and real parts should be equal too by some other argument

@Obliv yes

oh I know. Since $\text{Im}\langle \phi_2|A|\phi_1\rangle = -\text{Im}\langle\phi_1 | A | \phi_2\rangle$, we see that $\text{Im}\langle \phi_2|A|i\phi_1\rangle = -\text{Im}\langle i\phi_1 | A | \phi_2\rangle$

Now its not hard to see that by sesquilinearity of the complex inner product, this is actually $\text{Re}\langle \phi_2|A|\phi_1\rangle = \text{Re}\langle\phi_1 | A | \phi_2\rangle$

@Obliv

also $\langle \phi_1| A | \phi_2\rangle = \langle \phi_1, A\phi_2\rangle$ while $\langle \phi_2 | A | \phi_1\rangle^* = \langle A\phi_1, \phi_2\rangle$

from conjugate symmetry

right

so this tells you precisely that $\langle \phi_1, A\phi_2\rangle = \langle A\phi_1, \phi_2\rangle$, that is by definition and uniqueness of $A^\dagger$ that $A^\dagger = A$

I am working overtime trying to understand your physicist bra-ket notation

Yeah same, but I also don't know the math version to translate from so I have to learn the concepts and the notation at the same time :D

It's troublesome for sure.

it's free online if you wish to take a look. It's in the preliminaries in chapter 1.

what for

there are some typos tho.

for the "formal" notational definitions

I already understand the notation

I've learned it alongside with you as you were posting screenshots

So can all operators be represented as matrices?

And in this case A is a hermitian? because it is its own adjoint or something

@Obliv what do you mean by that

if your inner product space is finite-dimensional then any operator can be represented by a matrix

Ok so the bra vector is specified for the given matrix/operator right? If unspecified, then it's the bra vector for the identity?

On my phone now so writing out the latex is going to suck.

If you fix an orthonormal basis for your inner product space, say $(e_i)$, index by $I$, then $a_{ij} = (Ae_i, e_j)$ can be treated as entries of your "infinite" $I\times I$ matrix

Like what I mean is riesz theorem gives us a unique vector that we can define the inner product with to map onto the scalars, so the bra vector is that unique vector. If one applies an operator to that vector, it maps it to something else so we can represent langle psi | A | phi rangle as a new bra langle chi | phi rangle or something if we apply A onto the bra vector instead of the ket?

If $x = \sum b_i e_i$ then $Ax = \sum_i b_i Ae_i = \sum_{i, j} b_i a_{ij}e_j$

so in a sense we have multiplied the vector $x$ by the "infinite matrix" $A$

@Obliv so you mean to combine psi with A?

$\langle \psi | A|\phi\rangle = \langle \psi, A\phi\rangle = \langle A^\dagger \psi, \phi\rangle$

so essentially, $\langle \phi | A = \langle A^\dagger \phi |$

Oh right, and that's due to the properties of the inner product and not A right?

if you want to combine the thing on the left, its the same as to apply the conjugate of A onto that

@Obliv well, technically its the definition of the conjugate

Hmm, okay I'm starting to see the picture a bit.

I need to go to sleep, bye

Bye thanks for your help

Fell asleep in the History lecture

History used to be boring to me but now that I got more imagination at my disposal it's one of the more interesting aspects of life (especially non-anthropomorphic history)

but only interesting in my own pursuit of it since school puts a damper on things

$\displaystyle\sum_{n=0}^\infty\frac{n!(n+c)!}{(a+1+n)!(b+1+n)!}$ is of the form $\dfrac{\pi^2}6r+s$ for $r,s\in\Bbb Q$, where $r$ (the “pi part”) is $(-1)^c\dfrac{(a+b-c)!}{a!b!(a-c)!(b-c)!}$ if $a,b>0$ and $0$ otherwise.

I discovered this earlier

Is there any nice way to prove it?

Also, what's s?

@AkivaWeinberger playing with hypergoemetric functions?

does this only apply to non-finite fields

idk if a vector space over a finite field is ever even interesting but I can see that as being a counterexample

@Jakobian Is that what these are

I believe it… but I've never read about them so I don't know what's known about them already

@AkivaWeinberger Typo! “If $a,b\ge c$ and $0$ otherwise.”

@AkivaWeinberger it looks like an identity involving hypergeometric functions, yes

I believe a lot is known about them, they used to be largely studied

Also, it diverges if a+b<c (because it's roughly \sum 1/n^(a+b-c+1))

@Obliv This is also true for finite fields.

Note that for a map V->W to make sense, they must be vector spaces over the same field.

So if dimV>dimW, then V has more elements, so it can't inject into W.

can we construct a vector space over a direct product of finite fields?

something like $\mathbb{Z}_a\times \mathbb{Z}_b$ for some primes $a,b$?

obliv; usually the term "vector space" presumes a field as the "scalars." there are more general mathematical objects but i would not recommend investigating them until you have investigated vector spaces in more detail

@leslietownes ohh

wouldnt the scalars be the integers in that example tho

mod a and b

i don't know exactly what you are proposing, and i'm gently suggesting that maybe you don't either, and should spend your time on more focused questions

you don't have to follow my advice, but that's it (my answer to your question 'wouldn't the scalars be..' is 'no,' FWIW)

my general attitude is that vector spaces over fields present more than enough difficulty for someone to get lost, even when the field is fixed

So you can't have a vector space where $x_1 \in \mathbb{Z}_p$ for some prime $p$ and $x_2$ in some similar fashion?

you will only get more lost if you're like, well, what if i change the field? what if it's not a field but something like a field? what if it's, two fields?

at some point you're just randomly putting words together

obliv: a vector space over a field is a set together with various operations satisfying various conditions. to ask a question of that type you should be more specific about the set, and the field, and so on.

not just can you do a vector space with [string of symbols]

if p and q are distinct primes, there is no field F over which the cartesian product of sets Z_p x Z_q can be endowed with the structure of a vector space over F. if you were half drunk you might say that Z_p x Z_q "is not a vector space" or "cannot be made into a vector space," but i would prefer to say more precise things.

that's not remotely what you were asking, but it is an example of a clear question of the sort that you seemed to be getting near. until you can speak in the language of clear questions it is not helpful and maybe actively harmful to ask vague ones.

is Z_p x Z_q even a field

I didn't think that far ahead

I just mean

well yeah, you were already asking two things. you wanted a vector space "over" something that wasn't a field.

imagine this except instead of a ring with or without unity, we plug in Z_p x Z_q

ok then my mistake, I should have at least checked if it was a field

then you asked for a vector space "where X_1 in Z_p" (maybe meaning that at least one element of Z_p was an element of the vector space) "and x_2 in some similar fashion" (which is word salad but could be taken to mean, where some element of some other Z_p is also an element of the same space).

there's no logical relation between those questions at all, an LLM could have spit them out

yeah I just realized that's not even what the thing I posted earlier is about. It's the dimension of the set $M$ in that definition, not the field, that matters in linear mapping.

like the dimension of the vector space isn't related to the scalar field

What I should have asked is can a vector space have a finite span

those words seem randomly put together. can you be more precise? say V is the vector space and F is the underlying field. you're asking, can [what] be [what].

but i don't know what fills in those blanks

can $V$ have a finite span

given some basis B, can Span(B) be finite

I assume only if the scalar field in question is finite

okay, if B is a basis for V, then span(B) is just V. you're asking if V itself can be a finite set.

and yes, that can happen only if the scalar field is finite. it doesn't have to happen even then, but that's when it does happen.

i'm assuming you want V to be nonzero here. i suppose {0} is a perfectly good vector space over any field, even if it's an uninteresting one.

if V is a vector space and v is a nonzero vector in V, then the set {kv: k in F} has the same cardinality as F. under these circumstances, the vector space axioms force the multiples of v to be different elements of V whenever the scalars involved are different.

that's a good exercise, actually.

but can I have dim V > 1 with that property of being finite

I wanted to make an injection/bijection from different dimension vector spaces if possible

it's really helpful if you write more complete questions every time. you change what you're asking an awful lot, and it's putting a lot of burden on the reader to go back and piece together what "can I have dim V > 1 with that property of being finite" might mean now.

when you've just been told a couple of things, among them, that if V is a nonzero vector space over a field F, and V is a finite set, then F is also a finite set.

Okay, I guess that answers the question, I can just simply add another dimension by copying V so define P = VxV

@Obliv you should state this more carefully. if the vector spaces in question were required to be over the same field, and the bijection required to be a linear map, then the impossibility of that is a standard theorem, probably proved in any one of the dozen books you seem to be cycling through.

Yeah I'll mull this over and figure out where the flaw in reasoning is. I'll give it one last attempt though: define $V$ as a set of ordered pairs$\{(x_1,x_2)\mid x_1,x_2\in \mathbb{Z}/3\mathbb{Z}\}$ together with $+: V\times V \to V$ under which $V$ forms an abelian group, and the field action $\mathbb{Z}/3\mathbb{Z}\times V \to V$ denoted by $kx$, for all $k \in \mathbb{Z}/3\mathbb{Z}$ which satisfies $(k+j)x=kx+jx$ for all $k,j \in Z/3Z, x \in V$, associativity in a similar fashion,

then distributivity from V onto Z/3Z as well

basically, the module/vector space axioms. Then define a map $T: V \to W$ where $W$ is like $V$ except its made up of just $\{x\mid x \in \mathbb{Z}/7\mathbb{Z}\}$

er wait, had to change the scalar field of W

okay. your "V" is an example of the usual way of regarding the cartesian product F^2 as a vector space over F, when F is a field. in your case you are taking F = Z/3Z.

so far so good.

oh.. wait these are prime numbers. Obviously there won't be an injection/bijection between finite sets of prime cardinality..

derp

I think therein lies my problem

well, yes, if W is any set of 7 elements, there won't be any injection or bijection between V and W. for cardinality reasons.

so there is always something nonzero in the null of $T$ between different dimension vector spaces. at least 1 object

note that generally in linear algebra the maps most of interest are not arbitrary injections or bijections, but linear ones. and the usual definition of linearity requires vector spaces over the same field. but you can kind of forget about that in your toy example. it wouldn't matter what vector space structures you put on those sets, or whether or not you could do it. you couldn't find just a map of sets that did the job

I think you can force it to be linear, like for $a \in V$ and $b \in W$ we can define $a + b$ as $(x_1+y,x_2)$ for $x_1,x_2 \in \mathbb{Z}/3\mathbb{Z}$ y in Z/7Z

separately, you could ask, is there any way of "making" Z/7Z a vector space over Z/3Z, i.e. providing W with some kind of addition, and scalar multiplication by elements of Z/3Z, consistent with the axioms. the answer to that question is also no, not possible, but maybe it's more interesting as it might require deeper thought.

obliv: no, you can't "force" it to be linear. you can't make Z/7Z into a vector space over Z/3z.

some abelian groups just can't be vector spaces over some fields.

this is maybe surprising, but it's true.

I think that example I gave gets messed up because of inverses/identities or lack thereof

idk why it's not linear

all of this stuff gets really confusing if you forget to check the basics. e.g. whether or not a fixed candidate set of 'scalars' is actually a field. e.g. whether or not you actually have a vector space structure on the set you have in mind.

this stuff doesn't come automatically, so you can't just assume that you have these things. maybe it's counterintuitive, but it explains why so many examples in linear algebra books take the form of long and boring marches through axioms. you'd think "surely i can skip this and just assume it all works." well, as a matter of pedagogy you can, if someone else has marched through the axioms and checked them and given you some license to do that.

in particular, if they aren't going to ask you questions that require you to dive into those details. you can write a good book that skips those details, but it skips them at the cost of being able to get into that stuff.

but there isn't some general phenomenon that one can just grab any set one wants and any field one wants and presume that you've got some kind of vector space structure from those choices.

idk, aside from addressing the dimension of V (or M in the case of modules), I don't see how the set of 'vectors' is at all special other than by name. You could call them the set of fruit over a ring/field and still have the same structure

it can seem like that in the curated environment of some textbooks or treatments, but in reality, if someone makes exotic choices for what those things are, "is there a vector space consistent with these choices?" should be regarded as a live and open question until someone has marched through the axioms or otherwise convinced you that the answer is yes.

obliv: you're changing gears yet again, but indeed there is no special reason for elements of an abstract vector space to be called by any particular name, and you don't learn very much about an element of a set by learning that it is an element of an abstract vector space, or about a set by learning that it can somehow be made into an abstract vector space.

that's definitely true as a high level general statement. also true as a high level general statement, you can't arbitrarily assume that anything in sight has or could be given some vector space structure.

at a high enough level almost everything is true.

@leslietownes just to be clear, we can't define structure preserving maps between vector spaces of different underlying fields?

regardless of being able to make an injective/bijective one?

what do you want the definition to be? what do you get when you do that in simple examples?

you can't define "structure preserving maps" separate from "structure," really. you need to figure out what you want for all of that at the same time.

@leslietownes well, I'll explore it tomorrow and give you a break from my salads. I think since Ted doesn't frequent here anymore I've grown used to being able to ask vague-ish questions (I still like to believe I'm not being vague)

rolls pi^e eyes

as you said before, I must become the matrix in order to understand it, or whatever.

oh that reminds me, my senior year transfer CS teacher randomly joined in on a conversation b/t my friends and casually dropped a "do you guys know the matrix movies?" we're like yea.. and he said with 100% conviction "yeah I think it's real"

you can't be told what the matrix is, you have to see it for yourself

I can't tell if he was messing with us, but it was funny. He must have coded it.

@leslietownes yes that

I should have said "be careful, they're watching you" or something teehee

also jakobian please dont read my slop, I didn't define the "linear map" correctly anyway, it was some failed binary operation on $V\times W \to V$ which isn't what I wanted

when you say it that way, it kind of sounds like you want him to read it

I don't think it matters, it's not like his perception of me being a maths genius is in anyway tarnished.

don't think of an elephant, anybody. just forget about elephants entirely

Hi, is there a way to use citations in LaTeX by name so my reference appears as [DudeBro69] instead of [1]?

do you use bibtex? you can handle stuff like that with a bibtex style file. it does require hand-coding some of the abbreviations like DudeBro69, or at least, i recall tweaking that the last time i used it.

idk how to use bibtex but I guess chatgpt can give a template

it's not as straightforward as I would have expected. Do you have a working example I can copy from?

i recommend 'growing your own' as little as possible, but see if one of the options reed.edu/it/help/LaTeX/bibtexstyles.html works. if you do it 'right' you just maintain a database of papers (i.e., add the paper once, one time, and it's citable forever in all of your docs the same way) and the style does the rest.

there might be some hackish one-off ways to do it. the latex SE people would be horrified by the request but maybe also help you with that

:)

many places also export bibtex database style citation things so it's easy to add papers to a database without manually filling anything out. there are probably latex overlays that handle this automatically now. my knowledge in this area is 20ish years old.

@leslietownes Did anyone say that the elephants in consideration are pink and flying? Oh sorry, no one did. Please don’t think about them.

And they are definitely not doing circus tricks in the sky…

@Obliv okay I won't

@Obliv it isn't, you can be sure of that

Speaking about elephants, I wish mammoths were still alive

thanks for inviting me

Thanks for coming.

Just three guys here. Never seen that before.

@Jakobian I'm not from England either, but as an American citizen who now lives in Canada, some exposure to the history of England is unavoidable, plus as you alluded to, large sections of early Protestant history happened under the rule of a series of kings (Henry VIII, Mary Tudor, Elizabeth I, James I, Charles I).

It appears that British people love their King Arthur, a rather mythic figure. My point was that although Egyptians love their Pharaohs too, to the point of spicing up a historical king to mythic proportion, we don't have to discount the historical veracity of their other kingly succession (which in the Egyptian case seems to be solid). Similar to the history of Kings of ancient Israel.

@nsmon93 You should learn. If you are going to TeX a lot of papers, it would be good to learn some of the basic packages that get used a lot. Most journals will provide some template for TeX documents, which includes formatting information for BibTeX (so if you already have your sources in a .bib file, they will be automatically formatted per the journal's house style).

Also, once you figure out how to use BibTeX, you can (for your own work) tweak the way that citations are handled, including the tags used in-text.

were to vote?

@leslietownes Google Scholar, for example.

@FedericoRuck You can't, yet. The nomination phase has not ended.

ok sorry

when will I be able to do so?

math.stackexchange.com/election

... and to the history of Mesopotamian kings including the famous Akkadian King Hammurabi (c. 1750 BCE) and his famous Code of Hammurabi to which we owe tremendous assistance to understand the thought world behind the Torah.

@XanderHenderson why the mask on your profile picture? Is it from the pandemic times?

Yes.

@XanderHenderson You don't have to answer this, but seeing there are 8 candidates of the election, I wonder what's the incentive to be a moderator in Math.SE? I don't think there was much enthusiasm the last time we had election in Christianity.SE or Hermeneutics.SE. Or maybe I just read the candidate's statement, some of which show the reasons why they nominate themselves.

@GratefulDisciple I have no idea what motivates others. Personally, I see it as a way of being of service to my community.

But you also have to recognize that Math is a lot bigger than the other sites you mention.

@XanderHenderson Of course service to the community is an incentive. What I was wondering is the personal cost. Personally I would rather contribute in the many ways that non-moderators can already do, like doing the queues, flagging Q&A, welcoming new members in comments, answering questions in the chat rooms. I declined to be a moderator because I feel I'm in a hot seat when a controversial issue breaks out. Basically, I don't like to be in the midst of a conflict, especially when I'm "in charge"

@GratefulDisciple I don't mind being the target of controversy. :D

Honestly, when I first ran, I really didn't want to, but did so because I looked at the other candidates and thought "I think that I can do at least as well, if not better, than they would, and a couple of the candidates would be a disaster for the community."

@XanderHenderson Good for you. It's great that some are willing to do the necessary task. I'm just not cut out to be a moderator.

And I am only a moderator on Math Ed because I put my name forward to make sure that there were enough candidates to make the election run.

@XanderHenderson Yeah, again, that falls under the umbrella of "serving the community". I can relate to those in the American election cycle now, having to bear the brunt of unfair & unjust vilification.

@XanderHenderson Yes, I felt that too, before other qualified nominations appeared. And yet, you earned the community trust and you stayed. It's commendable.

@XanderHenderson Anyway, good luck to this Math.SE election, may the one who has the most serving heart and also the most worthy (mathematically) wins. Maybe by the next election I would have enough points to vote.

I'm curious to know what the Americans in this room think about Harris replacing Biden. Is she a good match against Trump?

@Sahaj I'm not sure that I understand your question.

How does she fare in competition against Trump according to you?

Polling indicated that Biden was nigh certainly going to lose. But current polling indicates that Harris is statistically even with Trump. So, from that point of view, Harris is a better candidate than Biden.

Polls show that they're head to head but I can't believe that at all

@XanderHenderson Her sudden surge in popularity is quite apalling. An article (on USAToday) shows she had an approval rating of 36% in March whereas now she has a lead over Trump

@XanderHenderson and it was true

@Sahaj What do you mean "apalling"?

I meant appalling. Something along the lines of "shocking" "surprising".

"Appalling" has a negative connotation, as in "it is terrible that her approval ratings have improved". This is a poor choice of word if you mean "shocking" or "surprising".

Personally, I don't find it all that surprising. As VP, she is relatively invisible, and her approval rating is largely tied to the approval rating of the administration as a whole (which is low, I suspect, because people have doubts about Biden's mental competence).

@XanderHenderson Wasn't aware of that. Thanks for your perspective though.

Once she was put forward as the de facto Democratic candidate, she became much more visible, and distinct from the current administration. Also, there is a kind of "honeymoon" bump or "sugarr ush" which is likely to wear off a bit over time.

Would be quite something to get to witness the first ever woman POTUS in US history.

The real questions are (a) can she sustain the "Oh thank g-d, it's not Biden!" honeymoon through to November? and (b) can she win in important swing states?

@冥王Hades US in POTUS stands for united states

Because I am generally kind of pessimistic, my prediction is that she wins the popular vote by a significant margin, but loses the electoral vote by one state (probably Ohio).

So you predict something like 2016 happening again. Interesting.

I'm not gonna be voting, just like last election.

But, if elected, would she make a better president than Trump if you were to compare them by their previous work record and major policies?

@冥王Hades why not

@Sahaj Any sane person would be a better president than trump

@Sahaj In my opinion? A warm stone would be a better president than Trump.

Oh damn was he really that bad as president? Unbelievable.

I think America is in the state that what "conservatism" and "right" is in reality "radical right"

So anyone reasonable would vote for the "left" or "republicans"

8 candidates

(sorry for interrupting)

@Sahaj How old are you, and how aware of American politics were you starting in 2016?

I'm in high school at the moment. I don't know much more of American politics than what's available on our media or on twitter.

Trump (and to a perhaps lesser extent, the modern Republican party) has a profound authoritarian streak. He clearly believes that the US president should have dictatorial powers, and has done quite a lot to ensure that if he is elected president again, he will be able to act in a much more authoritarian manner than has been historically possible.

@Sahaj Yeah, that makes sense. You are too young to have really experienced the first Trump administration.

@XanderHenderson So he's anti-democracy and YET there's voters who support him in the US, according to polls at least?

@Sahaj Yes.

@Sahaj voters care about feelings rather than facts and logic

Though, to be fair, I'm anti-democracy. I'm all for representative democracy, but I think that, for example, California's referendum system demonstrates the hazards that occur when you let anyone vote on anything. I am all for having a group of elected elites, who are paid well enough that they can focus on policy.

American politics is in the state of anti-intellectualism where facts don't matter and someone with no expertise should be treated on the same level as an expert

Trump can get away with anything by simply lying and appealing to feelings

I see. I'm sure that the US must have constitutional safeguards in place to prevent Trump, (or whoever) from concentrating too much power and acting in a authoritarian manner, though, no?

@Jakobian Oh yeah, I second that. The rampant anti-intellectualism present in American political spheres (and even outside of that) is simply appalling. It felt almost nauseating speaking to a group of such people.

They're under the delusion that Free speech implies that the opinion of a flat earther, for instance, must be given the same audience (and perhaps even same credence) as that of a scientist. And if that is what free speech is, then I would be against such insanity, obviously.

@Jakobian Do you think this is better in Germany ? Here , logic is not worth anything either. Politicians only speak about the climate change , but not about how the record inflation could be prevented.

@Sahaj In theory, sure.

In practice, the Republicans have worked very hard over the last 30 or so years to weekend and dismantle those protections.

Are you affiliated to the Democratic party?

I'm so very skeptical of your claims that Republicans are working to dismantle democracy, and that they have been doing so for 3 decades. It's possible to fool people in an election or two and win based on hate politics and lies, but it's surely impossible to discreetly and systematically remove a country's constitutional democratic safeguards without consequence. There were Democratic presidents and legislatures in the past 3 decades. Did they not do anything to prevent it?

@Sahaj Technically, I am a registered Republican. (At various times, I have been registered as a Democrat, a Green, a Democratic Socialist, an independent, and a Republican---I don't have much allegiance to any particular party).

I would suggest that you look into the history of Newt Gingrich and Mitch McConnell.

And Republican court packing practices over the last several decades.

Consider the absolutely disgusting manner in which RGB's successor was selected.

Alright; I will look into those. Thanks for the discussion though.

Take a look at the history of lynching blacks in the USA also...

3/5 amendment :(

Reading some Rudin :) In Theorem 3.3d), we have $s_n\to s$ as $n\to\infty$, so we can choose an $m$ such that $|s_n-s|<\frac12 |s|$ for $n\geq m$. Then he says $$\text{"...we see that }|s_n|>\frac12|s|,\quad (n\geq m).\text{"}$$Why is this true?

Let's see if this works, but here is part d) of Theorem 3.3: i.sstatic.net/N316Y.jpg

and here's the statement of the theorem: i.sstatic.net/6ryyK.jpg

Figured it out. Thanks for the help.

@SineoftheTime 3/5 amendment? Huh?

Do you mean the 3/5 compromise? It was not an amendment. It was baked right into the original constitution. It went away at the end of the Civil War, with the equal protection clause of the 14th amendment.

This dude is my new hero: tiktok.com/@thetnholler/video/7400097929437973802

Turkey's a veg, right?

What does 'external binary operation' means in general? Is it just a way to emphasize the fact that the codomain of the operation is different from the field $K$?

Not in Minnesota.

zawarudo: my guess is as follows. in many contexts a 'binary operation' is defined to be a map from S x S to S, for some fixed set S. when K and L are different, a map K x L to L does not literally meet this definition, but comes pretty close, and 'external' is a way of signaling that.

'external' might be signaling that the things coming in from the K part of the operation may not be members of the set L.

note that the addition in an abstract vector space (or more generally the operation of interest in any group) is a 'binary operation' in the restrictive sense. a lot of algebra books will define 'binary operation' that way, even though the urge to use 'binary operation' more generally for "some kind of map taking two inputs" is pretty strong, and sometimes people forget that it also has this restrictive meaning.

i kind of like the idea of using a word like 'external' to signal the use of the word in a non-restrictive sense, actually. what book is that?

i'm wondering if that's a work in translation. i'm not sure that i've seen a native english book that makes that distinction explicit, and it's a good one. of course there are probably some out there

I have another question about tensor products... All rings in my question will be commutative. Suppose $f:A\to B$ is a ring homomorphism, and $M$ is an $A$-module. The map $f$ allows us to view $B$ as an $A$-module, and so we can form the tensor product $M_B=B\otimes_{A} M$. In Atiyah and Macdonald, it is claimed that $M_B$ has a $B$-module structure: we set $b(b'\otimes x)=(bb')\otimes x$ and extend by linearity. My question is: why is this multiplication well-defined?

The generators of a tensor product might not be expressible in a unique way, i.e. $b\otimes x$ can equal $b'\otimes x'$ without $b=b'$ and $x=x'$, so it seems that there is some work needed in showing that this definition makes sense.

@leslietownes Thank you for your reply! The book is Kostrikin, Manin - Linear Algebra and Geometry

I had the impression that Ted was in chat for a millisecond

@SineoftheTime: Yes, I saw him too...

can someone help me find fundamental group of $\mathbb{RP}^2 / S$, where $|S| = 3$