Whelp... I now have a two pound yak tongue in my freezer. Looking forward to figuring out how to cook it up real nice.

Do the old Jewish beef tongue recipe.

@TedShifrin Which one is that?

LOL, my dad used to make it, but I never have, so I have no recipe.

Heh.

My mother made tongue once.

And it was good, but she has no idea what she did (I asked).

Weirdly, she is better at cooking Sonoran cuisine than traditional Jewish fare.

(Though all of her Sonoran food has a distinctly Jewish inflection, which is doubly strange.)

Beef tacos made with koshered beef; peppers fried in schmalz; etc).

The only tongue I've had is a filthy one.

@XanderHenderson what better to yak with than a tongue

@robjohn Oddly enough, I made that pun to my mother. :D

that's scary...

This post was asked in 2019, but answered in 2013...how? math.stackexchange.com/questions/3407917/…

@TedShifrin my mom makes my dad the tongue for rosh hashana, he loves it

it is the result of merging one question with another,

see math.stackexchange.com/questions/601900/…

so, the responses on the older post became responses to the newer post

@robjohn Especially to me!

@DavidRaveh Robjohn surely can explain the history.

Ah, leslie already did.

Beef Goulash

the competition went really well.

Hungarian. Mixing ethinicities.

@TedShifrin I was about to answer when I saw leslies's answer.

It was the target of a merge in 2020.

I imagine you would prepare yak in much the same way as beef?

Perhaps less fat than beef?

Based on what the farmer said, it sounds very lean---maybe more like elk or venison, in terms of prep. Low and slow.

Though tongue is tongue is tongue.

But I also got a brisket.

Corned yak.

no yak or duck sightings today, but we did see a great blue heron on a neighbor's roof. they are not rare to see in the more open areas around here, but i have never seen one in the middle of a subdivision on the roof of a house. i think it's a sign of the end times.

Why-sometry group

This might sound rather restrictive, but is there an axiomatic system that rejects existence of uncountable cardinalities?

'Cause, for computer-theoretical purposes, I tried to reject the Axiom of Power Set, and miserably failed.

i'm not a foundations person but there are intermediate paths between loving every occurrence of these things, on the one hand, and adopting axiom systems that "reject" them, on the other.

e.g. work in whatever set theory you want, but focus on, and give preference to, specific things within that theory, even if it allows for other things.

e.g. do "normal" set theory but back off of things that are uncountable, or care a whole lot about being able to prove that things are countable.

maybe it's clearer one level down, not at countable vs. not but at finite vs. infinite sets. you can be very "finitist" in your approach to things without adopting axioms that forbid infinite things or do not allow you to construct infinite things.

This might sound alien, but actually, I don't want my math to found over sets, but rather topological spaces. To be accurate, what I want to reject is existence of topological spaces that are not second-countable. (cf. pointless topology)

And I took this further. My math shall work over the category of zero-dimensional Hausdorff spaces. That will inevitably pull me into the realm of nonstandard analysis (because the definition of the reals became unusual), but let me see how far I can go.

peeeep

@DavidRaveh data.stackexchange.com/math/query/541284/quick-answers to find such answers by the way.

If anyone with some publishing/research experience and has a minute to talk about the academic publishing system plz @ me.

"History of Funcitonal Analysis" just got delivered to me, this diagram is in the introduction

Crypto-integral equations sounds like some modern maths

@s.harp that is a nice diagram

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$

Is this a tautological formula?

Hello, I am an undergraduate student and I recently spoke with an individual Rodolfo Nieves on my recent trip to Venezuela. He claims to have proven several Millenium Prize problems and his work is on Scribd. Could anyone provide their opinion or feedback on this character and/or his documents on Scribd?

Never heard of him …

Exactly.

I have been attempting to isolate the mistakes in his "proofs."

Found this gem after searching his name XD

OK. Throw him on the rubbish heap.

Got it.

Chicken Goulash

I need more chocolate. I can’t do math without it

Can anyone suggest a good book for learning ring theory? The thing is, I have learnt group theory from Topics in Abstract Algebra by I.N Herstein, but that book is not in the recommended book list in my univ, instead other books are there such as the one by Gil Strang, Fraleigh, Gallian, etc. Should I continue from Herstein or, anything else? This is to be noted, that I liked Herstein's way of explaining a bit too much, and I feel that his book's a treasure! What d ya all suggest?

Further I think Herstein has a good sort of exam type problems in it.

That what makes it so much ahead in the competition

if you already like the book and it covers the material you need, I see no harm in just continuing to read it

@Thorgott hmm, that's indeed a good suggestion.

I’ve learned so many colorful words in Russian just by spending hours in CSGO

I personally detest Herstein. But you're listing Strang — which is linear algebra only, Fraleigh, Gallian, which are much weaker books. I highly recommend Mike Artin's Algebra.

Artin writes algebra with a much more global, integrated viewpoint than Herstein. Herstein's problem style is much more trickery.

Are you talking about Topics in Algebra or Abstract Algebra? You mixed the titles.

The former is his classic book, which I know. The latter is an easier-going book written afterwards for more general market.

Ah, I made a typo. I am talking about "Topics in Algebra " by Herstein, @TedShifrin

also, Atiyah-MacDonald is a classic that's pretty good to use as a secondary reference (probably not as primary reading cause it's very terse on explanations); great exercises too

The British conquered the entire world, took control of spice trade, only to serve unseasoned fish and bland ‘chips’ for food

And beans

Right. That's the one I was taught out of in 1971. I stand by my comments.

Atiyah-Macdonald is a next course, assumes you've finished the year-long algebra course.

I think British food has changed over my lifetime. It's no longer so bland.

You’re that old? You’ve seen food evolve??

@TedShifrin ha ha, is there any personal reasons why you chose to detest it , he's a good writer, imo, but ofc , u can have ur own opinions, but that seemed a little strange

Atiyah-MacDonald starts right at the start

I disagree, Thor. I was taught out of it for a graduate course in commutative algebra.

steps in commutative algebra true start at the start

atiyah-macdonald world of pain and suffering

@Thomas I already said that he's much more about trickery. His approach to mathematics is formalistic. Artin is an algebraic geometer (as opposed to a technical ring theorist) and has a much better world view in his writing.

McDonald’s

@s.harp this actually doesn't have to be closed even if we add $0$

And Artin's exercises are much better.

well, we are in a situation where group theory and linear algebra has already been learned

what else is supposed to come in between

Your argument is flawed in some way that I'm not bothering to check properly

baby ring theory

Rings and modules (I agree that Galois theory is not needed), but Atiyah-Macdonald is super short and if you don't have experience with rings and modules, good luck learning all that fancy stuff. You don't need localization and going-up-down before you've had the standard course.

Get experience with quotient rings, Chinese remainder theorem stuff, etc.

Finitely generated modules over a Euclidean domain (or PID if you insist).

Chinese remainder theorem shows up in competitions too

I don't mean with modular arithmetic. I mean in the setting of ideals and quotient rings in general.

Gotcha

I almost got into an accident because I was texting and driving at the same time

I insist on PID

I don't ... :)

The row/column operations proof is much more intuitive with E.D.

@冥王Hades uh. Don't do that

@Jakobian yeah I learned my lesson. To be fair I was in a bit of a serious personal argument with someone

You are appalling, Hades.

How is that more important than not killing somebody

P

I lost anyway

edit: reworking question

This is nothing compared to the stuff I’ve said during Xbox 360 days

I don't do row/column operations, I do devissage

Think about this when you're killing a whole family of people that want to live just as much as you do but didn't argue on the phone

I'll stick to my approach for undergraduates, thanks.

Bold of you to assume I want to live :P

We're not concerned with your life, Hades, but with the lives of those you choose to endanger.

I do like the matrix proof where you calculate the SL_N(Z) orbits of Z^n

Don’t worry they’re all safe. I know how it feels to lose someone precious I have firsthand experience

(2) is an ideal of Z, the statement: "the Z-module (2)" is unambiguous, right?

what would be ambiguous about it?

to be perfectly honest, there are some ambiguities

does it refer to one and only one module

First, $(2)$ doesn't tell you that it's a principal ideal in $\mathbb{Z}$, and second no one said that the action of $\mathbb{Z}$ on $(2)$ needs to be by multiplication

the latter point would be overlooked by most and they'd choose what's most natural

for the former it'd be preferable to have a bit of context beforehand when saying such statements

excellent work

hmm but in what sense do we have (2) doesn't tell us it's a principal ideal in Z?

thanks btw

because we could just as well consider $(2)\subseteq R$ where $R$ is any unital ring

oh, understood as two times the unity

Yeah, in a lot of context people refer to $n\cdot 1_R$ simply as $n$, where $1_R$ is the multiplicative unit of $R$

well, (2) denoting the ideal of Z was included as a premise

I interpreted it as what's intended

I wanna try Indian food. I’m kinda worried it might be too spicy for me though

@冥王Hades just don’t get vindaloo.

There’s a nice Indian restaurant nearby serving butter chicken with rice and garlic naan

@TedShifrin What even is that? Sounds weird

@TedShifrin not a vindaloo fan, I take it.

here to what is the product of Dk and del k

i mean they explained del k as outer product

but said noting about Dk and del k

What is Vindaloo

please help

@robjohn

@robjohn Oh, I love it. Hades was afraid of spicy.

Afraid? I’m the God of Underworld!

@PrateekMourya $D^kf$ is a multilinear map. This is crazy notation. Just look at the classic formulation on wiki or any textbook.

I just heard about the burning of quran in Sweden

what's the proper definition of the ring whose elements are infinite dimensional vectors with integer entries?

addition is the natural one, multiplication is multiplication of the ith entry of vectors

hm this might be an issue, the direct sum is nice here because you're guaranteed to be able to compute multiplication and addition, but not in the case of the above mentioned structure

it might not even be well-defined

hm, the direct sum as i've encountered it has been defined to require finitely many nonzero elements, is there such a thing as an infinite direct sum?

there is, but we have to distinguish infinite direct sums from infinite direct products

suppose $A$ is a non finitely generated ideal, is there a surjection between $\otimes_{i \in \mathbb N} \mathbb Z$ to $A$?

or instead, is there a natural surjection sending the ith component $x_i$ of that direct product to $x_ig_i$, i.e., $x_i$ times the ith generator $g_i$?

whoops accidentally used tensor product notation, i meant infinite direct product

@Jakobian How so? If v≠0 then (R\ 0) x (X R) (at the appropriate index) is a neighbourhood of v that contains at most one of the points, so v is not a limit point. But yeah, 0 is definitely a limit point and the set isnt discrete.

@shintuku you're thinking of modules, not rings

also, there's at least two natural multiplications you can define on the abelian group $\mathbb{Z}^{\mathbb{N}}$

one is the multiplication making this into the ring $\mathbb{Z}^{\mathbb{N}}$ (the direct product of $\mathbb{N}$ copies of $\mathbb{Z}$), the other is the one belonging to the formal power series ring $\mathbb{Z}[[x]]$

oh right, where multiplication in the ring of formal power series is analogous to multiplication of polynomials, right?

yes

ty! i'm trying to figure out whether there is a surjection from $\mathbb Z^{\mathbb N}$ to a non finitely generated, but countably generated, ideal, given by a map to its generators

if we're in a ring $R$, clearly there is such a surjection from $R^{\mathbb N}$, right? what we'd think as the natural one

for that, you'd want the direct sum, not the direct product

couldn't we have the possibility of a non finitely generated ideal with countably many generators $g_i$, that contains an element $1*g_1+1*g_2+1*g_3+\dots$ and so on? i was thinking that this case excluded the possibility of a direct sum

@s.harp The only way this could be true if it was inconsistent that there exist measurable cardinals

so your argument must be false

since I really don't think you'd be able to prove that

oh, but then as you said we might have something more like multiplication in $\mathbb Z[[x]]$, right?

@s.harp can you write it in LaTeX? I'll try to find the mistake

@shintuku hm nevermind, I don't see why we would be forced into using that sort of multiplication if we're only looking for a surjection

you have a lot of explaining to do if you want to write down infinite sums

For any $\beta\in I$ $\displaystyle (\Bbb R\setminus \{0\})\times {\big\times}_{\alpha \in I, \alpha\neq \beta} \Bbb R$ is open in ${\big\times}_{\alpha\in I}\Bbb R$ and intersects only one of the vectors $\delta_\beta$, if $v\neq0$ then there is some $\beta$ so that $v_\beta\neq0$, so the above would be a neighbourhood of that vector.T

hm yeah...

@Jakobian I don't know if it compiles, my browser doesnt accept it when I run javascrip from the address bar

the command \bigtimes doesn't work

yeah now it definitely doesn't compile

lets see how i fix it

I mean it does prove that the set is discrete, but not that it's closed

Yes its not closed, 0 is a limit. But it seems that no other point is a limit.

I’ve never heard of \bigtimes.

Im here for a \goodtimes, not a \bigtimes

() tex.stackexchange.com/questions/28150/…

@s.harp if you give me a proof then I'll believe it

Can’t you write \prod instead?

@Jakobian right now im struggling to find a way to compile latex lol

but like I said, it's unlikely given that would prove existence of measurable cardinals is inconsistent

the definition of measurable cardinal is that there is a set $X$ of size $\kappa$ and a measure $\mu:X\to \{0, 1\}$ on all subsets of $X$ such that $\mu(X) = 1$ and $\mu(\{x\}) = 0$ for all $x\in X$, then $\kappa$ is called measurable.

ok, this compiles fine for me now: Let $v\in\prod_{\alpha\in I}\Bbb R$ with $v\neq0$, then $(\Bbb R\setminus\{0\})\times\prod_{\alpha\neq\beta}\Bbb R$ is an open neighbourhood of $v$. It intersects exactly one of the axis elements, and so $v$ cannot be a limit point. So the only limit is zero.

If you say its not a provable statement, I believe you, but whats the error?

This also seems like such measure should never exist - but no one has proved it

and it's suspected by set-theorists that we won't be able to disprove the existence of measurable cardinals

consider F: A \to B where there is some equivalence class E, of A. Let E=H (all else equal) where H is an equivalence class of B and obtained through F. What is this called?

@s.harp I'm not saying it's not provable, just that no one knows, if you really were to prove it, then you'd be famous in set theory

we can't prove that ZFC + "measurable cardinals exist" is consistent given consistency of ZFC

@Jakobian the argument i give has nothing novel in it, so if its equivalent to something hard then the argument is not correct

there's a possibility that this is inconsistent, but it's like the possibility that ZFC is inconsistent, unlikely

consider $F: A \to B$ where there is some equivalence class $E$ of $A$. Let $E=H$ (all else equal) where $H$ is an equivalence class of $B$ and obtained through $F$. What is this called?

@s.harp your argument only proved that the set is discrete, nowhere did you prove it's closed

I was asking you to provide an argument

@Jakobian Ok, I think we are talking about different things now, I know the set is not closed, since 0 is a limit. The only thing I am remarking is that there are no other limits.

No we are talking about the same thing

0 is clearly a limit, I was just overlooking this because it's a trivial detail

If you include 0 the set is closed, but not discrete

hmm that's true

The thing equivalent to measurable cardinals is existence of an uncountable discrete set in a $\prod_{\alpha\in I} \Bbb R$?

huh? no

there is a measurable cardinal iff there is a discrete space $X$ which can't be embedded as a closed set into any $\mathbb{R}^\kappa$

so if your argument were to provide such closed embedding for any discrete $X$, then you'd prove inconsistency of existence of measurable cardinals

I see, the original argument was totally wrong though because I don't have familiarty with infinite topological products

My original argument above there proved that discrete space $X$ with $|X|\leq\mathfrak{c}$ is realcompact, so this answers your question as to "why" would someone consider such argument that you might think as looking weird

the characterization of realcompactness as closed embedding into product of real lines is really unhandy when trying to prove something is actually realcompact

and yes, it was a cool argument!

I even still remember it

Hmm, I tried to read what you said again, but I have no idea what realcompact is lol

A space is realcompact if it can't be densely embedded into a larger space so that the continuous functions are the same

vaguely speaking

compare with bounded continuous functions and compactness

Wiki says: for all X, all ring morphisms (continuous?) C(X;R) --> C(Y;R) come from a continuous map Y-->X <=> Y is realcompact, is that a useful characterisation?

are you sure you didn't swap X and Y

I did, oops

it's an useful property, but as for characterization, idk

I think that generally continuous functions are just in abundance so it's impossible to tell things from them

when it comes to particular examples

also the ring homomorphisms aren't assumed to be continuous (with what topology would they be anyway?)

but they are assumed to be unital here

I have a proof in my lecture notes which I'm doubting the logical structure of. The proof is one sentence long, so very short, but uses another theorem. It's about differentiation and integration of power series. Anyone up for helping?

which is important because with non-unital homomorphisms we need to fix a clopen set and it gets more complicated

@Jakobian For C(X) you have pointwise and uniform on compacta as two topologies. There are probably more, but I get that C(X) sucks, I never worked with it only C_0(X) and C_b(X) (and the latter is already bad)

you also have uniform and m-topology on C(X), and you can probably find even more

C(X) doesn't really suck, it's just harder to find useful topologies on it for general Tychonnoff X

some properties of C(X) are much much nicer than for the ring of bounded continuous functions

@CassieSwett then * isn't even a bifunctor

What prompts the question is that I'm considering what it would mean for a category to act like a Kleisli category. Given a category with finite products and a monad T, and given morphisms a -> T(b) and c -> T(d), there are two different obvious ways to combine those morphisms to obtain a morphism a * c -> T(b * d). So, the product operation in the original category fails to make the Kleisli category into a monoidal category.

And yeah, you don't get a bifunctor C × C -> C; instead, you get two functors C × Disc(C) -> C and Disc(C) × C -> C, where Disc(C) is the discrete category whose objects are the objects of C.

@s.harp I must also mention that you'll probably never use the knowledge of what a realcompact space is. If you're working with metrizable spaces for instance then you already have nice properties like $C(X)\cong C(Y)\implies X\cong Y$, working with them is more when you're doing something wicked and totally unapplied (imo)

@Jakobian I can't remember the last time my research involved a space that wasn't a manifold or a Hilbert space, everything beyond those walls is just for fun for me

@Thorgott nonformal infinite sums are only defined on topologies, this changes everything

yeah, to make sense of something like that you often have a topology in the background. at least, you need some notion of how a collection of things can "converge" to something, whether or not it is convergence in a topology.

convergence spaces

it makes abstract sense, if you think about replacing + with just an abstract binary operation that you might not be as familiar with. you can use it to create various n-ary operations for any n, but it's not clear how you get anything for an infinite collection without making a lot of choices.

you even have to make choices at the n-ary realm, usually. it's only in the presence of things like commutativity and associativity that there is one meaning to "a * b * c * d" when * was at the outset only a binary operation.

are we guaranteed then that for any ideal, there exists some sort of direct sum (with more than one summand) ring-isomorphic to it?

Happy yak day to Munchkin.

@shintuku what's the context

@shintuku ideally, in such a way to represent linear combinations in that ideal as some sort of tuple

at Jakobian: am working on a proof of the isomorphism theorems for rings

then what you said seems irrelevant

Presumably you're trying to prove $(S+I)/I\cong S/S\cap I$

Any element on the right is of the form $s+I$ for $s\in S$

it suffices to send $s+I\mapsto s+S\cap I$

currently i'm at the lattice-isomorphism theorem/correspondence theorem

that one should be even easier

a general vibe, attempting to model what an ideal in an arbitrary ring must "look like" with a view toward proving these kinds of results, is far more difficult than working closer to the definitions.

note that this is why, when defining generating sets for modules, we only ever allow finite linear combinations

hmm... you said lattice isomorphism

so I guess you need to prove things like $(A+B)/I = A/I + B/I$ and $A/I \cap B/I = (A\cap B)/I$ as well

I really want a dorayaki for breakfast

Is that a Japanese croissant?

ah right the natural setting to think of this really is modules

at leslie: heheh yeah it's starting to seem so

Home again, home again.

Made good time today. 420 miles in about 7.5 hours. Would have been faster, but I got stuck behind a caravan of RVs in the mountains. :/

I didn’t realize you’d been abroad!

@TedShifrin more like a pancake, filled with whatever you like. Red bean, chocolate, and more

@TedShifrin Yup. Went up to my sister's place in Colorado for the 4th.

Ah.

They don't celebrate the 4th in her little town. But they do celebrate "Cherry Days" every year, and it just so happens that Cherry Days always seem to happen on the 4th.

That's where I got the yak tongue and brisket.

Hades, may I have savory instead of sweet?

savory Dorayaki??

@冥王Hades I'm going to make pancakes for dinner tonight.

Pancakes for dinner??

With mesquite flour.

Oh, did you have a freezer chest for the trip?

@TedShifrin Yup.

@冥王Hades Sure, why not?

Xander I’m the troll here

Übertroll

what’s next? Cereal for lunch?

Mesquite pancakes, with cheese.

Savory pancakes can be had at any meal.

For example, the Russians do amazing things with pancakes and caviar.

I’m sorry but my mere plebeian tastebuds have only known pancakes as a breakfast item

Well, I don't usually eat breakfast, and I love pancakes.

So sometimes, I have breakfast for dinner.

But, in this case, they are savory pancakes, which are already a lunch or dinner item.

I have had caviar, I had it on my flights as well, Emirates serves it.

People here think I’m buying pancakes for my entire family but in fact this is just for me

Hrm... it is 5 o'clock Colorado time, but only 4 here. Is it too early for a cocktail?

after that drive? it's 5 o'clock somewhere.

Heh.

I want a coffee that tastes like chocolate, I don’t know what it’s called

@冥王Hades Mocha?

(That is coffee with cocoa in it...)

Oh yeah that’s what it was.

So I had chocolate earlier during the night, I am having dorayaki with chocolate in them, and I’m going to have a Mocha.

My pancreas is working overtime

"Yaki" is a cooking method, right? What is "dora"?

@XanderHenderson I heard it meant “gong”

It looks like a gong

Huh. I googled "yakidora". They look good.

Stuffed pancakes are a pain to make. :/

i have been deviously misled by examples. now, instead of thinking of examples i simply thought of ideals only wrt to their containment relation, and the theorem made sudden sense. i've been stuck thinking about generators because i've been thinking in PIDs

we conclude from this that we must avoid examples, always

These people are experts at it, mine looks and tastes perfect

I need to go deal with food. Later.

you're joking, but the actual lesson is maybe just to be cautious about inferring general properties of rings, or general proof strategies about rings, from properties and strategies that are specific to examples. good examples will illustrate a theorem but often do not tell the full story. at least at this level of generality.

yeah, but on the plus side i've also been led naturally to the statements of most structure theorems and the actual niceness of noetherian rings

"prove a generality about objects X by first proving that every X 'looks like' an object Y, then use specifics of the structure of Y" is a useful meta technique to think about, but it is not all powerful, and in particular is not always the shortest route to proving something.