to digest everything in the first chapter. Some of the techniques are very clever.
Some sections you can skip if you want (covered in the commutative algebra class, not covered on the qual): 4, 9 onwards. I'm not even sure ch5 is covered.
Lots of canonical examples show up if you restrict to the k-algebra case. Here's a nice one: if $K/k$ is a nontrivial field extension, $K \otimes_k K$ is not an integral domain.
I was probably going to read them, if only because they're useful for algebraic number theory. But it's a relief to know I don't have to know them for the quals, which are intimidating enough at the moment.
@robjohn this was my idea $$\lim_{\large x\to i} \left(\frac{1}{2x^{4\infty}}-\frac{1}{2^2x^{4\infty-1}}+\frac{1}{2^3x^{4\infty-2}}- \cdots-\frac{1}{ 2^{4\infty-2}x^3}+\frac{1}{2^{4\infty-1} x^2}- \frac{1}{2^{4\infty}x}\right)$$
Using vector methods show that the distance between two non parallel lines $l_1$ and $l_2$ is given by $$d=\frac{|(\overrightarrow{v}_1 - \overrightarrow{v}_2) \cdot (\overrightarrow{ a}_1 \times \overrightarrow{a}_2)|}{||\overrightarrow{a}_1 \times \overrightarrow{a}_2||}$$ where $\vec{v}_1$ and...
@robjohn I didn't send that one to AMM. To AMM I sent the generalization of the problem with falling factorial only (all was nicely done, following the rules).
@robjohn OK. Let me know if the solution seems obvious to this last one (when you have time). Sometimes I create problems and I might miss at first sight some obvious way.
$$\phi(A^\circ)=\phi\left(\bigcup_{U\subseteq A}U\right)=\bigcup_{\phi(U)\subseteq\phi(A)}\phi(U)=\bigcup_{V\subseteq\phi(A)}V=\phi(A)^\circ $$ where $\phi$ is a homemorphism, $U$ ranges over open subsets of $A$, and $V$ ranges over open subsets of $\phi(A)$.
It's very sad to see all your high school friends doing so well in life while you are struggling with mental illness, when you were smarter than all of them.
@ᴇʏᴇs Can you understand how I feel? I could be doing so many things with my time but instead it has to be spent like that. No words can describe the pain.
First and foremost, @JasperLoy, you have to love and take care of yourself.
The "world" isn't going to do this for you-in fact, your personal experience should tell you otherwise, the world is often HARMFUL to you.
So you have to shelter yourself, whatever it takes for that to happen.
If you were smarter than your friends in high school, then you STILL are. Some people are lucky, they have good parents, they never encounter demons that devour them. It may be hard to do, but you have to shut that out. You are on YOUR path-one's individuality, one's very singular identity is a GIFT.
Just making a joke @MikeMiller. Pete (in the tradition of Bourbaki apparently) does this thing where he'll add "quasi" in front of anything that isn't necessarily Hausdorff.
of course it's the algebraists, who care about non-hausdorff spaces, that call compactness quasi-compactness; and the topologists, who care less about non-hausdorff spaces, who say 'compact hausdorff'
I heard from Pete, @Kaj. Congratulations. I wish he'd put one more challenging problem on there to distinguish a bit more. But that's my style. I digress, too.
No, he's working them plenty, @Mike. He's just copping out on testing. He tends to do that.
For example, teaching calculus, he can't be bothered to teach/test word problems. :(
Perhaps that's a slight exaggeration, but it's pretty much what he's told me.
@TedShifrin hey Ted! I'm still muddling through this exterior algebra stuff. Now that I've learned a bit, I want to know what I can use it for. Oh, and I learned what a ring is. That's stuff is awesome. I didn't know they were so cool
@Mike: My San Diego friends just sent me a Craigslist link to a gorgeous condo some friends of theirs are renting out. Just a bit too large (and rather too expensive) for me. Sigh.
@TedShifrin Yeah, the tensor product is really cool. And actually going through that has made me realize I don't know what a form is at all. I thought it was just a map $V \rightarrow \Bbb{R}$ but clearly that was not true at all. I think that's the disadvantage of reading a physicists view first.
If you want to see far-reaching consequences (a lot of Elie Cartan's handiwork), take a look at Exterior Differential Systems, by Bryant, Griffiths, Chern, Goldschmidt (in some order). Lots of partial differential equations and applications to differential and algebraic geometry.
oh, @Stan, there's all sorts of cool Euclidean geometry, even. Go get Pedoe's book Geometry: A Comprehensive Course (oh, and he uses exterior algebra bit in it).
I didn't realize they were fundamental theorems until I read that. So I felt like Apostol just didn't organize it as well. It's overall superb, but I think easier if you know some of it already.
no, Apostol's fine, but perhaps as a number theorist that stuff isn't his forte, @Stan. Check out Bamberg and Sternberg, A Course in Mathematics for Students in Physics.
i.e. that if you're working with block matrices whose block elements commute, taking the entire determinant is the same as taking the determinant block-wise and then again element-wise
@Semiclassical. Yes, if $A,B,C,D$ are $n\times n$ and $AC=CA$, then $$\det\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \det(AD-CB),$$ if I remember correctly.
I'll give you the hint for the proof when $A$ is invertible. You can then use various methods to deduce that it holds even when $A$ is not.
but i figure there has to be something more direct that just uses the abstract characterization of the determinant @Ted (i may, of course, be misguided in that hope)
They still have a Spivak freshman course. We finally abandoned ours ... Chicago may be the only university that basically does not accept AP credit. That makes a big difference.
@Semiclassical: But we've discussed that there is no abstract description of a determinant over a noncommutative ring, @Semiclassical. Not even something as nice as the quaternions.
@Semiclassical: My proof is based on the ideas of row operations, or basically, on the product rule for determinants. There is an exterior algebra interpretation of that if you don't like a standard proof of the product rule for determinants.
It's called a job, @Stan ... at a good department ... but the way times are changing in American universities, it's pretty clear to me that they'll never hire someone like me again.
No, no, I grew up in Berkeley and Boston, @Stan ... heading back to CA for the rest of my life.
@TedShifrin Peter Higgs won the Nobel prize last year in physics for predicting this thing called the Higgs Boson. And he said he doesn't think he would be hired in today's academic environment.
Let $R$ be a commutative subring of $F^{n\times n}$, where $F$ is a fi eld (or a commutative ring), and let $\mathbf{M}\in R^{m\times n}$. Then $\det_F(\mathbf{M})=\det_F(\det_R(\mathbf{M}))$.
