Mathematics - 2013-10-21 (page 1 of 3)

12:00 AM

rather, "if each $\alpha_1,\cdots,\alpha_n$ are algebraic over $F$ then $F(\alpha_1,\cdots,\alpha_n)$ is finite over $F$" (and this is proved by induction)

note that I replaced "algebraically dependent" with "algebraic."

leo

indeed

easy one

Pedro Tamaroff

@TedShifrin Hehe, this is interesting.

leo

I had a proof of my thing. It used the same ideas of the case $n=1$ (which is true). That is, using that $F[x_1,\ldots,x_n]$ is a PID. Then the prime are the irreducible there. Then there was a minimal polynomial associated to $(\alpha_1,\ldots,\alpha_n)$ and so on...

don't know where is wrong

Ted Shifrin

@Pedro, wot is?

Pedro Tamaroff

@TedShifrin Let $f:V\to V$.

Suppose $\lambda$ is an eigenvalue.

Then Halmos calls $\dim E_\lambda$ the geometric multiplicity of $\lambda$.

Ted Shifrin

12:07 AM

Yes, so do I :)

Pedro Tamaroff

While it calls ${\rm mult}(\chi_f,\lambda)$ the algebraic multiplicity.

Ted Shifrin

Yuppers.

Pedro Tamaroff

So we know ${\rm geometric}\;\leqslant \;{\rm algebraic }$. =O

Ted Shifrin

Yes. Nice proof of that.

Pedro Tamaroff

@TedShifrin I was trying to be funny.

Ted Shifrin

12:09 AM

Ohh :) pfeh.

You suckered me, you twerp.

Pedro Tamaroff

@TedShifrin Oh, noes.

Now I must go.

Ted Shifrin

Grr^3.

Pedro Tamaroff

Really, I should.

And tensors keep eluding me.

That's probably because I don't know them.

Ted Shifrin

Happy dinner.

Pedro Tamaroff

Haven't met them yet.

leo

12:11 AM

exactly :)

Pedro Tamaroff

@TedShifrin Not dinner.

Ted Shifrin

Oh ...

Pedro Tamaroff

@TedShifrin But probably in a while.

Yummers.

We had asado.

Now I jump.

Ted Shifrin

Jump?

Mexican jumping beans? Argentinian?

anon

@leo if those are indeterminates, then F[x,y,...] is not a PID

so the ideal of polys vanishing at (a1,...) is more complicated than "all multiples of some blah"

leo

12:26 AM

@anon yes they are. isn't it an euclidean domain? with degree as euclidean valuation? degree of $p$ the max of the powers of the indeterminates which appear in $p$

anon

if you think so, try to find a generator for (x,y) in Q[x,y].

leo

of course

again

anon

or in terms of being a euclidean domain, try to divide x by y using that valuation

leo

I've just thought about

it's none of the things I've just said

@PedroTamaroff enjoy

robjohn

@PedroTamaroff Convince yourself to learn General Relativity. You will learn tensors :-)

Pedro Tamaroff

12:46 AM

@robjohn I can try.

=)

Enjoys Math

Anyone know how to put more structure on this "semigroup"? math.stackexchange.com/questions/533835/…

Don Larynx

Hi @PedroTamaroff

Enjoys Math

Take ordered pairs of non-coprime integers $(a,b)$ and you can multiply them componentwise and get another member of that set. What other ops can you do?

anon

most generally any function $S\times S\to S$ is a binary operation under which $S$ is closed. but this is a boring answer.

so you want to narrow down some arithmetic properties you want the operations to have in order to narrow down the search and make it less boring.

Enjoys Math

How though

computer search?

anon

1:00 AM

in particular, some kind of arithmetic properties that your examples already have, which are flavored with ingredients from SL2 and divisibility-lattice theory

Enjoys Math

reason is that seeing if some of the operations form a ring-like structure with each other, gets really tedious on paper, especially when it seems like it will never work

anon

@EnjoysMath no, by thinking

trying to find qualities that make a problem exciting rather than boring is not a problem for a computer search

robjohn

@anon developing such a search would be exciting ;-)

anon

don't go all singularitarian now

robjohn

@anon I won't, but my computer will soon :-)

Don Larynx

1:14 AM

Denseness is transitive: Given three subsets A, B and C of a topological space X with A ⊆ B ⊆ C such that A is dense in B and B is dense in C (in the respective subspace topology) then A is also dense in C.. How exactly is this so? This implies $\bar A = C$.

That means that $B = C$.

But $\bar B = C$.

So...I'm lost here

Pedro Tamaroff

@DonLarynx Go back to the definitions.

Karl Kronenfeld

@DonLarynx You're working with two different topologies. In particular $\bar A$ is an ambiguous expression--it could be based on the topology of $B$ or it could be based on the topology of $C$.

Pedro Tamaroff

See Brian's answer here.

Don Larynx

1:35 AM

Oh so if we base the set A with the topology of C then that is what the proof is trying to say

?

agent154

1:49 AM

I'd like some input on a problem I'm having with this proof:

"If $(m,n)=1$, show that $m^{\varphi(n)}+n^{\varphi(m)}\equiv 1\mod{mn}$." I don't know if I need to show everything up to my final step to ask this question, but if you want to see my full working, let me know...

I've got $m^{\varphi(n)}+n^{\varphi(m)}-1=mn(m^{\varphi(n)-1}n^{\varphi(m)-1}-l)$... but I'm worried that I can't subtract 1 from $\varphi(m)$ and $\varphi(n)$ and guarantee that I'm getting a non-negative power.

Pedro Tamaroff

@agent154 Since $(m,n)=1$, the above is equivalent to something $\mod m,\mod n$. What is it equivalent to?

agent154

@PedroTamaroff OK, let me type out my whole proof. I was hoping to save time, but I guess I can't

Kevin Driscoll

@agent154 I think if your quesiton requires typing out the whole proof, it might be better to ask on main. But thats just my opinion

maybe your proof isnt long

agent154

It's not terribly long

And what's "main"? You mean to ask as a question on math.stackexchange.com?

Kevin Driscoll

Yes

Pedro Tamaroff

1:56 AM

@KevinDriscoll Sup?

agent154

Since $(m,n)=1$, by Euler's Theorem, we have $m^{\varphi(n)}-1=nk$ for some $k\in\mathbb{Z}$, and $n^{\varphi(m)}-1=mj$ for some $j\in\mathbb{Z}$.

\begin{align}
(m^{\varphi(n)}-1)(n^{\varphi(m)}-1)&=(nk)(mj)\\
m^{\varphi(n)}n^{\varphi(m)}-m^{\varphi(n)}-n^{\varphi(m)}+1&=(mn)(jk)\\
m^{\varphi(n)}+n^{\varphi(m)}-1&=(mn)(-jk)+m^{\varphi(n)}n^{\varphi(m)}\\
m^{\varphi(n)}+n^{\varphi(m)}-1&=(mn)[jk+m^{\varphi(n)-1}n^{\varphi(m)-1}]\\
\end{align}
Hence, $m^{\varphi(n)}+n^{\varphi(m)}\equiv 1\mod{mn}$

Pedro Tamaroff

@agent154 Perfecto.

In fact, you can mod out by $mn$ in the second step to conclude.

Do you see?

agent154

Again, I'm not confident I can factor out an $m$ and $n$ from $m^{\varphi(n)}$ and $n^{\varphi(m)}$. What's to guarantee that $m$ or $n$ isn't 1? What's $\varphi(1)-1$?

Pedro Tamaroff

@agent154 You needn't factor anything.

Kevin Driscoll

@Pedro Basically nothing

agent154

2:02 AM

@PedroTamaroff I'm not sure I see what you're pointing out. How can I get what I want when there's an extra term on the left, and the factors are negative?

Pedro Tamaroff

$\varphi(a)$ is always greater or equal to $1$.

Thus when you write this $m^{\varphi(n)} n^{\varphi(m)}-m^{\varphi(n)}-n^{\varphi(m)}+1=(mn)(jk)$$

You can look at that modulo $mn$ to get $1=m^{\varphi n}+n^{\varphi m}\mod mn$.

@KarlKronenfeld Man. I will take a step further to the Dark Side.

And avoid using parenthesis for evaluation.

@KevinDriscoll Then read the title of this quickly.

agent154

What happens to the term $m^{\varphi(n)}n^{\varphi(m)}$?

If I don't move it to the right?

Pedro Tamaroff

@agent154 Nothing, really.

It is $=0\mod mn$.

Kevin Driscoll

@pEDRO Looks like menstruation

Karl Kronenfeld

@PedroTamaroff :)

robjohn

2:07 AM

$$
m^{\varphi(n)}\equiv1\pmod{n}\\
n^{\varphi(m)}\equiv0\pmod{n}
$$
and
$$
m^{\varphi(n)}\equiv0\pmod{m}\\
n^{\varphi(m)}\equiv1\pmod{m}
$$
add each

Pedro Tamaroff

@robjohn Well, I wanted to say that.

But I do like his proof.

robjohn

@PedroTamaroff the one without mods?

Pedro Tamaroff

@robjohn Yes.

Fernando Martin

@PedroTamaroff NNNNNNOOOOOOOOO

robjohn

@PedroTamaroff Each one is suited for a different point in the process.

Pedro Tamaroff

2:08 AM

►

Fernando Martin

I was thinking something along the lines of this

Pedro Tamaroff

@FernandoMartin Heh! I still have to see the original trilogy.

robjohn

►

Alec Teal

You know what I like?

robjohn

I use that no for the ringtone for my boss

Alec Teal

2:11 AM

Jar-Jar Binks, he really made the films for me.

robjohn

@AlecTeal has your mother had you tested?

Pedro Tamaroff

@robjohn HAHAHAHAHAHA

agent154

@PedroTamaroff OK... I see what you're saying. But I feel more comfortable being more explicit anyhow, so I looked up the definition of $\varphi(n)$ again, and $\varphi(1)=1$, so $n^{\varphi(m)-1}$ is indeed still an integer.

Fernando Martin

@AlecTeal what the

Alec Teal

@robjohn for several things, yes, what specifically? Also I was obviously not being serious. no one has ever said that.

Pedro Tamaroff

2:12 AM

Why don't people like Jar Jar?

Alec Teal

Everyone knows the most awesome bit of starwars is in the third film, where he replaces Jar-Jar with droids that have a crap sense of humor.

@PedroTamaroff watch the films.

robjohn

@AlecTeal I was just quoting Sheldon from Big Bang Theory. When anyone says he is crazy, he says, "no, my mother had me tested."

►

agent154

@robjohn Just to refresh my memory... if $a\equiv b\mod{m}$ and $a\equiv b\mod{n}$, then is $a\equiv b\mod{mn}$?

Fernando Martin

@PedroTamaroff: I don't know much about SW but IIRC most people complained that Jar Jar was just introduced to make the films appealing to children

robjohn

@agent154 if $(m,n)=1$

@agent154 actually, $a\equiv b\pmod{\mathrm{lcm}(m,n)}$

Alec Teal

2:15 AM

Anyone here know what a "regulated" function is? Because I'm pretty sure my university picks names just to make topics impossible to find.

It's similar but not the same as bounded variation.

robjohn

@AlecTeal I don't think I've used that term for anything like that...

Alec Teal

@robjohn I can imagine, I'm pretty sure they make things up.

Pedro Tamaroff

@AlecTeal Yes.

Alec Teal

I am surprised they haven't started using random dictionary words

robjohn

Regulated function

In mathematics, a regulated function (or ruled function) is a "well-behaved" function of a single real variable. Regulated functions arise as a class of integrable functions, and have several equivalent characterisations. Regulated functions were introduced by Georg Aumann in 1954; the corresponding regulated integral was promoted by the Bourbaki group, including Jean Dieudonné. Definition Let X be a Banach space with norm || - ||X. A function f : [0, T] → X is said to be a regulated function if one (and hence both) of the following two equivalent conditions holds true : * for eve...

Pedro Tamaroff

2:17 AM

It means it is the uniform limit of a sequence of step functions.

Alec Teal

"Help, I'm trying to prove the fish lemma"

@robjohn that's wikipedia, written by people who have forgotten more than I know. Very little of that makes sense. I am really searching for some examples.

Bitrex

Regulated?

►

Alec Teal

books.google.co.uk/…

probably the best so far.

robjohn

@AlecTeal give someone a lemma and they will graduate high school, teach someone how to prove a lemma and they will graduate college?

Alec Teal

@robjohn good one, I did not expect that!

/me means that.

►

Bitrex

2:24 AM

to isengard to isengard

Alec Teal

what did you say?

agent154

Am I correct with this one?

Suppose that $p$ is an odd prime. Show that $1^{p-1}+2^{p-1}+\dots+(p-1)^{p-1}\equiv -1\mod{p}$.

Proof:
Note that by Fermat's Little Theorem, we have $a^{p-1}\equiv 1\mod{p}$ when $a\nmid p$. Therefore, $a^{p-1}\equiv 1\mod{p}$ for all $1\leq a\leq p-1$.
\begin{align*}
\sum_{i=1}^{p-1}i^{p-1}&\equiv\sum_{i=1}^{p-1}1\mod{p}\\
&\equiv p-1\mod{p}\\
&\equiv -1\mod{p}
\end{align*}

Pedro Tamaroff

@agent154 Yiss.

agent154

I feel like I'm on a roll tonight

Pedro Tamaroff

@agent154 Keep it up! =)

agent154

2:35 AM

Now, to tell you the truth, I have absolutely NO idea where to start with this one... "Prove that $\frac{1}{5}n^{5}+\frac{1}{3}n^{3}+\frac{7}{15}n$ is an integer for every $n$."

Pedro Tamaroff

@agent154 Heh.

Well, not factoring.

agent154

Ohhh.. am I going to use FLT here too?

I think I see it

Pedro Tamaroff

But at any rate what you want to do is show $3,5\mid 3n^5+5n^3+7$.

agent154

If I post a PDF of my assignment somewhere, would anybody be willing to proofread it? It's only 6 questions, at most 3 pages (once I typeset the last two questions in latex)

Karl Kronenfeld

@anon It may be preferable to take the cue from the author of that paper and not work with that specific $T$ in proving the containments.

anon

2:39 AM

IOW to prove that A is equal to the intersection of prime-localizations containing it

Karl Kronenfeld

For instance, the set of all denominators occurring in $A$ should be a saturated multiplicative set.

(That's just what I gathered from reading on)

anon

how is the set of denominators formally defined?

Karl Kronenfeld

Just invertible elements of $R\subseteq A$.

Pedro Tamaroff

@agent154 Note that $\mod 3$, $3n^5+5n^3+7n=5n+7n=12n=0$.

And $\mod 5$, $3n^5+5n^3+7n=3n+7n=10n=0$.

anon

I have considered $(R\cap A^\times)^{-1}R\subseteq A$, but I haven't been able to do the other direction.

agent154

2:43 AM

@PedroTamaroff Oh, ok.. I see how this works

Karl Kronenfeld

@anon Oh right, I was having trouble with that too. Getting the other direction would make the forward containment really easy in the starred equation.

agent154

@PedroTamaroff I'm not sure I see how $5n^3\equiv 5n\mod{3}$.

Pedro Tamaroff

@agent154 We aways have $n^p=n\mod p$.

agent154

Ah yes, FLT again

Pedro Tamaroff

If $p\mid n$; this is trivial. If not $p\not\mid n$, and we use Little Fermat.

agent154

3:10 AM

Prove that $m>1$ is prime if and only if $m$ divides $(m-1)!+1$.

I have the "If $m$ is prime, then $(m-1)!+1\equiv 0\mod{m}$" part done, but I'm not sure how to do the reverse. I get $(m-1)!\equiv -1\mod{m}$, which is the consequent of Wilson's Theorem, but I can't deduce that $m$ is prime from this. Any tips?

That is, I'm trying to assume that $(m-1)!+1=mk$ for some $k\in\mathbb{Z}$, and showing that $m$ is prime.

anon

if m is composite then (m-1)! is no longer coprime to m

Enjoys Math

How would I deduce the form of these polynomials: texpaste.com/n/kgtz4d44

I know they exist, for example $(a c - b d, a d + bc) \in S$.

I think it's of the form, let $R = \Bbb{Z}[x_i : i =1..4]$, $acR + bdR + adR + bcR$

Karl Kronenfeld

@anon Let $\mathfrak q$ be a prime ideal of $A$ and let $\mathfrak p=R\cap\mathfrak q$. Don't we have $R_\mathfrak p=A_\mathfrak q$? Thus set $T=\{\mathfrak q\cap A:\mathfrak q\in\operatorname{Spec A}\}$.

Damn, we'd still have to first show $A\subseteq R_\mathfrak p$.

Pedro Tamaroff

3:25 AM

@KarlKronenfeld What has you and the @anon coming and going like this?

Karl Kronenfeld

@PedroTamaroff This question

Pedro Tamaroff

People sholdn't be allowed to use names of famous mathematicians as usernames! ARGH!

Karl Kronenfeld

@anon I was trying earlier to construct a variant of $I=\{a\in R:a/b\in A\text{ for some }b\in\mathfrak p\}$, noting that $I^n=(r)$ for some $r\in R$, $n\in\mathbb N$.

Pedro Tamaroff

I am still lost at what to study to further my knowledge, @KarlKronenfeld, @anon. Of course there are lots of options, but I am either not getting hooked on stuff or not finding a path to follow.

anon

go read some ams notices, google around for cool stuff

I am out of brain for the night, cannot into localizations

Fernando Martin

3:33 AM

@Pedro: Do you have anything in mind to study in the summer?

Pedro Tamaroff

@anon I'm sure you'll get around it. =)

Fernando Martin

I usually don't have much time to wade through a book unrelated to my courses during the academic year

I think I'll study some commutative algebra this summer

Pedro Tamaroff

@FernandoMartin Try to keep up with Ireland and Rosen, and Algebra II, so maybe pick parts from Lang and Rotman.

Fernando Martin

Nice!

Pedro Tamaroff

@FernandoMartin I hope so.

@FernandoMartin I think I am a bit more zen with linear algebra now. =)

After reading parts of Halmos.

I will keep reading a little though.

The only downside is the gurrdarmn tensor product. =D

Fernando Martin

3:40 AM

Have you read about extension of scalars?

I don't know much about tensor products (actually I know close to nothing) but that's a nice motivation

Pedro Tamaroff

@FernandoMartin For example, complexification?

Fernando Martin

Yes, that's an example

Pedro Tamaroff

@FernandoMartin What is the general situation?

Fernando Martin

Well, in general, if you have an $R$-module $M$ and a ring morphism $f:R\rightarrow S$, then $S\otimes_R M$ is an $S$-module

actually I think it's a bimodule but I don't remember

and that is, in some sense, the "most natural way" to extend scalars from your ring $R$ to $S$

When dealing with general rings, things can get ugly

Stuff like $M\neq 0$ but $S\otimes_R M = 0$ can happen

Pedro Tamaroff

@FernandoMartin What do you mean "general rings"? What are the assumptions on $R,S$?

Fernando Martin

3:46 AM

Just rings

Pedro Tamaroff

"When dealing with general rings, things can get ugly."

Fernando Martin

The theory is much more well-behaved if we're dealing with vector spaces (though I'm not 100% sure on this)

If you'd like to read about this, I studied it from Dummit-Foote

Great book btw

Pedro Tamaroff

@FernandoMartin Guille me paso ese libro.

Fernando Martin

Yep, he likes it too

I know it because of him actually

There's another nice characterization via some universal property regarding bilinear functions, but I think you already know that one

Pedro Tamaroff

@FernandoMartin Yes, I do.

Fernando Martin

3:53 AM

Universal properties are nice.

Pedro Tamaroff

That is why Halmos defined it as the dual of the space of bilinear forms.

@FernandoMartin Aha.

Pero se tiene que demostrar la existencia de la estructura en cuestion, no?

Fernando Martin

Claro, pero se hace

Es un cociente de un libre horrendamente grande

(la construcción standard)

agent154

I just thought of something...

"Suppose that $p$ is an odd prime. Show that $1^{p-1}+2^{p-1}+\dots+(p-1)^{p-1}\equiv -1\mod{p}$."

Does this imply that I need to show that this doesn't work for $p=2$?

Pedro Tamaroff

@agent154 Not necessarily. But it does work for $2$, since $1=-1$.

@FernandoMartin Bueno, en el libro de Jacobson leí un poco sobre estructuras libres.

Grande en que sentido?

Fernando Martin

Era un cociente de un grupo abeliano libre enorme, dejame buscar mis apuntes porque no quiero inventar

Para fabricarte el producto tensorial entre $M$ y $N$, usás un cociente de $\mathbb{Z}^{(M\times N)}$

es monstruoso ese grupo

Pedro Tamaroff

4:01 AM

@FernandoMartin La base son los elementos de $M\times N$?

Fernando Martin

claro

suponete que llamo $(m,n)$ a los elementos básicos (es un poco de abuso de notación)

Pedro Tamaroff

No se si Jacobson lo define para el caso infinito.

Fernando Martin

con esa construcción es bastante fácil probar que eso verifica la PU

Pedro Tamaroff

@FernandoMartin Claro.

Fernando Martin

Edité sin querer el mensaje anterior y no se qué decía :/

Pedro Tamaroff

4:03 AM

@FernandoMartin Jajja, las relaciones.

Fernando Martin

Ah sí

Pedro Tamaroff

Jacobson lo hace para finitos generadores.

Y enuncia la PU.

Fernando Martin

Basic Algebra?

Pedro Tamaroff

@FernandoMartin Si.

Fernando Martin

Creo que lo "tengo" pero nunca lo miré demasiado

A ver

Pedro Tamaroff

4:05 AM

Para cualquier abeliano $G$, $g_1,\ldots,g_r$ en $G$ existe un unico homomorfismo $\Bbb Z^{(r)}\to G$ con $x_i\to g_i$.

Fernando Martin

Ah, claro, pero eso vale con cualquier conjunto de generadores

Pedro Tamaroff

@FernandoMartin Ah.

Que un homomorfismo queda unívocamene determinado sobre los generadores.

Fernando Martin

Digo, si tenés $\mathbb{Z}^{(X)}$, hay un único morfismo que manda $x_i$ en $g_i$

Pedro Tamaroff

@FernandoMartin Esa es o no es la PU?

Fernando Martin

con $i$ indexando $X$

Esa es la PU de los grupos abelianos libres

Pedro Tamaroff

4:07 AM

El homomorfismo es $\sum n_ix_i\to \sum n_ia_i$, o si preferis $\prod a_i^{n_i}$ (aditivo o multiplicativo)

Fernando Martin

En general hay una muy similar para objetos libres en cualquier categoría

Claro, eso mismo en el caso de grupos

Pedro Tamaroff

@FernandoMartin Claro, claro.

@FernandoMartin Chau, che. Mañana curso.

Fernando Martin

Nos vemos!

agent154

4:39 AM

@anon I'm having problems understanding why $(m,(m-1)!)>1$ if $m$ is composite, and that this leads to the conclusion that $(m-1)!\equiv -1\mod{m}$ is not possible... Looking at the proof at primes.utm.edu/notes/proofs/Wilsons.html and it just says this is clearly the case...

anon

what does it mean for m to be composite?

agent154

I'm afraid that doesn't help. I did try to consider that but I still can't see the connection.

anon

answer the question

what does it mean for m to be composite?

this is not a trick question

agent154

It means that m=jk, where $2\leq j,k<m$

anon

$\color{Red}{j,k<m}$ and $(m-1)!=1\cdot2\cdots(m-1)$ hmm...

for example if $m=\color{Red}{2}\cdot\color{Red}{3}$ then $m-1=5$ and $(m-1)!=5!=1\cdot\color{Red}{2}\cdot\color{Red}{3}\cdot4\cdot5$

agent154

4:43 AM

OK then, I see that now... but I can't follow why this means that $(m-1)!\not\equiv -1\mod{m}$. (how do you typeset not equivalent or not congruent?)

anon

if something shares a factor with m then it is not invertible mod m

whereas $\pm1$ are clearly invertible mod $m$

\not\equiv

oops, \equiv not \cong :)

agent154

I did read up about the invertable property of integers coprime to $m$ $\mod{m}$... but that isn't helping me see the answer.

anon

if A is invertible mod m but B is not invertible mod m then A and B are not congruent mod m

do you agree?

agent154

Not congruent, meaning there does not exist any $x$ such that $A\equiv x\mod{m}$?

anon

meaning $A\not\equiv B\bmod m$

agent154

4:50 AM

I'm not sure I follow what you mean by "not congruent"

OK, let me explain what I do know on this subject.

anon

do you know what it means to say two things are congruent? saying two things are not congruent means saying they that are not congruent...

agent154

$\mathbb{Z}_m^*$ is a set of all invertable integers $\mod{m}$. So each member $a$ of $\mathbb{Z}_{m}^{*}$ has a multiplicative inverse $x_0$ in the set such that $ax_0\equiv 1\mod{m}$

anon

if you want to bypass the whole invertible business, then multiply $(m-1)!\equiv-1\bmod m$ by any proper nontrivial divisor $d\mid m$ and get $0\equiv -d\bmod m$, a contradiction

this works because $m/d$ is a factor in the product $(m-1)!$, so when you multiply the two factors $d$ and $m/d$ you get $0$ mod $m$

any comments @agent154 ?

agent154

The reason I didn't follow you regarding "not congruent" is because you can even have two invertable integers A and B $\mod{m}$ but they're not congruent. For example, in $\mathbb{Z}_5^*$, $\overline{2}\neq\overline{3}$...

anon

so?

agent154

4:57 AM

@anon I'm still trying to make sense of it. Maybe I just need to have my prof show me on paper because I can't make these huge bounds in logic. I'm getting better at this material, but I'm not that great.

anon

where are these huge bounds in logic you speak of?

agent154

Well, huge for me because I'm not comfortable with the theorems that bridge one concept to another.

so I need to see the intermediate steps and explanations as to why the steps apply

anon

if one complex number is zero and the other is not zero, then the two numbers are not equal. if you told me this and I said "I'm not sure I understand - can't there be two nonzero numbers that are not equal," how would you respond?

I have already been including almost all of the intermediate steps. it feels like if I provide more steps, I will have done literally all of the work for you. it is your responsibility to point out which steps you don't follow.

If person A has red hair and person B has black hair, then they are not the same person. Because if they were the same person, they'd have the same color hair. Of course there are people who have the same color hair but who are different people, but this is irrelevant.

agent154

Well, in any event... thank you for your effort, but it's not penetrating my brain. It could very well be because I'm tired, or that I need to review the material more.

anon

do not wait for math to penetrate you. you must go out of your way to penetrate it.

7

agent154

5:04 AM

That's why I ask questions here. I read proofs and don't understand some of the steps. I really do want to understand them well so I can do well on my upcoming test.

Pedro Tamaroff

@anon

anon

yeah

Pedro Tamaroff

MacLane calls (0) an improper ideal.

Birkoff &

anon

le sigh

Pedro Tamaroff

=D Just saying, I had read it somewhere.

Now I sleep.