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00:00 - 22:0022:00 - 00:00

00:12
@PVAL-inactive Yo I have a question for you
is there a degree formula for integral cohomology instead of de Rham?
Like I want to find $[M]\cap f^*\alpha$ in terms of the degree of $f$ and $[f(M)]\cap \alpha$
That's more or less the homological definition of degree, innit
gah found a silly mistake in my proof, fixed now
this reasoning is correct, right? If a Hilbert space has $H = M \oplus M^\perp$ and $M$ is closed, then $M^\perp$ is closed?
it "feels" obvious
if its obvious, you ought to be able to prove it pretty quickly
lol
"obviously true" != proof is obvious
00:28
obviously true == terry tao told me himself
in this forum, == ted is so smart, he told me last night
I think if $S$ is a set, then $S^{\perp}$ should be closed
yeah I dont even need that $M$ is closed looks like
Like, if $x_n\in S^{\perp}$ and $x_n\to x$, then for $s\in S$, you have $\langle x,s\rangle = \lim_{n\to\infty} \langle x_n,s\rangle = 0$, so that's just continuity of the inner product
^
LATEX
what is this garbage
what 0ce
00:32
the $\cong$
"obvious"
@0celo7 Are you asking for $f_*([M] \cap f^*\alpha) = f_*[M] \cap \alpha = deg(f) [f(M)] \cap \alpha$?
$H_1 = \pi_1^\text{abel}$
I am, and I'm pretty sure it's true @loch
so you can use that, but he should justify it
00:34
@GFauxPas where did I say obvious
wuz joking
@loch I can't find a reference. It seems to be kind of folklore...
are you asking why though 0ce
00:36
of course not, the damn latex is not wrapping
¯_(ツ)_/¯
@0celo7 Oh! Well it's stated e.g. on the top of p.241 of Hatcher (not as a proposition of anything though..) / on Wikipedia - but in any case the first equality is a consequence of the definition of the cap product, and the second equality follows from $f_*[M] = deg(f) [f(M)]$ by definition of $deg(f)$.
Yeah, the definition of degree I'm working with here is the smooth one. My proof that they're equivalent is contrived. What's the best proof?
I agree if you define degree that way it's obvious, but I haven't done so
how is degree defined in your setting?
$\sum \sign \det df$
00:41
$\displaystyle \lim_{x, y \mathop \to \infty} \dbinom {x + y} y = \sqrt {\dfrac 1 {2 \pi} \left({\frac 1 x + \frac 1 y}\right)} \left({1 + \dfrac y x}\right)^x \left({1 + \dfrac x y}\right)^y$ neat
don't know how useful it is but it looks cool
@0celo7 oh. I don't know the proof off the top of my head - but I think you can find your answer in mathoverflow.net/questions/5001/… - think the upshot is to realise that the topological definition can be computed using local degrees using excision, and from there on it's not too hard (but I haven't thought about it very carefully)
Yeah, I finally found a reference for S^n, trying to write it in general
Ok it's basically an exercise in Hatcher
I was hoping to be able to cite something concrete
proof: if Hatcher gives it as an exercise, it's true
except for the book has like 20 pages of errata
00:52
Just check the errata then
inb4 errata of errata
@GFauxPas that can't possibly be true
What can't
I mean it's not really a limit, it's saying they're asymptotic
this analysis thesis has turned into poincare duality time
01:59
[Random]
The punctured long disk
Take a long line, rotate about $\omega_1$ in a circle
The result is a topology where the origin is unreachable from the outside
02:16
math.stackexchange.com/questions/2780032/… can you guys take a look at this, somebody commented but then deleted for some reason.
What is definiton of dense matrix? @Mozzie
a matrix where 75% of all elements are not 0
user131753
03:13
@PraphullaKoushik Thanks. (Sorry for the late reply)
03:31
@Mozzie I am on my way to bed, and my basic linear algebra is a bit rusty, but if the columns are linearly independent, then you should be able to do better than put it into some block form, and actually diagonalize it, no?
@Xander. No.
 
2 hours later…
05:25
Hi all
hi
please, what is equal to $\{x, (x-1)^2+(y+1)^2<1\}$ it is $\{x, (x+1)^2<1-(y+1)^2\}$ or $\{x, (x+1)^2<1\}$
@skull ?
06:01
someone here ?
 
1 hour later…
07:29
If x, y, z are distinct positive real numbers, the minimum value of
{x^2 (y + z) + y^2 (x + z) + z^2 (x + y)} /xyz is ?
this is a classroom problem and the solution provided by my teacher is evaluate the expression @ x = y =z . I do not get it because the question states distinct positive real numbers .
So is my teacher right ? He is not able to satisfy my query.
 
3 hours later…
10:10
Hi, I have the following problem:
Given a finite size 2D lattice determined by the nearest neighbour links among sites, I need to generate all possible ways to divide the lattice in connected regions of at most C sites. Though I would like to be able to enumerate all the possible subdivisions, I would be content also with some programmatically way to generate them.
Could someone point me to some article dealing with this problem?
Thanks
 
2 hours later…
11:53
Problem: Show that $\mathcal{C}[a,b]$, normed by the maximum norm, has a bounded sequence that fails to have any strongly convergent subsequence. Attempt: Consider $\{f_n\} \subseteq \mathcal{C}[0,1]$ defined by $f_n(x) = x^n$. Since $f_n'(x) = nx^{n-1} \ge 0$, $f_n$ is increasing and therefore $f_n(1)=1^n=1$ is the max. for every $n$. I.e., $||f_n||_{max} = 1$ for every $n$, so that it is a bounded sequence. Note that $||f_n-f||_{max} \to 0$ if and only if $f_n \to f$ uniformly in $\Bbb{R}$.
But any subsequence of $\{f_n\}$ converges pointwise the function $f$ that is $0$ on $[0,1)$ and $1$ at $x=1$, which is not continuous. Hence, no subsequence converges strongly in $\mathcal{C}[0,1]$...Does this sound right?
12:46
Dear folks, please consider removing on-hold from math.stackexchange.com/questions/2779956/… .
I am not sure how that is a mathematical question
@mercio Normal, non-maths folks would not be sure what we are talking about.
13:06
@MdAyquassar then so much the worse for them? this is a site for math Q&A, so to the extent that it's a question about writing in general (not mathematical writing in particular) it's simply off topic
@Semiclassical I don't think it's about writing in general. To draw an analogy, in maths I saw lots of folks avoiding the second comma and (depending on their style) even the period in "Therefore, x=[huge formula] (,) and y=[another huge formula] (.)", though the two clauses are independent and require a comma, and all English sentences require punctuation at the end.
13:34
How do you guys feel about a period after an expression thats at the end of a sentence, especially if it's its own line
Recall that $A\sim \pi^{n/2}r^n$.
It reminds me of that xkcd lemme find it
2004: Ian Stewart: Galois Theory (3rd ed.):
"I have come to the conclusion that eliminating visual junk from the printed page is more important than punctuatory pedantry ... everything is much cleaner and less ambiguous without punctuation."
Lol "Whichever rule you follow, the journal you send it to will want the opposite."
@GFauxPas Note that the ellipsis is considered a punctuation mark. The sentence of Ian Stewart should apply to itself, to stay completely truthful.
You're assuming that the ellipses was in the original material.
But I imagine that it's just GFauxPas omitting parts of the quotation that are unnecessary.
13:49
@Semiclassical I assumed that the quotation of @GFauxPas is verbatim there. If it were not, I presume he would wrtie [...] instead of ... .
I like, commas.
I think you have to stop assuming that everyone follows precisely the same conventions as you do.
@Semiclassical I think you have to stop imagining the opposite.
Indeed, some folks might follow the same convention as mine, some might follow a different one.
I was just copy and pasting, you can't infer anything lol
who cares
just put commas
13:52
@GFauxPas So, the ellipsis is verbatim there, right?
moreover, if you actually bother to google the quote, it becomes immediately obvious that the ellipses aren't verbatim
@Semiclassical Ok, thanks. Though, that contradicts what GFauxPas has just said.
And Stewart's point is, i think, about mathematical expressions mixed with English words in a sentence.
Depends on what he was copy-pasting from.
(from ProofWiki, I'm guessing. huzzah for google-fu)
Yes, I was looking up their conventions for said issue
13:56
yep. and the way they cite the comment is without those []'s
Stewart's point is probably that you can omit punctuation if you want to to make the math symbols clearer
Anyway Md your question probably belongs in academia s.e.
There no "best way" for periods terminating sentences containing (block) formulas at the end. Without periods at the end of block formulas, you might not know where your sentences end, and with periods you might mis-take them for multiplication dots. That's why authors, publishers, and even whole states have their own styles regarding terminating periods. My question is about commas, and periods was an example of punctuation deviating from normal English.
can anyone remind me why $\int_0^{\pi/2} \log(4(\cos x)^2) dx = 0$
@GFauxPas I wouldn't mind if anyone migrates. This should be someone with way, way higher reputation than me, since otherwise ac.se is likely to say it's not really about academia, but about maths and/or languages.
As of now, I'm only asking for getting the on-hold cleared.
14:24
1
Q: Converting Matrix Into Desired Block Form

MozzieIf I have a linearly independent set of vectors say \begin{bmatrix}0&0&9&0&0&0&0&0\\1&2&0&0&0&2&0&0\\0&0&0&0&20&0&13&0\\0&20&0&20&20&18&4&0\\0&4&0&8&5&0&0&19\\0&20&0&12&9&7&0&5\\0&4&0&0&0&0&0&0\\0&0&0&0&0&11&0&6\end{bmatrix} is it possible to use row reduction to convert the matrix into two no...

ideas are very appreciated
@XanderHenderson
the reason I don't want to diagonalize is because in a large matrix there becomes a point where the extra effort to diagonalize is inefficient.
@MdAyquassar just ask it again in academiase
@Mozzie do you know about block diagonalization
@GFauxPas yes
it's kinda sorta related to what youre saying, maybe you can play with that
oh
14:28
well block diagonalization already assumes you're matrix is in block form
no, theres a theorem that every real/complex finite matrix can be written in block diagonal form
but let's call the space spanned by your columns $M$
you agree that, in the block form you wrote above, the vectors in one block are all orthogonal to the vectors in the other block?
yes I agree
so you want to find a decomposition $M = M_{\text{top}} \oplus M_{\text{bottom}}$
so that's a starting point if you were looking for more things to Google
let me see if I can figure out more
hmm
14:31
I've tried for a while but again I always seem to get stuck with 1 or two non zero entires where a zero should be
certainly if $M$ is diagonalizable you get what you want just by diagonalization, right
after a change of basis
correct
but diagonalizing the matrix is not computationally sound for larger 1Mx1M matrices, therefore if the process could be lowered and only form sub blocks of size nxn a gradient decent algorithm could be used to solve sub blocks.
well every square matrix is similar to a block-diagonal matrix
oh it's a computational problem
bleh
lol
rref is easy
but terrible compute wize
This ill conditioning makes it very hard to develop a robust numerical algorithm for the Jordan normal form, as the result depends critically on whether two eigenvalues are deemed to be equal. For this reason, the Jordan normal form is usually avoided in numerical analysis; the stable Schur decomposition[14] or pseudospectra[15] are better alternatives.
well calculating the Jordan normal form of the matrix shouldn't be that difficult, I would think? It's the change of basis matrices that are computationally intensive
I am not a computer scientist
14:37
interesting Ill look into it, LU and QR decomposition i've looked at but no Schur or pseudospectra
(ianacs)
I am computer scientist and a mathematician
nice to meet you cscientist and mathematician, I'm dad
hi dad
lol
well LU gives you what you want
"Block LDU" is what wikipedia calls this
that i'm looking at
oh that gives you a matrix equivalence, not a matrix similarity
14:41
yea
they also all already assume that your matrix is in block form
well I mean you can always arbitarily make your matrix into block form by just... drawing blocks
I mean where the block are all orthogonal to each other
as to your jordan normal form comment, finding the det of matrices is a lot of work because first you must convert your matrix to reduced row form then multiply the diagonals
doing that along with solving $(c_1x^n + c_2x^(n-1) ...)$ is difficult
there are determinant algorithms out there
that are faster than doing that
maybe block form like I would like is impossible
converting a matrix into increasing block sizes (1x1, 2x2, 3x3, 4x4, ....) might be easier, although that means that blocks would approach the size of the matrix
"can one do this" and "can one do this in a computationally efficient way" are different questions\
your q doesnt mention the compsci aspect of it so I'd mention in the question somewhere that efficiency matters
15:06
@GFauxPas No thanks; I've had enough of academia.se personally; I consider it even more toxic than here. If someone would like to ask there or to migrate, I wouldn't mind.
I'm sure you wouldn't
3
Q: Hilbert function of a fat point

konoaI'm reading Geramita's lectures on fat points, but I found a passage a bit confusing. (I'll post the reference as soon as I'll find the link) I read that a fat point (of order $t$, defined by the ${p}^t$-primary ideal $p^t$) in $\mathbb{P}^n$ behaves like $\binom{n+t-1}{n}$ distinct points (in th...

I know this question is a lot confusing, but maybe someone can help me clarifying things
15:23
hmm, I wonder if the closest-point theorem for convex sets in Hilbert spaces can apply to star shaped sets
I will examine this after my exam if I remember and still am interested
What's the closest point theorem?
for every point $v$ in a convex subset of a Hilbert space, there is a unique point $P(v)$ in the Hilbert space such that $||v - P(v)||$ is minimized
call the Hilbert space $H$ and the convex subset $C$
this proof I have in my notes does this
let $(c_n)$ be a squence in $C$ satisfying $d \le ||v - c_n || \le d + 1/n$
then the theorem is proved if you can show $c_n$ is Cauchy
That's false even in $\Bbb R^2$ for star shaped sets then
oh, you're right, i can think of a counterexample in my head
you have a closest point but not a unique one
good call
anyway the proof invokes $||c_n + c_m - 2v||^2 = 2||(c_n+c_m)/2 -v||^2$ that's where convexity is used
extra credit that I didn't do: $v \mapsto P(v)$ is a linear map
and I am not going to do it because im studying for an exam that is today
okay, time to study ~*homology groups*~ and ~*cellular/simplicial decomposition*~
15:49
what the heck is $\mathbb RP^\infty$ :/
how do I think about $\lim_{n \to \infty} \mathbb RP^n$, what does that even mean
16:01
Hello there, i want to get a comment on my approach to the extension of the problem in math.stackexchange.com/questions/2777056/… but as of now i could not had the time of the experts who have replied the question therein. Kindly help out!
.
Cannot this be solved without using Fourier transform?
2
Q: $f'(x) = f(x-1)$ then $f$ is not bounded

JohnJJLet $f:\mathbb{R} \rightarrow \mathbb{R}$. Then consider the following delay equation : $$f'(x) = f(x-1)$$ Let $S$ be the set of solution ot this equation. Then I would like to prove that : $\forall f \in S -\{ x \mapsto 0\}$, $f$ is not bounded. What I've done so far is that : The set of...

16:49
what about proving it doesn't have a Laplace transform
$$f'(x)=f(x-1)$$
$$f(x)=\int f(x-1) dx = \int f(u) du$$
So $f$ is an eigenfunction of the integral operator
Oh neat observation
one of these is $e^x$
Int is bdd on L2 though isnt it
But then if this integral is definite, then we have a problem of the integration interval being shifted by 1

uh, all L2 integral functions only, as I would expect an L3 can be unbounded?
16:58
hi.. I asked this prob/combinatorics question a couple of days ago and it isn't getting much love. Any suggestions? math.stackexchange.com/questions/2779061/…
right
17:16
@BAYMAX I think I proved that it doesn't have a bilateral laplace transform, which is sufficient ot show it's not bounded
I think
I'm actually not sure what you can conclude about boundedness by it not being integrable for $([-\infty..\infty], e^{-st} \, \mathrm d\lambda)$
17:42
Let $a,b,c$ and $d$ be positive real numbers , with $\frac{a}{b}<\frac{c}{d}$, then $\frac{a}{b}<\frac{a+c}{b+d}$. I shlould find the flaw in the following proof
If $\frac{a}{b}<\frac{a+c}{b+d}$, then $a(b+d)<(a+c)b$, so $ab+ad<ab+cb$, hence $\frac{a}{b}<\frac{c}{d}$
And the assertion follows.
Where is the flaw in this proof? I cant find any, all division are justified because everything is positive.
18:02
the flaw is that it's not a proof of the statement
Why?
I mean all the deductions runs backwards as well.
I can repeat the same step backwards. I would have the same result.
in that case you should write the proof in the correct way
Ok. So it is just the problem with the direction of the implication?
you are asked to prove that if A, then B. And you are saying "if B, then computations, then A."
which is completely irrelevant to the problem
that you are asked to prove
you could say
Yes, I see. In this direction, I should use contradiction then.
18:06
@BAYMAX actually i'm not sure it doesnt have a laplace transform now, Because maybe $f(x) = \displaystyle \sum_{n=0}^\infty \frac{e^{W_n(1)}}{e^{W_n(1)}+1}$ up to a constant scalar
"let a b cd be positive real numbers. Then we have a chain of equivalences A ... <=> computations ... <=> B"
i have no idea if that converges
which gives you a proof that if A then B
and also that if B then A as a bonus
@mercio Right, makes absolutely sense.
Thanks.
if you use a backwards reasoning you should make it extremely clear
and especially not start with "if"
18:08
Backwards works only if I use contradiction, isn't it.
maybe
oh actually it would be $f(x) = \displaystyle \sum_{n=\infty}^\infty \frac{e^{W_n(1)}}{e^{W_n(1)}+1}$
if that converges (???)
are you talking about that delay differential equation ?
doesn't $W_n(1)$ diverge to infinity ?
18:11
aaah I forgot to multiply the integrand by $e^{st}$
I dunno but I have to go, ttyl
though you take the exponential.... idk if their real parts diverges
to + infinity
yeah and you're also multiplying by $e^\text{something}$ each time and I dont know if that helps it or hurts it
well my thinking was that for the thing to converge, the term inside has to converge to $0$, so $\exp(W_n(1))$ also has to converge to $0$, so $\Re(W_n(1))$ has to diverge to $- \infty$, but then I have to admit I don't know by heart how that one behaves
18:27
hi chat, a veru easy (and fast) doubt came to my mind
Hello
please i need help here
0
Q: Direct image of a set

Poline Sandrafor a function definded on $\mathbb{R}^2$ to $\mathbb{R}$ defined by $f((x,y))=x-9$ i want to find $f(A)$ where $$A=\{(x,y)\in\mathbb{R}^2, (x-1)^2+(y+1)^2<1\}$$ How to do ?

can i say that $f(A)=]1-\sqrt{1-(y+1)^2}, 1+\sqrt{1-(y+1)^2}[, -2<y<0$
suppose you have $S=k[y_0,\ldots,y_n]$, and you consider $S_d$, i.e. the homogenous polynomials of degree $d$. Now we know that $S_d$ is isomorphic to $\mathbb{P}^N$, where $N=\binom{n+d}{d}-1$. How doeas this map associate a polynomial to a point in the projective space? I think I create a monomial basis of $S_d$, and then map the coefficient. Can anybbbody tell me if it is correct?
Isa
Isa
How to calculate the maximum of $max_{-1<x<1}\vert\frac{1-(-x^2)^{n+1}}{1+x^2}\vert $?
18:46
can someone speak with me
If x, y, z are distinct positive real numbers, the minimum value of
{x^2 (y + z) + y^2 (x + z) + z^2 (x + y)} /xyz is ?
this is a classroom problem and the solution provided by my teacher is evaluate the expression @ x = y =z . I do not get it because the question states distinct positive real numbers .
So is my teacher right ? He is not able to satisfy my query.
19:33
How can I show that if $a$ divides $n$ and $a$ divides $m$, then $a$ divides $3n-m$?
19:48
@quallenjäger What does it mean to say $a \mid m$?
Taking the closed points of a topology gives a function from topological soaces to T1 spaces right
but unfortunately it is not a functor right
I don't think so - the image of a closed point is not necessarily closed.
20:05
@ÍgjøgnumMeg There exists a integer $k$ s.t $a=k*m$
@quallenjäger No, there exists an integer $k$ such that $m = ka$
@quallenjäger perhaps that's where your confusion was! The exercise should be easy now :)
Right
So $m=k*a$ and $n=k'*a$
Thus $3m-n=(3k-k')a$
20:09
Only need to prove $3k-k'$ is an integer
Well, if $k, k^\prime \in \Bbb Z$ then $3k - k^\prime \in \Bbb Z$ because $3 \in \Bbb Z$ and $\Bbb Z$ is a ring, but I suppose that is unnecessarily pedantic
Ah right, it is a ring.
Thanks!
no problem!
20:41
@LeakyNun if you want to construct a funtor from topological spaces to T1 spaces, here's what you can do: Let $X$ be a topological space, then define an equivalence relation by saying that $x \sim y$ iff for every continuous surjection $f:X \to Y$ where $Y$ is a T1 space, we have $f(x)=f(y)$
this actually can be made into a functor
I heard a functor
@LeakyNun surprisingly, the section of nlab on this is readable: ncatlab.org/nlab/show/separation+axioms#Reflection
@MatheinBoulomenos well quite a lot of sections of nlab is readable
that's not just any functor, it's left adjoint to the forgetful functor
true
and glad to see you
I'm recently fascinated by the beauty of algebraic varieties
20:45
we're doing some stuff with varieties in my algebraic groups course
like?
well, an algebraic group is a group object in the category of algebraic varieties
the proper name would be algebro-geometric group
what is that?
basically you write down all the axioms for a group as diagrams. You have morphisms for multiplication, inversion and unit element
then if you require that all these are actually morphisms in the category, that's called a group object
in the category of sets, these are just groups, in the category of topological spaces, topological groups, in the category of smooth manifolds Lie groups etc.
can I have morphisms between varieties over different fields?
20:52
I haven't seen that. If you want to compare different base fields, then you probably want to work with schemes
I have a very neat map from S^2 to CP^1 that you already know
it's one of the things I seen before I touched algebraic geometry, hence my affinity
which increased when it turns out to be a regular map
S^2 is affine, CP^1 is projective
$(x,y,z) \mapsto \begin{cases} [x+iy:1-z] & z \ne 1 \\ [1+z:x-iy] & z \ne -1 \end{cases}$
some of my affinities can't really be explained
@Daminark hi
21:18
possibly pedantic: but I think it is slightly confusing to say $S^2$ is an affine variety (w.r.t. common terminology).. I think the term variety is always considered over an algebraically closed fields in the classical setting (of course you can define varieties w/ schemes, then we don't need to be working over alg. closed fields)
Historically, Weil did varieties over general fields before schemes
I was confused for a different reason: there is a notion of affine structure in differential geometry, and S^2 does not have one
Do you also get confused by 1000 unrelated things being called normal? :P
no
the point is it's not clear to me any sense in which one can say "S^2 is affine"
The Zariski topology makes sense over any field. So you can look at closed subsets of $\Bbb A^n(k)$ for any field $k$
21:28
interesting - I didn't know that. Did he define varieties literally as zero set of polynomials in affine space over $k$, or did he implicitly also work with extensions though..
an affine real variety, ok. The word "real" never seemed to appear in the discussion above
@loch Weil's foundations are now considered inelegant and there doesn't seem to be a reason to learn them except for understanding the history. So they're not treated in any modern literature afaik. But iirc, he worked embedded all fields involved into some large algebraically closed field
he did definitely work with extensions. The statement of the Weil conjectures involve the number of (closed) points on a variety over a finite field as you base change to larger finite fields
21:42
I apologize for the mass confusion. have been afk.
I was confused because I thought an affine variety surjecting to a projective thing (not a point) seemed weird (but I just realised there's nothing wrong with that lol TIL)
@LeakyNun No need to apologize.
May 1 at 11:32, by Silent
I know that every polynomial with complex coefficients has a solution in complex field. is it true that every 'multivariate polynomial' has solution in complex field?
@Silent you asked this a long time ago
my answer was incomplete:
May 1 at 12:13, by Leaky Nun
@Silent any multivariate polynomial can become a monovariate polynomial by substituting 1 to the other values
the objection is "what if I get 0 everytime I substitute the values?"
so the correct answer is this: let $f \in \Bbb C[X_1,\cdots,X_n]$ be a multivariate polynomial
If there is no $\vec v \in \Bbb C^n$ such that $f(\vec v) = 0$
Then, $V((f)) = \varnothing$
So, $I(V((f))) = (1)$
if you substitute 1 to the other values and you get zero, then that is a solution to the polynomial!
@MikeMiller I mean, the zero polynomial
oh wait
so my answer was complete?
21:57
yup
anyway, let's finish my nuke
because why not
im for that
However, by Hilbert's Nullstellensatz, $I(V((f))) = \sqrt{(f)}$
So $\sqrt{(f)} = (1)$
So $(f)=(1)$
So $f$ is a unit, i.e. a non-zero constant polynomial
QED
@MikeMiller is this right
21:59
yeah
yay
@MatheinBoulomenos you'll be proud
00:00 - 22:0022:00 - 00:00

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