@ÉricoMeloSilva I suppose the more proper statement of Poe now would be: even if the author says otherwise, there is no way to distinguish between genuine conviction and parody
The author remains dead even when they start jumping up and down :P
@Semiclassical maybe you can get quantative: The more satirical and obviously crazy a claim is the more likely it is that the author actually holds that position unironically
that's not always true but it's true way more often than it should be
Okay so you're thinking some kind of orthogonal projection. That would seem helpful since the image of $\exp$ is $\mathbb{C}^{\times}$. But then the question is, if you take a variety and you project it linearly, do you get a variety?
@Liad I mean you just answered it, $V(xy-1)$ is a variety in $\mathbb{A}^2$, but if you project to the first coordinate you get $\mathbb{A}^1\setminus \{0\}$, which isn't a variety
So the projection idea isn't exactly what you're going for. But is there something you can definitely do to a variety that gets you a 1-dimensional variety?
Eh, it could be better. We're beginning to do schemes now though and I like this part more than all the sheaves and prevarieties business we've done so far
(Also with our definitions $\Bbb A^1\setminus\{0\}$ is a variety, but it's not closed in $\Bbb A^1$ so it still works to show that the projection isn't closed)
The dimension of a variety is the degree of the Hilbert polynomial, and the degree of a variety is $n!a_n$ where the leading term of the polynomial is $a_nx^n$
We went the abstract route, for us a variety is a separated prevariety and a prevariety is a topological space with a sheaf of $k$-valued functions $(X,\mathcal O_X)$ with an open cover $U_i$ such that $(U_i,\mathcal O_{X|U_{i}})$ is isomorphic to an affine irreducible set in $k^n$ for some $n$ with its structure sheaf
And for example, we showed that given a finite set, the Hilbert polynomial is a constant, and defined, for $p\in X\cap Y$, $I_p(X,Y) = h_J$ where $J$ is the $I(p)$-primary component of $I(X) + I(Y)$
Now, in our most recent pset we had a problem to show that if $X$ and $Y$ are curves in $\mathbb{P}^2$ and $p\in X\cap Y$, then $I_p(X,Y) = 1$ if and only if $p$ is a non-singular point of each, and the tangent planes coincide.
That's very intuitive but... how do you even compute any of this with the Hilbert polynomial? No one I talked to had a clue how to approach this problem
Also just like, idk, I feel it's simultaneously kinda hard to work with and also doesn't really /feel/ like it should say all that. Like, dimension kinda means something going in, as does degree, and while we computed some examples that it's plausible it lines up... Having defined degree this way makes it feel like the only content of Bezout's theorem, for example, is that a finite set has Hilbert polynomial at least its cardinality
Lol, you see this is exactly why I'm of the belief that one should be willing to spend as long as it takes working with algebra and proving that multiple definitions are equivalent, and then you can work with the most convenient one at any given time as opposed to being stuck with one and running around in the dark
We also proved that this is equivalent to taking the sup of the length of chains of irreducible closed subsets which makes waaaay more sense intuitively to me
This is how I'd define it and works for arbitrary rings (agrees with Krull dimension of a ring) - so you can also talk about $\operatorname{Spec} \mathbb{Z}$ being one-dimensional, $\operatorname{Spec} \mathbb{Z}[x]$ being two dimensional and so on --- and then you can really start to see more analogies between working over number fields and function fields (if you care about number theory)
that said though this definition is quite hard to compute - so it's good that there's some theorem (transcendence degree) allowing you to compute it, in the case of varieties over a field!
Loch: So, let's say one has a vague idea of how varieties work. A class' worth but where the definitions are kinda awkward so there are definitely holes. Next quarter I'm taking grad commutative/AG (I can send the notes that were used last year, though apparently even the prof felt it was a disaster so he may change it up this time)
What would you think is a good idea? Doing a more a standard treatment of old-school AG? Just jumping into Hartshorne/Vakil?
You might know this already but in case you don't -- you can define the degree of a variety $X\subset \mathbb{P}^n$ to be the following: say the dimension of $X$ is $k$, then its degree is the number of points you get if you intersect your variety with $k$ general hyperplanes
(This clearly agrees with what you expect in the case of curves in $\mathbb{P}^2$ !)
You'll have to think a bit to see why it agrees with the hilbert polynomial definition though
Speaking of $\operatorname{Spec}(\Bbb Z[x])$ being two dimensional if I looks at the morphism $\operatorname{Spec}(\Bbb Z[x])\to\operatorname{Spec}(\Bbb Z)$ induced by the inclusion $\Bbb Z\to\Bbb Z[x]$ I have that it is sending all the points of the form $(p,f(x))\mapsto p$ for $p$ prime and $f(x)$ irreducible (similarly for the preimage of $(0)$).
So I imagine $\operatorname{Spec}(\Bbb Z[x])$ as a "line" corresponding to $\operatorname{Spec}(\Bbb Z)$, and over every $(p)$, going in the second dimension, there are all of the $(p,f(x))$ so that this morphism from earlier actually looks like a projection.
Though it was kinda weird. There were two parts, first part defined it as $\max \{|X\cap L|\}$ where $L$ is an $n-\dim(X)$ dimensional linear subspace of $\mathbb{P}^n$
One direction is easy by Bezout, the other I couldn't figure out honestly. And the other part was to show that the space of appropriate linear subspaces attaining that max is Zariski open
It's very heavily commutative algebra - which I guess makes sense since it's a commutative/AG class :p
I guess it depends on whether you want to spend more time on commutative algebra (which is very important - but I actually forgot how I learnt these things back then..)
I think Vakil's text is intended for people with not that much commutative algebra background - and you can start play with fancy words like sheaves and schemes so that might be appealing to you
@AlessandroCodenotti Yeah - have you every looked at "Mumford's treasure map?" - it's a picture of Spec Z[x], and it's basically what you described
in fact you can actually interpret your fibers over (p) to be Spec F_p[x]
(you can see this set-theoretically now - but once you learn fibre products you'll see why the scheme-theoretic fiber should correspond to Spec F_p[x])
I was like yeah, thinking along the lines of Sard's theorem we can see that you're gonna intersect transversally except for a "null set of lines", and intuitively transversal intersections are multiplicity 1, so that'll do it
But that's kinda cheating, couldn't honestly connect to Hilbert polynomial
@loch hmm that's where my intuition breaks down. If I look at "horizontal" lines, that is I fix $f$ irreducible and vary $p$ in $(p,f)$ this looks like $\operatorname{Spec}(\Bbb Z)$ because of the projection. But what if I project on the other axis? That is I fix $p$ and vary $f$, that's weirder
I mean, you already told me what that is, I don't see (yet) why it makes sense
@Daminark Yeah intuitively you're absolutely right --- kind of the only thing missing is to how to realise this "null set" as a Zariski closed set (cut out by polynomials). But yeah the argument is probably something along the lines of - argue that being tangential to your variety is an algebraic condition (this is probably the harder part i think?), and argue that once the intersections are transverse then the multiplicity is 1
Let's be slightly careful here by what you mean when you're looking at fixing $f$ and looking at "horizontal lines" - perhaps you meant the set of prime ideals which are of the form $(p,f)$ -- but I don't think this is Zariski closed! What you can think of has "horizontal lines" (which are the ones drawn in Mumford's picture) are $V(f)$, where $f$ is some irreducible polynomial in $\mathbb{Z}[x]$. For example you can take $f=x^2+1$.
The closed subscheme corresponding to an ideal is $\operatorname{Spec}A/I$, so in this case it is $\operatorname{Spec} \mathbb{Z}[x]/(x^2+1) = \operatorname{Sp…
@MatsGranvik Ah, your n can become larger than k for some values, thus you also plotted some rationals > 1. I am assuming that the left end is close to zero though?
@MatsGranvik Ok so what you plotted there are the rationals of the form n/k up to denominators 12 in the interval [0,12]. As you may have noticed, in that form, there aren't many rationals > 1
