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?

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

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

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

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

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

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

no

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

@Ted!

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

@TedShifrin well that's a theorem...

No, that's my notation.

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

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

That's not much of a theorem.

It's a rudimentary observation/exercise.

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

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

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

you can always use \dfrac

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?)

Oh yeah that's better lmao

I was thinking localization with my notation

no, it's displayfraction

@Daminark it's also localization

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

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

Not ambiguous

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

Actually that's ambiguous

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

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

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

@Daminark localization at 2

Oh, right, never mind.

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

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

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

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)

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

right

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

Ah, it's the ideal. Neat.

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. $3$

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

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

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 $(2x-3)$.

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

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

(I've seen it used both ways)

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

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

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

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

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

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

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.

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.

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

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

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

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

I've never heard of such a thing.

it's a mostly algebraic-geometry thing