Mathematics - 2019-01-25 (page 2 of 2)

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Rithaniel

23:21

Question: Is there a way to denote "The set of rational numbers of the form $\frac{a}{2^n}, a\in\Bbb{Z},n\in\Bbb{N}$" in just two or three symbols?

Daminark

$\mathbb{Z}_2$ is one way but it's unfortunate because of the 2-adics

Leaky Nun

@Rithaniel $\Bbb Z[1/2]$

Rithaniel

Ah, that's the syntax for polynomials over a ring, right?

Leaky Nun

that's also the syntax of adjoining element to a subring

Rithaniel

Ah, right, like the Gaussian integers $\Bbb{Z}[i]$

Though, wouldn't that only allow me elements of the form $a+\frac{b}{2}|a,b\in\Bbb{Z}$?

Leaky Nun

23:30

Ted Shifrin

No, @Rithaniel, it's the smallest ring containing $\Bbb Z$ that you get with $1/2$ in it.

Leaky Nun

@Ted!

Ted Shifrin

Think of it as the set of all polynomials with integer coefficients (so $\Bbb Z[x]$) with $x$ set equal to $1/2$.

Leaky Nun

@TedShifrin well that's a theorem...

Ted Shifrin

No, that's my notation.

Leaky Nun

23:32

I mean, the fact that the smallest ring containing $A$ and $\alpha$ is the image of the map $A[x] \to B$ evaluating at $\alpha$

Ted Shifrin

Where $B$ contains $\alpha$ of course ...

That's not much of a theorem.

It's a rudimentary observation/exercise.

Rithaniel

Okay, so it's essentially $\frac{a_n}{2^n}+\frac{a_{n-1}}{2^{n-1}}+...\frac{a_0}{2^0}$

Ted Shifrin

With various values of $n$ and $a_i$, of course ...

Rithaniel

(Sorry if the latex is a difficult to read there. It's easy to make things too small with subscripts and superscripts)

Ted Shifrin

you can always use \dfrac

Rithaniel

23:35

Okay, then yeah, $\Bbb{Z}[\frac{1}{2}]$ seems a good option.

\dfrac? I don't know this and must google it (is it diagonal fractions?)

Daminark

Oh yeah that's better lmao

I was thinking localization with my notation

Ted Shifrin

no, it's displayfraction

Leaky Nun

@Daminark it's also localization

Ted Shifrin

Demonark, better to write $\Bbb Z_{(2)}$.

Leaky Nun

@TedShifrin sure, it's trivial to us, but not to the year 1's I have been teaching

Ted Shifrin

23:36

Not ambiguous

Leaky Nun

(now they understand it it's trivial to them)

Daminark

Actually that's ambiguous

Ted Shifrin

Leaky, in my algebra book, I introduced the polynomial definition and made the "smallest ring..." as a comment.

Leaky Nun

@TedShifrin $\Bbb Z_{(2)}$ is a differnet ring (localization at 2)!

Daminark

It's localization at the complement of the prime ideal (2)

Leaky Nun

23:37

@Daminark localization at 2

Ted Shifrin

Oh, right, never mind.

Daminark

To me, localize at $f\in A$ means $S = \{1,f,\ldots\}$ and taking $S^{-1}A$. Anyway I think we meant the same ring, whatever we call it

Leaky Nun

@Daminark the geometric interpretation is that when you localize at something, other things become invertible because you're only focusing on something

Rithaniel

So $\Bbb{Z}_{(2)}$ means "the set of rational numbers of NOT the form $\dfrac{a}{2^n},a\in\Bbb{Z},n\in\Bbb{N}$?"

Leaky Nun

e.g. at $0$, $X-3$ becomes invertible

@Rithaniel no, it means the set of rational of the form $\frac{a}{b}$ where $b$ is odd

@Daminark that's localizing away from $f$

(so that $f$ becomes non-zero and hence invertible)

Rithaniel

23:41

Ah, not divisible, and not just "not a power of two."

Leaky Nun

right

Ted Shifrin

cuz the ideal $(2)$ is all even numbers.

Rithaniel

Ah, it's the ideal. Neat.

Leaky Nun

elements of rings are functions on the space of all prime ideals

at the prime ideal $(2)$, every odd number is non-zero, so they become invertible

so you can divide by e.g. $3$

Daminark

Okay, question: can you localize a non-integral domain to get an integral domain?

Leaky Nun

23:43

@Daminark sure, you can localize $\Bbb Z/6\Bbb Z$ to get $\Bbb Z/2\Bbb Z$ or $\Bbb Z/3\Bbb Z$

Ted Shifrin

Didn't you do my exercise a while ago, Demonark, about $\Bbb Z_6[x]/(2x-1)$?

I stole it from Artin. I think there was also modding out by $(2x-3)$.

Daminark

I guess what I had in mind was more, can you do it to any ring

Also I don't recall that exercise immediately @Ted

Rithaniel

Is that $\Bbb{Z}_6$ the localization of $\Bbb{Z}$ at 6 or $\Bbb{Z}$ mod 6?

(I've seen it used both ways)

Ted Shifrin

No, no, $\Bbb Z/6\Bbb Z$.

Rithaniel

Okay, that's what I refer to when I say "$\Bbb{Z}$ mod 6." What was the exercise?

Daminark

23:54

I prefer denoting the latter as $\mathbb{Z}/n$, $\mathbb{Z}/(n)$, or $\mathbb{Z}/n\mathbb{Z}$ because of the potential for confusion with p-adics and/or localization

Leaky Nun

@Daminark If you localize any ring $A$ at any prime ideal $\mathfrak p$ you get a local ring $A_\mathfrak p$ with maximal ideal $\mathfrak p A_\mathfrak p$

Daminark

A lot of folk don't interact with p-adics in which case $\mathbb{Z}_n$ works

Ted Shifrin

Demonark, in a first course there isn't confusion. I agree that there may be subsequently.

Daminark

This is true, I guess I've found notational habits hard to break which is why I prefer to use something that will always be clear from the start if possible.

Ted Shifrin

I ~~vary~~ varied my notations depending on the course I ~~teach~~ taught. It's not that hard to do that ...

It's harder for me to use past tense, however.

Leaky Nun

23:58

15 mins ago, by Daminark

Okay, question: can you localize a non-integral domain to get an integral domain?

the roughly corresponding question for topological space is whether there is always an open set of a space that is irreducible

Ted Shifrin

I don't know what irreducible means in point-set topology

Leaky Nun

it means every non-empty open set is dense, alternatively that every two non-empty open sets have non-empty intersection

Daminark

Huh, I've found that usually I stick somewhat rigidly by habit with the first thing I see. Unless I see multiple right from the start in which case it's random what I use

Ted Shifrin

I've never heard of such a thing.

Leaky Nun

it's a mostly algebraic-geometry thing