Karl Kroningfeld

woah

I think the method is fine. The division algorithm steps can be used to write $1$ as a linear combination of $4a^2+1$ and $2a+1$. Thus $4a^2+1$ and $2a+1$ must be relatively prime.

Yes, yes.

I started writing down an example in the margin of my textbook, but ...

:)

@Shaun I might have said a few things about transcendence degree that seem to be useful, or were useful in similar problems. Likewise for P^n (i.e. stating whether you know and understand the standard affine cover would be helpful). It's tough (for me) to answer a question in which the asker does not state what they know already.

Authors these days

Not that they had to prove it as part of the statement of Positive Definiteness :P

Why?

These particular authors start the proof of (i) with a proof of $|a|\ge 0$ :)

Why did they include P at all?

Note that Q is stronger than P. So, it would have been more efficient to simply state Q iff R

Wouldn't you actually want to show the inequality holds?!

Indeed, this disagreement would be resolved with a case in which $P$ is not true.

The statement $|a|\ge 0$ applies regardless of whether $a=0$. A different example of the same wording: artofproblemsolving.com/wiki/index.php/AM-GM_Inequality

I would normally parse P with Q if and only if R as $P\land(Q\iff R)$

one might want $K_n\subset K_{n+1}^{\circ}$ for exhaustions in general

Readability is my standard as well :)

hey Ted

doing fine thanks

hey how's it going?

"I am a Bitcoin enthusiast, with no background knowledge in computer science and cryptography." :)

I am connected to that person via Blockchain, somehow

Everything on the right of the tensor has to stay a multiple of $3$ if you're computing in $\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$.

I'm guessing you were computing in $\mathbb Z/3\mathbb Z\otimes \mathbb Z$ and not $\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$

Because $3\mathbb Z\cong \mathbb Z$ as $\mathbb Z$-modules.

$\mathbb Z/3\mathbb Z\otimes 3\mathbb Z$ is cyclic of order $3$.

@Wave Math is hard lol. I jumped on it mainly because I remembered making a similar mistake before.

I'm just saying, what's in your notes is false.

You have to localize at $m$ first to apply Nakayama's lemma.

If $N$ is finitely generated and $R$ is Noetherian, then the tensor product is $0$ if and only if $mN=0$ already, by Nakayama's lemma.

In general, you have $R/I\otimes_RM\cong M/IM$ for ideals $I$ and $R$-modules $M$, and this allows you to figure out what $R/m\otimes_RmN$ is. (E.g., consider the case $N=R$).

If you want to transfer scalars across $\otimes$ to get $0$, then you are not computing in $R/m\otimes_RmN$ but rather $R/m\otimes N$.

I think you're trying to prove something false @Wave, generally $R/m\otimes_RmN$ is nonzero.

Atiyah-Macdonald say more about the approach I suggested (Corollary 3.4) though it may be equally hard to digest.

@Wave The first answer here approaches it in a more direct way.

I doubt I can fully explain the "fun fact", maybe I can find a decent reference.

It's probably easier to see that $S^{-1}p$ is $pR_p$ and then identify $S^{-1}p\cong p\otimes S^{-1}R$.

You can localize $R$-modules, and that preserves exact sequences of $R$-modules. It is the same as tensoring with $S^{-1}R$.

@Wave Briefly, you can localize the exact sequence of $R$-module $0\to p\to R \to R/p\to 0$, using the multiplicative set $S=R\setminus p$. That leads to $0\to pR_p\to R_p\to S^{-1}(R/p)\to 0$. Then, one shows that $S^{-1}(R/p)$ is the field of fractions of $R/p$.

Also, one can take $p$ to be any prime ideal. Fun fact: the quotient $R_p/pR_p$ is the field of fractions of the integral domain $R/p$.

Are there any reservations about what you said @wave ?

Yes :)

Hold up, it is a direct application of the maximum principle. The real part of $f$ is constant, so $|e^f|$ is constant and a fortiori attains its maximum on the connected open domain. Hence, $e^f$ must be constant. This implies $f$ is discrete-valued and thus locally constant. @Wave

@BalarkaSen the usual, making false assertions, etc.

@BalarkaSen hey yo

emphasis on "more-or-less thinking"

i was more-or-less thinking to reprove liouville lol

ah, open mapping

Have you tried taking the exponential of $f$ @Wave?