...that's why one Daniel Huybrechts has been spotted in the logic seminar here, or why one Peter Scholze is interested in formal verification absolute nobodies, right? :^)
but yeah, in pretty much all other mathematical disciplines there's enough courses here to feed the appetite of even a very "hungry" bachelor's student
and then you can reevaluate your choices when you decide where to do your masters
if you want to convince yourself that what I'm saying is true, just look at the course list
keep in mind that pretty much everything under the master's header can also be taken by bachelor students, although you can take only a limited amount of master's exclusive courses for credit
(but whatever you don't take for credit you can carry over into the masters' if you stay here)
i am naively coming into algebraic geometry. to my naive understanding, the historical motivation of the subject is to solve for common zeroes of a set of functions. so, why do I not see algebraic geometry applied to minimizing functions $f(x) \geq 0, \quad \forall x$, i.e. whose minima are zeroes?
@BenSteffan While I agree with your overall point about "solve" vs "study", I see no reason why you cannot "solve for the zeros of a multivariable polynomial".
@BenSteffan A solution an equation $f(x_1,x_2,\dotsc,x_n) = 0$ (where the $x_i$ are variables) is an $n$-tuple $(z_1, z_2, \dotsc, z_n)$ such that $f(z_1,z_2,\dotsc,z_n) = 0$. Solving an equation means describing the set of all solutions.
Solving $x^2 + x + 1 = 0$ means finding all of the solutions to that equation.
The fact that the set of solutions is a discrete (2 element) set isn't really relevant.
@XanderHenderson Say you have a multivariate polynomial in $n$ variables. You view this as a function $\mathbb{R}^n \to \mathbb{R}$. If 0 is a regular value it follows that the zero set is an $(n - 1)$-dimensional manifold. What description of this manifold could I give that counts a "solution"?
@XanderHenderson $\{f(z_1, z_2, \ldots, z_n) = 0\}$ is a description of the set of solutions.
Let $\tau$ be the Zarinski topology. Suppose $Y \in \tau$. Then, if $Y$ is irreducible (w.r.t. $\tau$), $Y = Z(T) \neq Z(T_1) \cup Z(T_2)$, so the zero set corresponding to $Y$ cannot be "factored" into the union of two distinct zero sets
@BenSteffan Yes. And I don't have a problem with that.
The point is that actually giving "nice" description of the solution set can be a pain in the ass, or no different than just saying "Well, it is the set of solutions. *shrugs*"
as i read the beginnings of hartshorne, i get the feeling that the machinery being built is motivated by very simple ideas, but I am trying to understand what those simple ideas are. the zero set definition for instance makes sense.
@BenSteffan To solve an equation (or system of equations) is to describe the set of all possible solutions. Sometimes, those sets defy "nice" descriptions.
But I don't see how you can draw a line at "discreteness". Does it not make sense to solve a system of linear equations, even if the solution happens to be some $n$-dimensional hyperplane?
The set of solutions to $$\begin{cases} x+y+z = 0 \\ x+2y-3z=0 \end{cases}$$ is some line in $\mathbb{R}^3$. "Solving" this system comes down to providing a better description of that set.
This is the core of a lot of the introductory material in a basic linear algebra class---how does one "solve" a system of linear equations?
My point is that the verb "to solve" is kind of useless. It still has meaning in more complicated situations, it just isn't the thing that you are actually trying to do. I was agreeing with you that "to study" is the better verb (for precisely this reason).
the thing about linear systems of equations is that you can describe a "solution" concisely: at most you have to describe a hyperplane, for which we have standard forms
Honestly, I don't think that we actually disagree all that much. Neither one of us wants to use "to solve" in the context of AG---I just think that the term is useless in that context, while you seem to think that it is actively wrong.
@BenSteffan Okay, but that is part of my definition: "nice" includes "concise". The intersection of a plane and a sphere is a circle, which can also be described "nicely".
@BenSteffan Again and again, I have argued that "solving" an equation (or system) is not a trivial or easy thing to. In order to "solve" something, you are looking for a concise, "nice" description of the set of all solutions.
Often, you aren't going to get anything more simple that the equation itself. Again, consider a circle---is there a nicer way of describing the set of solutions to $x^2 + y^2 =1 $ than to just write down that equation?
@BenSteffan Sure, but discreteness clearly does not match the way that the word is actually used, since I can cook up examples of "solving" some equation or system of equations where the set of solutions is not discrete.
@BenSteffan Okay... though even in the case of discrete examples, what are the solutions to $x^5 + x^4 + x^3 + x^2 + 1 = 0$? That polynomial does not have roots which can be expressed in radicals---so we can either rely on numerical methods to give approximate solutions, or we can just say that the set of solutions is the set of values which satisfy that equation.
and even there it's technically a misuse, because there are discrete solutions to things that cannot easily be described using a finite amount of data (although in context it is ok because this never happens for polynomials)
So, again, I'm perfectly happy to have a fairly general and vague definition of what it means to "solve" an equation. But then the point is that AGers don't "solve" equations, they study the solution sets.
Suppose a single logic binary relation (i.e., any of those in $\{\lor, \land, \to, \leftrightarrow\}$) were to take the place of addition in a continued fraction, with $\neg$ as taking multiplicative inverse and the values in the fraction being propositions. For example, for $\lor$, a "continued ...
Once I got here, I evaluated this integral and found that the term involving $\sin(\theta)$ it cancels out when integrating over $\theta \in [0,2\pi]$, but so at this point I can immediately say that the integral involving $\sin$ will be 0?
this works because the bounds of $r$ are not a function of $\theta$, in fact you can bring $\sin \theta$ out and you get $\int_0^{2\pi} \sin \theta d\theta \int_0^1 \dots dr=0$
@Malka: An alternative to Alessandro's suggestion would be to split $\mathbb N$ up into the powers of $2$, the powers of $3$, the powers of $5$, etc., together with all numbers that are not prime powers.
@Malka: If I am understanding Alessandro's suggestion correctly, I think he is saying that you should first try writing $\mathbb N\times \mathbb N$ as a disjoint union of countably infinite sets. Then, given a bijection $f:\mathbb N\to \mathbb N\times \mathbb N$, you can try lifting the decomposition of $\mathbb N^2$ back to a decomposition of $\mathbb N$.
> Proposition 0.14: If $\mathrm{card}(X)\leq\mathfrak{c}$ and $\mathrm{card}(Y)\leq\mathfrak{c}$, then $\mathrm{card}(X\times Y)\leq\mathfrak{c}$.
> Proof. It suffices to take $X=Y=\mathcal{P}(\mathbb N)$. ...
Why does it suffice to take $X=Y=\mathcal{P}(\mathbb N)$? I understand that the injections from $X,Y$ to say $\mathcal{P}(\mathbb N)$ induce bijections to subsets of $\mathcal{P}(\mathbb N)$, yet I don't see why it suffices to take $X=Y=\mathcal{P}(\mathbb N)$.
@psie: I think that the authors are just trying to concentrate on the case where $\operatorname{card}(X)=\operatorname{card}(Y)=\mathfrak c$. In the general case, if $X_0$ and $Y_0$ have cardinality $\le\mathfrak c$, then we can inject $X_0\times Y_0$ into $\mathcal P(\mathbb N)\times\mathcal P(\mathbb N)$.
If you prove that $\mathcal P(\mathbb N)\times\mathcal P(\mathbb N)$ has cardinality $\le\mathcal c$, then you are done.
Sorry if that was formulated confusingly. Does that make sense?
@psie: Let $f$ be an injection of $X_0$ into $\mathcal P(\mathbb N)$, and let $g$ be an injection of $Y_0$ into $\mathcal P(\mathbb N)$. The function $h$, given by $h(x_0,y_0)=(f(x_0),g(y_0))$, is an injection of $X_0\times Y_0$ into $\mathcal P(\mathbb N)\times\mathcal P(\mathbb N)$.
ok, it makes sense now. Thanks a lot. I see now how we are done by showing $\mathrm{card}(\mathcal P(\mathbb N)\times\mathcal P(\mathbb N))\leq\mathfrak{c}$.
@SineoftheTime I'm trying to solve some surface integrals from exam test, can I show you if I did it right? Even just the steps, not the calculations etc.
$\int_S \frac{y}{\sqrt{1+x^2+y^2}} d\sigma$ , where $S$ is the portion of the surface of equation $z = xy$ that projects into the semicircle $D = \{(x, y) : y ≥ 0, x^2+y^2 − 2x ≤ 0\}$.
Would it end up being a semicircle in the first quadrant?
Oh ok, in fact if I used the translated polar coordinates, I would have found $0 \leq r \leq 1$, so only in this case I just had to solve $\sin(\theta) \geq 0$ !
Now I am doing this: Calculate the flux of the vector field $F(x, y, z) = (x, y, z^4)$ through the surface S of the circular cylinder of equation $x^2+y^2 = 4$, bounded by the planes $z = −1$ and $z=1$.
I'm using the divergence theorem
since the vector field is defined in a closed space
Let $\mathbf{x} \in \mathbb{R}^n$. I must prove that $\lim_{\mathbf{x} \to 0} \frac{x_1}{\|\mathbf{x}\|}$ doesn't exist. My idea is to consider the restriction $x_i = 0$ for each $i \in \{2,\dots,n\}$ and so when $x_1 \to 0^+$ and $x_1 \to 0^-$ we have $\|\mathbf{x}\| = |x_1| \to 0$ but $\lim_{x_1 \to 0^+} \frac{x_1}{\|\mathbf{x}\|}=1$ and $\lim_{x_1 \to 0^-} \frac{x_1}{\|\mathbf{x}\|}=-1$. Is this right?
@SineoftheTime $\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = z \end{cases}$ , $r \in [0,2] , \theta \in [0,2\pi] , dV = r dr d\theta dz, z \in [-1,1]$
$x\ge 0 \implies \cos \theta \ge 0 \implies \theta \in [-\pi,\pi]$, on the other hand from the first relation you get $2\sin \theta \ge r\ge0$, hence $\sin \theta \ge 0 \implies \theta \in [0,\pi]$. Since both must hold, $\theta \in [0,\pi/2]$
