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| bio | website | simpletonsymposium.wordpress.… |
|---|---|---|
| location | Evanston, IL | |
| age | 21 | |
| visits | member for | 2 years, 7 months |
| seen | 5 hours ago | |
| stats | profile views | 659 |
I'm a first-year grad student at Northwestern interested in homotopy theory.
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May 4 |
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Canonical bundle and Möbius bundle @ArthurStuart: a section of the canonical bundle over $\mathbb{R} P^1$ is a continuous choice of a point on each line through the origin in $\mathbb{R}^2$, which is a certain kind of map from $S^1$ to $\mathbb{R}^2$. You have to show that any such map must pass through the origin. |
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May 4 |
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Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? @rschwieb: that's totally fine! Thanks! |
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May 3 |
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Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? You're totally right! I don't know what I was thinking. In the case where every zero-divisor is nilpotent, they do form a proper ideal, though, since the sum of two nilpotent elements is nilpotent. |
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May 2 |
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Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? @MarianoSuárez-Alvarez: Thanks for the clarification. I only specified commutative rings because I don't know anything about noncommutative rings and didn't want to say anything false. |
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May 2 |
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Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? @rschwieb: All I mean by 'prime ideal of zero-divisors' is that the subset of $A$ consisting of the zero-divisors is always a prime ideal, which you can easily check. |
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Apr 10 |
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Regarding the definition of vector field flow Oh, sorry, I thought $X$ was the manifold in question. If you replace all the $X$'s in my above comment by $M$'s, you should get a true statement. The above $\Phi^X_t$ is precisely the flow you're talking about in your second comment. |
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Apr 9 |
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Regarding the definition of vector field flow The Roman d is the same as the italic d. Each $\Phi^X_t$ is a diffeomorphism of $X$, which is a map $X \to X$; thus, its derivative maps $T_pX \to T_{\Phi^X_t(p)}X$ for each $p \in X$. Since this is a diffeomorphism, it's injective and surjective, and so starting with a smooth vector field on all of $X$ gives you a new vector field on all of $X$; you can check that this is smooth. Also, your title has nothing to do with your question - would you mind changing it? |
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Apr 3 |
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Product of two trivial sigma algebra is trivial No, the problem is correct. $X \times \emptyset = \emptyset \times Y = \emptyset$. |
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Apr 3 |
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Rings with finitely many finitely generated free modules, up to isomorphism Thanks, this is just what I was looking for! |
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Mar 31 |
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Integral basis for an ideal Amusing side fact: every ideal in a ring of integers can be generated (over the ring, not over $\mathbb{Z}$) by at most two elements. This is actually true in any Dedekind domain. |
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Mar 22 |
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Calculus and Category theory Your question about category theory, though, is interesting! I've been looking around for things about categorical ways to talk about calculus, and none of them really seem convincing. The only thing I can think of is this categorical description of the unit interval. |
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Mar 22 |
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Calculus and Category theory Of course you can do this! You can just define it in terms of the chain rule: $df/dg = (df/dx)/(dg/dx)$. When the functions are differentiable, I'd argue this is the only way it should be defined -- in particular, when you have a number of physical quantities that depend on each other simultaneously, this is the correct notion. |
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Mar 22 |
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A question concerning compactness - Topology Looks good to me! |
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Mar 21 |
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Show that if $Y$ is contractible, then the canonical projection $\pi: X\times Y\to X$ is a homotopy equivalence What have you tried so far? Where are you stuck? Also, if this is homework, it should be tagged with the homework tag. |
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Mar 20 |
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degree of the extension of the normal closure Remember that 'separable' is a property of field extensions, not fields. The primitive element theorem holds for $K/F$ if $K$ is separable over $F$, not if '$F$ is separable'. |
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Mar 19 |
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algebraic distance of an element of a ring from an ideal (cont'd) For example, if $X = \mathrm{Spec}\, k[x]/(x^2)$, then a function on $X$ is something of the form $a + bx$, but $0$ is the only point of $X$. Restricting a function from $\mathbb{A}^1$ to $X$ tells you its value at $0$ but also its first derivative in the $x$ direction. For this reason, $X$ is also drawn as a point with a tangent vector sticking out. Likewise you could have schemes that capture second-order information at a point and so on (Spec of a completion will give you the Taylor series!) Eisenbud and Harris's Geometry of Schemes is a good place to learn about this stuff. |
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Mar 19 |
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algebraic distance of an element of a ring from an ideal Re nilpotents: I'm not totally sure, but one way people think about it is as follows. Let $k$ be an algebraically closed field. If $X = \mathrm{Spec} k[x_1,\dotsc,x_n]/I$ is an affine variety over $k$, then giving a function on $X$ is the same as giving one on $\mathbb{A}^n_k$ but only caring about its values at the points of $X$. If we now allow $X$ to have nilpotents in its ring, functions on $X$ contain slightly more information than just values at its points, and I guess you could think of this as 'analytical'. |
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Mar 19 |
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algebraic distance of an element of a ring from an ideal The main problem I have with Hartshorne is that it starts with varieties as 'motivation', but then abandons all pretext of motivation as soon as it starts doing schemes, precisely when motivation is most needed! It also seems a little less useful if you're more interested in e.g. arithmetic geometry than varieties, which I'd wager is true for a lot of people who read that book. Vakil starts you out in the language of schemes and has an EGA-ish table of contents, but there's always enough intuition to get you used to the language. |
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Mar 18 |
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algebraic distance of an element of a ring from an ideal Thanks! The problem is that it's really telling you how deep $x$ is inside $I$, rather than how far away $x$ is from $I$. Which is roughly the opposite of what you want. |
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Mar 15 |
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Topology: Homeomorphisms Some hints: Onto just means that every point of $(-2,0]$ is in the image of $f$. Given $x \in (-2,0]$, what point in $[0,1)$ hits it? For continuous and open, recall that the preimages/images of elements of some basis are open. Do you know bases for the two topologies you're looking at? (You can also look at subbases, if you're familiar with them.) I second Neal's idea to draw a picture. Even if it's not part of the formal proof, it can help you come up with it. |
