If your vector isn’t a differential operator, you’re probably not doing differential geometry. That’s too sharp a cut, but I think it’s not far off the mark
Now an ambiguity arises, since if I write 3(u-1)^2 = (3u^2 - 6u + 3) and integrate that instead, I am missing the constant term that would arise by integrating into (u-1)^3
I'm asking if the last equality holds, since I can integrate to two different values while evaluating into an indefinite integral. Do you require more elaboration or is it clear what I mean?
So we let R be a commutative ring with identity and let S = 1 + I be a multiplicatively closed set. I'm trying to establish a homeomorphism between m-spec(S^{-1}R) and m-spec(R/I). It seems relatively clear that there is a bijection between these sets, but I'm struggling with defining such a function so that I can show the homeomorphism. So it seems that they both share a bijection with V(I) intersect with m-spec(R), where V(I) denotes all the prime ideals containing the ideal I. I think I can construct a homeomorphism from this intersection to m-spec(R/I), but the struggle is really showin…
@User203940 So you want to show that a maximal ideal contains $I$ iff it has nothing in common with $1 + I$. One direction is clear to me, but not quite the other
@TedShifrin Yeah, sorry. I was working through the whole thing, I guess it's fine. I indeed missed a factor in the first part of the integral that would cancel. I'm a bit overworked. Thanks.
"John has 46 bookshelf meters of books in his library. The width of the books can be seen as independent random variables with expected value 1.8 cm and standard deviation 0.7 cm. Determine approximately the probability that John has more than 2 500 books!"
I want to use a sum $Y=4600=\sum_{i=1}^{n\in\mathbb{N}} X_i\sim N(1.8n,0.49n)$. I don't think that it can be solved that way. Any ideas or suggestions?
@Silent: So I had a counterexample when I thought about $p\in E\cap\bar A$. But if $p$ is a limit point of $E\cap A$, then of course it's a limit point of $E$, hence belongs to $E$ and of course it belongs to $\bar A$, too.
@BalarkaSen If we have a field $K$ of characteristic $p$, then for all $a \in K$, $x^p-x+a$ is irreducible iff it has no roots in $K$ and it has a cyclic Galois group
@TedShifrin My reasoning for asking it on here was precisely that: It has to work, I'm not sure why it doesn't work now, talking through it with somebody can help understanding the matter
@User203940 Ahh, so if a maximal ideal $m$ does not contain $I$ then $m + I = R$ so there is some $x\in m$ and $y\in I$ with $x + y = 1$, so $x = 1 - y$ is an element that $m$ and $1 + I$ has in common
OK, @user55789. LOL ... If you had started off saying you knew it had to work on general principle and you needed help finding a silly error, I would have responded differently.
"The introduction of the cipher 0 or the group concept was general nonsense too, and mathematics was more or less stagnating for thousands of years because nobody was around to take such childish steps..." - Grothendieck
I formulate the proof as "if we have a real closed field, then the algebraic closure is obtained by adjoining a 4th root of unity", that is purely algebraic
you still need to use analysis to show that $\Bbb R$ is real closed
@LeakyNun as an exercise: Compute the degree of the splitting field of $x^6+x^3+1$ over $\mathbb{F}_p$ for $p \equiv 1 \pmod{9}$, $p \equiv 2 \pmod{9}$ and $p \equiv 7 \pmod{9}$
I like the following: by Laurent series, $f$ has a removable singularity at infinity, thus we can lift it to a holomorphic map from the Riemann sphere to itself, this is an open map by the open mapping theorem, but also a closed map as the Riemann sphere is compact Hausdorff, thus the image is clopen, hence $f$ is surjective by connectedness
But there's a difference between the statements "any non-constant holomorphic function P^1 -> P^1$ is surjective and "any holomorphic function P^1 -> C is constant"
