Lozansky

Hmmm

So this basically proves that $I$ maximal ideal $\Leftrightarrow$ $R/I$ is a field

Okay makes sense I guess

No?

Ah, the correspondence is $0 \to 0 + I = I$

My book and my lecture notes use different notations so maybe I'm just mixing things up though

$\{0 \}$ to be precise

$(0) = 0$ no??

Which makes no sense

So $(1)$ corresponds to $R$ and $(0)$ to $I$?

It follows from the correspondence theorem I guess?

If $I$ is a maximal ideal of $R$, why does this imply that $R/I$ has exactly the two ideals $(0)$ and $(1)$?

Ah, so I should require the $\alpha$'s to be irreducible in $S$?

@Thorgott Let's assume $\alpha_k \neq 0$ for all k, and not a unit so the ideal is proper

Hmm well $M$ should be a proper ideal I guess

If $R$ is a commutative unitary ring and $M$ is an ideal in $R$ generated by elements $\alpha_1,..,\alpha_m$, is it true that $M$ is maximal in $R$ iff I can find a composition of surjective homomorphisms $\pi_1 \circ... \circ \pi_m$ where $\pi_1$ has domain $R$ and $\pi_m$ has codomain $S$ such that $R/M \cong S$ with $Ker(\pi_1 \circ... \circ \pi_m) = M$?

Hey guys

I think it follows from the degree theorem or whatever it's called

Does this result have a name?

The product is elementwise

If $E$ is a finite extension of $F$ and $K$ is a finite extension of $E$ and we have bases $\beta_E, \beta_K$, then can we be sure a basis of $K$ over $F$ is given by $\beta_E \times \beta_K$?

Induction is probably a more proper term than "track back" :P

Yeah

One can "track back" the factors of $x^n$

@Thorgott Yes, I see what you mean (I think)

Oh it does

Ah, a prime ideal needs to be a proper ideal?

@Thorgott Ah I was thinking about the counter-case when every power of $x$ is in $P$ but then the statement (all nilpotent elements are containted in every prime ideal of a commutative ring) is trivially true

If $P$ is a prime ideal and $x$ a nilpotent element, then I can be sure $x^n = 0 \in P$ for some $n \geq 0$. How can I be sure there exists $k$ such that $x^k \in P$ but $x^{k-1} \notin P$?

We haven't touched upon categories so I suspect a superficial understanding of it is enough:P

But the converse isn't necessarily true

@Thorgott Are all actions of commutative groups natural?

Is there some convention for what "naturally" means in terms of group/ring actions?

@Thorgott Good point

$x^p-x$ has p distinct zeros in $\mathbb{Z}_p$ for p prime. I know this follows easily from Fermat's, but I want to show it using the fact that the ideal generated by a polynomial $p(x)$ in a field is maximal iff $p(x)$ is irreducible. This should be possible, yes?

@MathGeek Try here chat.stackexchange.com/rooms/3229/the-periodic-table

@Sonal_sqrt The divergence operator is invariant under coordination transformations

Abel equation I think

I meant $(k^4-(k_x^4+k_y^4-k_z^4))/2$

It can be noted that $k_x^2k_y^2+...= k^4/2 - (k_x^4+k_y^4-k_z^4)$ but I don't know if this helps

I have expression of the form $f(k) \sim \sqrt{B^2k^4+C^2(k_x^2k_y^2+k_y^2k_z^2+k_z^2k_x^2)}$ where $k^2 = k_x^2+k_y^2+k_z^2$ and I want to find $\partial^2 f/ \partial k^2$, is there a neat way to do this?

@Semiclassical Are you any savvy at numerical analysis?:P

What could $\chi[-1,1](t)$ signify in the context of averaging kernels?

Yes, looks good

@MrAP Well, not necessarily but yes that's useful. Basically you are then adding "point masses" on the complex unit circle that are evenly spread out so that their COG is at origin.

Oy

Gotta look at them band gaps

In two dimensions

How do you translate distance in real space to reciprocal space for a hexagonal lattice?

Lowest free energy in the n:th Brillouin zone according to band theory corresponds to the shortest distance to that zone in reciprocal space, yes?