Mathematics - 2023-07-07

Jakobian

00:16

I had tomato soup today

Xander Henderson

@Jakobian Too hot for that.

I just ordered some kvass from Amazon, so that I can make some okroshka next week.

Ted Shifrin

What dat?

Jakobian

we don't eat anything like that here

Xander Henderson

@TedShifrin Kvass is fermented bread-water. Kind of like kombucha, but made with rye bread.

Jakobian

I recall one cold soup but I need to remember the name

Xander Henderson

00:28

Okroshka is a cold soup made with kvass as its base.

Ted Shifrin

No, okroshka?

Xander Henderson

@Jakobian Vicheysoisse?

Ted Shifrin

Ah, what else?

Borscht?

Xander Henderson

@TedShifrin Borsh' *

No "t".

Unless you are Ukrainian, I guess.

Jakobian

Who eats borsch cold

Xander Henderson

00:29

@Jakobian Lots of people. It's good.

Ted Shifrin

Lots of people

Jakobian

in my language we spell it barszcz

Xander Henderson

Jink.

Ted Shifrin

So what else goes in your soup, Xander?

Xander Henderson

@Jakobian I mean, I spell it борщ. :P

Ted Shifrin

00:31

Schch instead of scht

Xander Henderson

@TedShifrin Okroshka? I like to do it with boiled potatoes, and sausage, and cucumber (all chilled).

Ted Shifrin

Cold sausage? Hmm.

Xander Henderson

@TedShifrin Cooked, first.

Then chilled.

Ted Shifrin

I get it, but ..

Jakobian

ah, the soup I was thinking about is chłodnik

Xander Henderson

00:31

I've also seen it done with hard boiled eggs, and other veggies.

Ted Shifrin

Caraway seeds, too?

Xander Henderson

@Jakobian That's, like, the Polish version of cold borsh, right?

@TedShifrin You could, sure. I like to use lots of dill.

Ted Shifrin

I love dill in lots of things.

Xander Henderson

Oh, and kefir. I almost forgot the kefir.

Jakobian

I never actually ate it. But wikipedia lists chłodnik litewski as cold borscht

Ted Shifrin

00:33

Ah.

Xander Henderson

@Jakobian Yum.

Jakobian

oh, chłodnik I think just means a cold soup

or a type of could soup at least, but it includes vichyssoise and окрошка

Xander Henderson

Interesting. In English (such as it is, as I understand it), "khlodnik" is specifically a cold beet soup. But then "chai" means a specific kind of spiced tea, whereas the original Indian word just means "tea".

Jakobian

I love white borscht with lots of eggs and white sausage

I love red borscht too

the tomato soup I ate today was pretty tasty though

Xander Henderson

"White" borscht? What you have Poles done? :P

Are you sure you don't mean щи?

:P

Ted Shifrin

00:46

Cabbage borscht.

Xander Henderson

Щи да каша, пища наша.

Jakobian

well I mean barszcz but those are English rules

Ted Shifrin

My one PhD student was from Poland.

copper.hat

My one PhD advisor was from Poland.

Xander Henderson

... I know someone from Poland?

Ted Shifrin

00:48

I’m 1/4 from Poland?

copper.hat

My GP is of Polish descent.

Jakobian

@TedShifrin does that actually exist

Xander Henderson

I need ice cream. But I have none. :(

Ted Shifrin

Polish medicine?

copper.hat

:-)

Ted Shifrin

00:49

@Jakobian Sure.

Xander Henderson

@TedShifrin Isn't that was Lister was going for with Listerine? Floor polish, recast as medicine?

冥王 Hades

Forgot my umbrella again

why do I do this to myself

Xander Henderson

No ice cream. But... I have peaches from a farmstand in Colorado. That might be better than ice cream.

Jakobian

I ate something called mexican soup here (apparently not done by mexicans, couldn't find it on wikipedia), it was pretty good

冥王 Hades

@XanderHenderson Are you sure? I’ve eaten some of the most exotic ice cream here

Japan can’t be beat when it comes to sweets

Xander Henderson

00:58

@冥王Hades "Exotic" ice cream does not exist where I live.

冥王 Hades

Too bad

Xander Henderson

And I don't have the ingredients to make my own.

Jakobian

minced meat, beans, tomatos, corn
apparently it's based on chilli con carne?

冥王 Hades

@XanderHenderson is it even possible to make one’s own ice cream?

Xander Henderson

@冥王Hades Yes?

@Jakobian Yeah, that sounds right.

Though the corn would be a little unusual in chili.

冥王 Hades

00:59

Wait, what? Isn’t it supposed to be made in giant machines and whatnot

Xander Henderson

Was it maybe something like a TexMex tortilla soup?

@冥王Hades No?

Ice cream is made by freezing cream and sugar while agitating it so that it doesn't form crystals.

冥王 Hades

Well who am I kidding. I’m the guy that can’t even boil water, I’m not making ice cream any time soon

Xander Henderson

It is possible to make ice cream with two ziplock bags.

冥王 Hades

I’ve seen people freeze slushes and create “ice cream” out of that

I had one in orange flavor. It was nice

Jakobian

I mean, they have those large machines in professional kitchens but it doesn't mean you can't deal without them for various dishes

Xander Henderson

01:01

Fill a one- or two-pint bag with an ice cream base (milk, cream, egg yolks, sugar, whatever flavors you want), seal it up real good, and put it in a gallon bag with salt and ice. Shake it for 20 minutes. Ice cream.

You can also do it with a kichenaid mixer and liquid nitrogen.

冥王 Hades

Where’s the chocolate and all that other delicious stuff?

Xander Henderson

Or you get a small countertop ice cream maker, which consists of a cold bowl and some kind of agitator.

@冥王Hades "whatever flavors you want".

冥王 Hades

What are the odds I’ll be catching diabetes soon given my addiction to sweets

Jakobian

I saw those cool mini pizza ovens, for people to make pizza at home more realistically

Xander Henderson

@Jakobian No need. A pizza stone and a normal oven are fine.

Not ideal, maybe, but more than good enough for most people.

Jakobian

01:05

Yeah, but still if you want a more oven-like pizza then you can grab one of those

冥王 Hades

Have you ever seen one of those flameless heaters used by military? It’s fascinating

Jakobian

ah, the ones used for rations, yeah I saw a few videos

Xander Henderson

Mmmm... MREs.

冥王 Hades

Yeah those.

shintuku

does the map from an element to the partition containing it have a standardized name

冥王 Hades

01:08

I just realized, my girlfriend’s death anniversary is tomorrow.

Jakobian

@shintuku you could call it choice of representants for the equivalence classes

Ted Shifrin

You don’t catch diabetes.

冥王 Hades

@TedShifrin develop?

shintuku

at jakobian: ty

Ted Shifrin

Yes … succumb to.

冥王 Hades

01:12

For now I seem to be fine. My pancreas works harder than a single mother

Kinda wish I was back in the states right now

shintuku

in $\mathbb Z/(3)$, how do i formalize the fact that the canonical quotient map applied to $2$ sends it to the "natural" representative $\overline 2$, while this is not the case for $5, 8, \dots$?

Lemon

01:27

i was doing some experiment. For $S^1$, the unit normal in Cartesian is $x^j/|x|\partial_j$. Converting this into spherical coordinates (is there a short cut other than just writing out the vector transformation for each $\partial_j$?), I got $N =\sin^2 \phi \partial_\rho$. Now I am thinking, with only one component, what tangent vector to the sphere is normal to this other than $0$?

Jakobian

Eric Wofsey always has answers for the most interesting questions

I am a bit of a fan

Ted Shifrin

@Lemon First, the unit normal is $\partial/\partial\rho$. Your last question is crazy,

I think Eric quit Harvard grad school before his degree, but he knows a lot, yes.

@Lemon. What vectors in $\Bbb R^3$ are normal to $\partial/\partial x$?

Jakobian

@shintuku well, when you divide you want something like Euclidean domain. And then you want to choose which element by say, which one is positive

So don't you want a totally ordered Euclidean domain?

What I'm saying is that Z is a pretty special ring, and this is why those choices feel natural

shintuku

yeah i get that, i'm trying to find this for modding by ideals in general. for instance, for ideals A,B, we have A + B (understood as A mod B) = A if and only if the quotient map is a bijection

冥王 Hades

Jakobian

01:43

@shintuku I don't really understand

shintuku

am cogitating, thanks for attempt

residue class may not have "canonical" representation

Lemon

02:04

@TedShifrin oh yes i forgot to add a missing cosine in my calculation. So the normal is just $\partial_\rho$. Normal to $\partial_x$ are just forms of $(0,y,z)$

Oh i see

Its all forms of $(0,\theta, \phi)$ since only the $d\theta^2 + d\phi^2$ part of the metric is what it describing $S^1$

robjohn

Hey, @Ted! how was your 4th?

Ted Shifrin

@Lemon So, yeah, $\partial/\partial\phi$ and $\partial/\partial\theta$ span the tangent space.

@robjohn unremarkable :)

robjohn

There were a lot of non-sanctioned displays here even though they were forbidden by LA

Ted Shifrin

I hope no fires

robjohn

None that I heard about.

冥王 Hades

02:20

I’m hungry again. My metabolism is too quick

Ted Shifrin

03:09

Wait til you hit 30.

shintuku

04:00

@Jakobian found it: if $R$ is a ring and $A$ is an ideal, we have the canonical map $\phi: R \to R/A$ and the canonical map $\psi: R/A \to R$. i was looking for the property $\phi(x) = \psi \circ \phi(x)$

i mean, $x = \psi \circ \phi(x)$

Ted Shifrin

What canonical map?

There is no map.

shintuku

argh there is no canonical $R/A \to R$ in general?

no of course not

Ted Shifrin

There is no map, period.

shintuku

oh... i see

is there a canonical set-injection $R/A \to R$?

for a quotient of $\mathbb Z$ it would seem, so, but in general?

Ted Shifrin

No.

shintuku

04:10

i see, so that's a special property of $\mathbb Z$ in particular

Ted Shifrin

This is not something even to be thinking about.

shintuku

thanks a lot!

leslie townes

difficult to parse what you are even thinking of there.

shintuku

consider the following set-map: $\psi: \mathbb Z/(z) \to \mathbb Z: \overline x \mapsto x$

leslie townes

no, that's crap.

shintuku

04:13

even as a set-injection?

leslie townes

you can choose representatives of each element of R/A, and thus think of elements of R/A as represented by particular elements of R. but no 'canonical' (ugh) way of making those choices. it's just an arbitrary set of choices.

and the ring operation changes, so it's definitely not a ring homomorphism.

shin: for example, does \overline 1 go to 1 or to 3? or to -535? in your "set-map"

in Z/2Z

you can definitely choose a representative of each element of Z/2Z and think of the map sending the coset -> the representative that you get from those choices as a set map from Z/2Z into Z.

shintuku

to 1

leslie townes

but different choices of representatives give you different maps, and the map is not a homomorphism.

shintuku

we can give a formal definition maybe by considering how many unities away from unity $\overline x$ is

leslie townes

so there's a way in which you can think of any R/A as "a particular set of elements of R, with funny new operations," the way that you can think of Z/nZ as the subset {0,1....,n-1} with funny new operations, but that's not particularly helpful.

as you already see in the Z/nZ case, when sometimes it feels more natural to take representatives in some set other than {0,...,n-1}, and in how just making sense of the arithmetic becomes more difficult if you require everything to be done in terms of representatives.

shintuku

04:23

yeah i guess there is no such canonical map, but there does seem to be a class of rings that have a set injection $R/A \to R$ given by sending an element that is $n$ times the unity to $n$

then that class of rings has an unambiguous minimal representative for any residue class

leslie townes

step back for a minute and think about what the existence of a set injection tells you. just that one set has cardinality no greater than that of the other set.

it is sometimes useful and important to choose representatives of cosets in some nice way, but that is not this.

shintuku

it would have made for a nice proof of the lattice-isomorphism theorem: ideals which are mapped to themselves under this map are in a sense "unacted upon" by the quotient map, meaning that the action of moving from one ideal to another within the lattice is the same in both lattices

leslie townes

something kinda related and more useful is that for any unital ring R there is a unique unital ring homomorphism from Z into R. the kernel of this homorphism (which is either {0} or a set of the form nZ for some positive integer n > 1) tells you a tiny bit about R (it is called the "characteristic" of the ring).

shintuku

e.g., $(6), (3), (2)$ in $\mathbb Z/(18)$, but not those generated by $5,7,10$, etc.

leslie townes

there are no unital ring homomorphisms from a ring of characteristic n (like Z/nZ) into a ring of characteristic 0 (like Z).

it's really helpful to try to step away as much as possible from particular choices of representatives at this stage of getting used to these notions.

shintuku

04:56

ty for the comments!

Juliamisto

07:21

Hello. I have an English question about mathematics. Is it okay to speak of "a greatest common divisor" (of two elements in a commutative ring)?

leslie townes

hm, how would you define it in an arbitrary commutative ring? you might be able to come up with something, but i don't think it will resemble the integer gcd

you get something similar in rings with slightly more structure, e.g. integral domains

ooh, wikipedia has it en.wikipedia.org/wiki/…

per wikipedia, you get the nicest properties in UFDs, or something called "gcd domains" which might just be defined so that you get those nice properties

Juliamisto

Suppose that x, y are two elements (in a commutative ring R). If there exists some z in R such that x = yz, we say that y is a "divisor" of x.

If c is a divisor of x and c is a divisor of y, we say that c is a "common divisor" of x and y.

If g is a common divisor of x and y, and every common divisor of x and y is a divisor of g, we say that g is a "greatest common divisor" of x and y.

Here are some definitions.

I am late; you have found them in Wikipedia.

For two integers, it is okay to speak of "the greatest common divisor" because of the order; for two elements in an arbitrary commutative ring, "the greatest common divisor" does not seem to make sense.

leslie townes

right. whether or not a 'greatest common divisor of a and b' exists can depend on a and b, or there can be more than one of them, so you wouldn't say "the" (implying uniqueness) in reference to it

i guess there's already more than one of them in Z, but because Z only has two units this is only a sign thing, and we just pick the positive one by convention

i don't know what would replace 'pick the positive one' in more general situations

Juliamisto

I do not know, either.

I have never seen "a/an + superlative form" before, so I raised such a question.

leslie townes

07:42

yeah, it's a little weird, and maybe some contexts people wouldn't use it. wikipedia does indicate that it is sometimes used in abstract order theory, though (see en.wikipedia.org/wiki/Greatest_element_and_least_element and its reference to the notion of "a greatest element" in a preordered set, which becomes the unique "the greatest element" in a partially ordered set)

the introductory/summary section in the first paragraph actually gives the definition in terms of a partially ordered set (where it's unique) before discussing it in the context of preordered sets (where it might not be)

Juliamisto

I see. Hence, in the ring of gaussian integers, it could be okay to speak of "the greatest common divisor" of two gaussian integers, if the lexicographical order is used.

The order is not compatible with multiplication, though....

Juliamisto