> “If you lack background knowledge about the topic, ample evidence from the last 40 years indicates you will not comprehend the author’s claims in the first place,” wrote Willingham, citing his own 2017 book.
just looked at the scores my students got on their first quizzes. two of my sections are part of one lecture group, the other two in the other. hence there's two different quizzes between them and two different graders
the first problem is supposed to be an easy one from the HW. I graded one of the quizzes, and my most common score was 25/25 (though that wasn't a plurality by any means)
intro physics. (the problem was on kinematics, specifically)
in my book, 21-25 = only minor mistakes, 17-20 = at least one major mistakes, 10-16 = answer was substantially flawed, 1-10 = "I can't tell if you've ever sat in class."
I only gave one score lower than a 10 on this quiz problem, and that was to someone who only did 60 km/hr = 0.025 km/s.
So my threshold for getting at least 10/25 is not very high.
Hence a 6/25 plurality is to my eyes simply mental
The first one is a hyperbolic space resembling a Smith Chart
In the dream, only the square with the stairwell is visible (and all other squares are implied to be the exact same stairwell tiled all over).
Walking in any direction slows to a crawl. A later scene reveals the reason is because the whole world rotates as you walk, and the world rotates only slightly slower in the direction you walk, thus the net velocity that is perceived as movement is very small
The 2nd and 3rd are a ring (unknot) and a strange knot like object called a tubloid. A tubloid when seen from the side looks like a sine wave, unlike the ring which is a straight line. Both look like circles as seen from the top.
However, compared to a ring, a tubloid behaves differently in the following way:
1. When you pull one half of the ring over the other half as shown, that loop will get stuck and you make a knot
2. But for a tubloid, doing the same thing you cannot make a knot, because its twisted nature means the two overlapping strands shown can rotate out of the way by switching place, thus the loop will not encounter an obstacle and simply fall through like a loose loop of strand does
Thus, a tubloid has the weird property that you can do a full 360 rotation of the half loop over another strand without get stuck
if you have infinitely many non-zero terms with negative exponents you have an essential singularity, if you have at least one non-zero term with a negative exponent you have a pole and if you have no negative exponent terms you have a removable singularity
and for meromorphic functions, you only allow poles as singularities
if you work with meromorphic functions $\mathcal{M}(U)$ on a domain $U$ you can also apply a lot of algebra: meromorphic functions on $U$ are a field, in fact the quotient field of holomorphic functions $\mathcal{O}(U)$ on $U$. And for each $z \in U$, you can define a discrete valuation on $\mathcal{M}(U)$ by defining $v_z(f)$ to be $n>0$ if $f$ has a zero of order $n$ and $v_z(f)=-n$ if $f$ has pole of order $n$ and $0$ else
you can use analytic results from complex analysis to show algebraic results about $\mathcal{O}(U)$ which is a fun exercise. For instance Weierstraß factorization implies that $f \mid g$ in $\mathcal{O}(U)$ iff $v_z(f) \leq v_z(g)$ for all $z \in U$
(that's not stuff you need just neat things about complex analysis from an algebraic pov)
using this result regarding divisbility, it's easy to show that elements in $\mathcal{O}(U)$ admit a gcd and a little more work even shows that you can always write $\gcd(f,g)=af+bg$ for some $a,b \in \mathcal{O}$, so we have a Bezout domain
but $\mathcal{O}(U)$ is never Noetherian
so we have naturally occuring example of a non-Noetherian Bezout domain
lmao, it seems that for my thesis I need to work out some stuff for which according to Kevin Buzzard "the only useful reference seems to be Brian Conrads's brian"
@BalarkaSen I saw the question, but I'm not sure about the answer
the map t->(t^2,t^3,t^5) is a bijective morphism which is not an isomorphism of affine varieties
that alone doesn't really tell us about the vanishing ideal being 2-generated though, if we look at t->(t^2,t^3,t^6), then the image is V(x^3-y^2,y^2-z)
$f_{3}: \mathbb{Z} \rightarrow \mathbb{Z} \text { with rule } f_{3}(x)=\lfloor x / 2\rfloor+\lceil x / 2\rceil$
Can you please tell me whether this is identity function on its domain ?
I guess yes because $\lfloor x / 2\rfloor$ = $x /2$ $-$ {$x$/2} and same for ceiling function so when we add ...
@BalarkaSen let $\zeta_3$ be a primitve third root of unity and consider the point $(\zeta_3,-1,\zeta_3)$, this point is contained in $V(x^3-y^2,x^5-z^2)$, but not in $V(x^3-y^2,x^5-z^2,y^5-z^3)$
Good morning guys, suppose we have a $\sigma$-finite measure space $(X,\mathcal A,\mu)$ and let $\delta:=\{A\in\mathcal A:\mu(A)<\infty}$
$\delta:=\{A\in\mathcal A:\mu(A)<\infty\}$
then my textbook says that because of $\sigma$-finiteness of $\mu$ is easy to see that $\sigma(\delta)=\mathcal A$
I understand that the sigma finiteness of $\mu$ implies that there's an exhausting sequences with finite measure, but I don't quite follow the reasoning to make that assertion
@MatheinBoulomenos If I let $I = (x^3 - y^2, x^5 - z^2, y^5 - z^3)$ and $m = (x, y, z)$, then $x^3 - y^2, x^5 - z^2, y^5 - z^3$ are independent in $I/mI$ as a $\Bbb C$-vector space, right? If $c_1(x^3 - y^2) + c_2(x^5 - z^2) + c_3(y^5 - z^3) \in mI$ then there can be no pure term of $x$ with degree $3$, nor a pure term of $z$ with degree $2$, which forces $c_1 = c_2 = 0$. But this would force $c_3 = 0$ likewise.
@RScrlli if you take any $A \in \mathcal{A}$ and you have an exhausting sequence $X=\bigcup_{n=1}^\infty X_n$ such that $\mu(X_n)<\infty$, then $A=\bigcup_{n=1}^\infty (A \cap X_n)$ where $\mu(A\cap X_n) \leq \mu(X_n) < \infty$
Eg that's why $(y - x^2) + (y) = (x^2, y)$ in $\Bbb C[x, y]$. The hyperusrfaces $y = x^2$ and $y = 0$ do not intersect transversely at $(0, 0)$, which is why we get something smaller than $(x, y)$.
Ideal-theoretic intersection seem to encode "which direction" the intersection is not transverse.
@MatheinBoulomenos I guess scheme-theoretically you think of $V(I)$ in $\text{Spec}\, R$ as the morphism of schemes $\text{Spec}\, R/I \to \text{\Spec}\, R$, and intersection is then a fiber diagram, which is $\text{Spec} R/I \otimes_R R/J$, as you say?
Which does correspond to $V(I + J)$ like you mentioned
I am never entirely comfortable with this viewpoint. You throw away the notion of subschemes entirely, right? Instead subobjects of a scheme $X$ are somehow morphisms $Z \to X$ from other schemes, and you work in this "etale category"?
but it's not determined by its set-theoretic image
you don't need the étale site
closed immersions are stable under base change and composition, so if you have two closed subschemes $Y \to X$ and $Z \to X$ where $X$ is a separated scheme, then $Y \times_X Z \to X \times_X X \to X$ is a closed immersion, so we get another closed subscheme
so it makes sense to speak of $Y \times_X Z$ as the "intersection"
I agree that it's a bit weird at first that subobjects are some kind of morphisms $Z \to X$, not subsets with an induced structure
but the only thing that this allows us is to have non-reduced things: for any closed (or even locally closed) subset of a scheme there's a unique reduced closed subscheme structure
Let me work this out. A closed immersion $f : Z \to X$ would be a morphism of schemes such that $f^\# : \mathcal{O}_X \to f_* \mathcal{O}_Z$ is surjective, i.e., every regular function on $Z$ extends to a regular function on $X$, basically.
If $C \subset X$ is a closed subset, it inherits a scheme structure, yes? I restrict the structure sheaf $O_X$ to $C$. Then the inclusion $C \to X$ would be a closed immersion.
Yeah, that's what I mean by restriction. Hm, I have no clue. Locally a scheme is isomorphic to $\text{Spec} A$, and closed subsets are then like $V(I)$.
It seems to me that the $C$ will be locally $V(I) = \text{Spec}(A/I)$
suppose that we take a closed point in an affine scheme $\mathrm{Spec}(A)$, corresponding to a maximal ideal $\mathfrak{m}$, then the restriction will be (I think) the locally ringed space with one point $x$ and global sections the stalk at that point $A_{\mathfrak{m}}$, but this is not a scheme if $A_{\mathfrak{m}}$ is not zero-dimensional
what you want to do is take the quotient by a sheaf of ideals
for example in the affine case, for the closed subset $V(I)$ we have a sheaf of ideals given on basic open subsets $D(f)$ by $I_f$, then we can take the quotient of the structure sheaf by this sheaf of ideals and we get the structure sheaf for $\mathrm{Spec}(A/I)$
you can prove that this works out globally as well, for a morphism of sheaves of rings you get an ideal sheaf as the kernel and you can define the quotient sheaf and this turns out to be a scheme
and you have an analog of the homomorphism theorem
if $(X,\mathcal{O}_X)$ is a ringed space, then by a sheaf of ideals on that we just mean a $\mathcal{O}_X$ submodule of $\mathcal{O}_X$
OK, let me see what I really want. I'll fix the case I understand: $\text{Spec}(A)$ with it's structure sheaf $\underline{A}$. For a closed subset $V(I)$, I can identify that with $\text{Spec}(A/I)$ set-theoretically, and pullback it's own structure sheaf $\underline{A/I}$. What does it correspond to inside $\text{Spec}(A)$, is I suppose my question. Let me think for a while.
I take a induced basic open subset $D(f) \cap V(I)$ in $V(I)$. The sections over $D(f)$ in $\text{Spec}(A)$ is $A_f$.
@BalarkaSen I think one punchline is that you can't just construct $\mathrm{Spec}(A/I)$ from the structure sheaf of $\mathrm{Spec}(A)$ and the mere set $V(I)$ since different ideals can give rise to the same closed subset
it's maybe not that weird if you think about what your functions on $V(I)$ are supposed to be: if two functions on $\mathrm{Spec}(A)$ agree up to an element of $I$, they should give rise to the same function on $V(I)$
I guess it makes sense if I think of quasiprojective varieties.
Surjectivity of $f^\# : \mathcal{O}_X \to f_* \mathcal{O}_Z$ makes sense, in light of this conversation. It's just saying that we're quotienting by some sheaf of ideals in $\mathcal{O}_X$.
Does there exist a unital ring $R$ such that the group of upper triangular matrices over $R$ does not have the infinite conjugacy class property? That is, there exists a nontrivial invertible upper triangular matrix $A$ such that the conjugacy class of $A$ is finite?
Obviously if $R$ is a finite field, then this should work (all conjugacy classes will be finite)...
I guess I'm interested in whether there exists an infinite ring or field....Does $R = \Bbb{Z}$ work?
@user193391 Surely you want to say "non-central" instead of non-trivial, because say $cI$ for every $c \in R^\times$ is such an element. But it does work for upper triangular integer matrices, where I compute by hand that the conjugacy class of [1 a] [0 1] only includes itself and [1 -a] 0 1]
We don't need it, in fact usually we use first order logic because there are very nice theorems that hold for first order but not second order logic (notably completeness and compactness)
However second order logic also has its perks, for example the second order Peano axioms have a single model, which is the standard natural numbers, while the first order Peano axioms have a lot of nonstandard models
Haven't spoken in a while. Your analysis teachings are serving me long time; not having to study for three analysis courses in a row counting this semester.
How does one show that the inverse image of an open set of the Riemann sphere under a Möbius transformation $\varphi$ is open (as a proof that Möbius transformations are continuous... so don't say "cuz they're continuous!" lol). If I pick a $z_0 \in U \subseteq \overline{\Bbb C}$ where $U$ is open then $z_0$ is an interior point so there's an $\varepsilon$-ball of $z_0$ contained in $U$, can I somehow show that there is a $\varepsilon$-ball of $\varphi^{-1}(z_0)$ contained in $\varphi^{-1}(U)$?
I don't really know what you mean by thinking of the Riemann sphere as $\Bbb C P^1$, I'm just working through a first course in complex analysis to refresh my memory :P
Anyhow, that's the most "elegant" solution. But, no, I don't see any easy $\delta$-$\epsilon$ argument. If you know these map lines/circles to lines/circles, maybe there's an ad-hoc geometric way to do it.
Hi all I'd like to discuss a really illegal manipulation
So I got a double product over signum functions
With discrete arguments, so the expression under the product is something like $\operatorname{sgn}(x_i-y_j)$, where the $i$ and $j$ run from $1$ to some integer
The argument is never zero, so the discontinuity is never "felt"
Now, the system in which I'm considering this has $x_i-y_i$ very small, so I figured what the heck, I will do a series expansion (illegal of course because of the case $x=0$ for $\operatorname{sgn}(x)$)
What are the axes of the 3-space in which the surface representing the general solution to a first-order ODE lives? My first guess was that the horizontal axes are the independent and dependent variable, and the vertical axis is the constant of integration, but this has bizarre effect of "dearbitrizing" the constant. For example, the general solutions $x^2 + y^2 = C$ and $x^2 + y^2 + 1 = C$ have different graphs, but are the same general solution.
