Mathematics

12:21 AM

Let $K[x,y]_{\langle x,y\rangle}$ be the localization of $K[x,y]$ at the maximal ideal $\langle x,y \rangle$. Is $K[x,y]_{\langle x,y\rangle}[1/y]$ a local ring?

Ted Shifrin

Gee. I wonder why it’s called localization.

shintuku

is it obvious to you it is local or nonlocal?

Ted Shifrin

Have you done simple examples, like not with polynomials?

shintuku

i'm studying valuation rings and that's the simplest example i could think of of extending a ring that is local but not a valuation ring

Ted Shifrin

You’re in over your head. Start with $\Bbb Z$.

shintuku

12:29 AM

problem is all localization at prime ideals there are valuation rings

i want the behaviour of extending local, nonvaluation rings

am slowly working through the $K[x,y]_{\langle x,y\rangle}[1/y]$ example step by step

Ted Shifrin

So why is $\Bbb Z_{(5)}$ local?

shintuku

i lied

$\mathbb Z_{(5)}$ is local because any element of $\mathbb Z_{(5)}$ who has a numerator divisible by $5$ is a nonunit

and that set is an ideal

Ted Shifrin

So is there an obvious maximal ideal?

shintuku

yes: any nonunit is contained in this single ideal

if you extend then this ideal by any other element not already in it, we get the entire ring

Ted Shifrin

Is there a good way to describe it more succinctly?

shintuku

12:40 AM

let me think

$(\mathbb Z - \langle 5\rangle)^{-1}\langle 5\rangle$?

Ted Shifrin

Yeah, just the ideal generated by 5.

shintuku

the issue is, in my above mentioned example the extension doesn't make it a localization anymore and we're making elements who were previously nonunits into units

Ted Shifrin

Huh? I have no idea what you’re talking about.

shintuku

the example $K[x,y]_{\langle x,y\rangle}[1/y]$

Ted Shifrin

Oh, I missed the 1/y.

shintuku

12:47 AM

previously, an element of $K[x,y]$ who's numerator was in $\langle y \rangle$ was a nonunit, now it becomes a unit

Ted Shifrin

So how is that related to localizing just at $(x)$?

shintuku

it's not a localization anymore, or at least, not a localization given by the natural homomorphism for localization at a prime ideal

Ted Shifrin

So answer my question.

shintuku

i'll think about it 2 seconds

i'm not sure, i need to think on it a while

shintuku

1:24 AM

i have the $K[x,y]_{(x,y)}[1/y] \subseteq K[x,y]_{(x)}$ direction, I'm trying to get the other one

shintuku

1:57 AM

any tips?

shintuku

2:47 AM

@TedShifrin Notice $\frac{1}{x+y} \in K[x,y]_{(x)}$. Suppose we have $\frac{1}{x+y} \in K[x,y]_{(x,y)}[1/y]$. Then $\frac{1}{x+y} = \frac{p_0}{p'_0} + \frac{p_1}{p'_1y}+\cdots+\frac{p_n}{p'_ny^n}$ with $p_i, p'_i \in K[x,y]$ and $p'_i \notin (x,y)$. But since $x+y$ is not divisible by a power of $y$, we must have $\frac{1}{x+y} = \frac{p_0}{p'_0}$ and therefore $p'_0=x+y$, which is clearly in $(x,y)$, a contradiction.

so $K[x,y]_{(x,y)}[1/y] \subsetneq K[x,y]_{(x)}$

shintuku

2:59 AM

anyhow, we have that it is a subring of a local ring and that's probably local

thanks for the help

Sourav Ghosh

4:51 AM

A bilinear form $B$ on $\Bbb {R}^n$ is an inner product iff matrix of $B$ is positive definite.

Let $A$ is positive definite and $B(X, Y) =Y^tAX $

Then positivity and definiteness of $B$ follows from the positive definiteness of $A$

Linearity in the first slot is also obvious.

Conjugate symmetry of $B$ follows the the symmetry of $A$

gewbDog5

@TedShifrin If I were to watch your lecture series for review

do you have like a curated problem set you would recommend to accompany watching the lectures?

I'm talking about the calc 3/lin. al. series

D Left Adjoint to U

8:37 AM

0

Q: There is a twin prime counting formula that is a direct generalization of the $\pi(x)$ inclusion-exclusion formula found on Wikipedia

I can prove on paper, using elementary techniques that if $f(t) =$ the total number of twin prime averages in the interval $(\sqrt{t+1}, t]$, then we may write: $$ f(t) = \sum_{d \mid p_n \#}(-1)^{\omega(d)} \sum_{r^2 = 1 \pmod d} \lfloor\frac{ t - r}{d} \rfloor $$ for all real numbers $t\in \Bbb...

one potato two potato

10:24 AM

peep

Mad

10:51 AM

$\Pi_i (A - \lambda_i \mathrm{Id}) (\psi) = \Pi_i (A - \lambda_i \mathrm{Id}) \left(\sum_i c_i v_i\right) \Leftrightarrow \Pi_i \left( A(\sum_i c_i v_i) - \lambda_i \mathrm{Id}(\sum_i c_i v_i) \right) = \Pi_i \left( \sum_i c_i (A(v_i) - \lambda_i \mathrm{Id} v_i) \right) = \Pi_i \left( \sum_i c_i (\lambda_i v_i - \lambda_i v_i) \right) = 0
$

is this argumentation correct

i am bit unsure due to the use of index i in both the sum and the product

A id matrices lamda eigenvalues psi is avector vi are the basis vectors and are eigenvectors of A.

technically speaking, i should denjote the sum with j?

Farhad Rouhbakhsh

11:07 AM

@TedShifrin Hello. I have a question regarding my question of a smooth simple closed curve as a union of finite number of arcs, each of them being a smooth function of one variable. I was thinking of an example (although not closed) : Define your curve $C$ by C(t) = (t^3,t^2) for t between -1 and 1. This is infinitely differentiable but has a corner at t=0. At this corner the derivative is (0,0) . So the curve is non-regular.

I observe that by "cutting" the curve at the point (0,0), you can consider the left part and the right part of the curve as a smooth function of y, where the derivative of each function (one variable, regarding to y) is 0 at point x = 0.

Mad

nvm i found my mistake

Koro

12:44 PM

Let Z be the set of all irrational nos. How do I show that $\mathbb R^2\setminus Z^2$ is connected?

one potato two potato

project to one of the axis and connect to the origin (for given point)

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$Mathematics$ Mathematics