PseudoNeo
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Commutative non Noetherian rings in which all maximal ideals are finitely generated
56 votes

Let $A = C^\infty(S^1)$ be the ring of smooth functions on the circle (if you prefer, you can see it as the ring of smooth $2\pi$-periodic functions $\mathbb R \to \mathbb R$). First, $A$ isn't ...

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Geometric motivation for negative self-intersection
36 votes

I'll just add a remark to Georges's (excellent) answer. The real plane $\mathbb R^2$ has two freely substitutable orientations but the complex plane $\mathbb C$ has a canonical one (as a real vector ...

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The set of functions which map convergent series to convergent series
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34 votes

I'm quite late on this one, but I think the result is nice enough to be included here. Definition A function $f : \mathbb R \to \mathbb R$ is said to be convergence-preserving (hereafter CP) if $\sum ...

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Presentation $\langle x,y,z\mid xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2}\rangle$ of group equal to trivial group
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30 votes

This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof. The proof I'm ...

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What is the math behind the game Spot It?
19 votes

Here's an article (in French) that aims to explain the mathematics behind the game to a wide audience. In the interest of link rot prevention, here are two diagrams from the article that may be of ...

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Reduced homology
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12 votes

There is a map $\epsilon : C_0(X) \to \mathbb Z$. It is indeed a part of the complex for reduced homology, but it exists independently. Because $\epsilon \partial_1$ is zero, the value of $\epsilon(z)$...

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Classical presentation of fundamental group of surface with boundary
11 votes

OK, my bad, Fulton's Algebraic topology: A First Course only deals with the closed case. I'll suppose that you know this case quite well. Let's do the bounded case by hand. First case: one boundary ...

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Possible cup product structures on a manifold
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11 votes

I will denote by $LH^*$ the free part of the cohomology ring. (That means that $LH^r(M)$ is $H^r(M)/\mathrm{torsion part}$.) For a connected 4-manifold, the symmetric form $I : LH^2(M) \times LH^2(M) \...

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Open subsets in a manifold as submanifold of the same dimension?
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11 votes

Yes. In general, $N \subset M$ is a submanifold if you can find for every $x \in N$ an open neighbourhood $U \subset M$ of $x$, an open neighbourhood $V \subset \mathbb R^{\dim M}$ of $0$ and a ...

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Show that $S_4/V$ is isomorphic to $S_3 $, where $V$ is the Klein Four Group.
10 votes

Just a “visual” construction of the isomorphism between $\mathfrak S(4)/V_4$ and $\mathfrak S(3)$... It's quite well known that $4 = 2 +2$. Concretely, it means that if you have four things, you can ...

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Construction of Exotic Spheres
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10 votes

Yes. This procedure is called clutching (and the resulting spheres are clutched spheres or twisted spheres In this procedure $g$ is a diffeomorphism. If $g$ extends to a diffeomorphism of the whole ...

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Does there exist a double cover with trivial deck transformation group?
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9 votes

Look at the classification of covering spaces (e.g. Hatcher's Theorem 1.38 and Proposition 1.39): A two-sheeted cover $p : (\tilde X, \tilde x_0) \to(X, x_0)$ is determined by the subgroup $\Gamma = ...

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Original article on the Grothendieck group
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9 votes

What seems to be clear is that the origin of the Grothendieck group is Grothendieck's work on Riemann-Roch's theorem around 1956. According to Weibel's The Development of Algebraic K-theory before ...

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Historical basis and mathematical significance of Riemann surfaces
9 votes

You have to understand that the notion of function as it is used nowadays is quite recent. During a long time, analysts were perfectly happy to work with so-called multiform functions. For example, $\...

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If F is a field, then $F[x,y]$ is a Principal Ideal Domain?
9 votes

If $A$ is a commutative ring, a classical result states that the polynomial ring $A[x]$ is a PID if and only if $A$ is a field. It is a good exercise. In your case, as $F[x]$ isn't a field, $F[x,y] \...

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Intuitive reason of why the negation of $P \Rightarrow Q$ is $P \land \neg Q$ instead of $P \Rightarrow \neg Q$
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8 votes

Here's the example I give to my students: there's a rule that says that "If a traffic light is red, then you must stop". Recall that this is what implication means: $P \implies Q$ means "if $P$, then $...

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Non-trivial example of algebraically closed fields
8 votes

In On Numbers and Games, Conway defined a field structure on the set of all ordinals, and he calls the result $\mathbf{On}_2$. It is an algebraically closed field of characteristic two, if you are ...

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Structures on torus
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8 votes

You're mostly right. The relation between complex structures and metrics comes from their common passion about angles. Basically, a complex structure on a Riemann surface is just a procedure for ...

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Is there such kind of theorem saying two homotopic ways of attaching handle result in diffeomorphic manifolds?
8 votes

Your question can be subject to interpretation, but here's a proof that you cannot hope for too strong a result. Take the simplest example you can think of: $M'$ is the $n$-ball, and you'll add a ...

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Can $x^2+y^2,y^2+z^2,z^2+x^2$ and $x^2+y^2+z^2$ all be square numbers?
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7 votes

It's a famous open problem : the perfect cuboid problem.

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Reference request in contact geometry.
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7 votes

Hansjörg Geiges's Introduction to Contact Topology seems to be the only textbook-style reference on Contact Geometry. (At least it was three years ago, but I'm unaware of a more recent book with this ...

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What is wrong with this proof on ring of integers being finitely generated
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7 votes

It is not true that if $A$ is not finitely generated, you can find $\mathbb Z$-independent elements. For example, $\mathbb Q$ is a non finitely generated $\mathbb Z$-module, but, clearly, you cannot ...

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Twisted Cech cohomology
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7 votes

Local coefficient homology is a particular case of sheaf cohomology (cohomology of a locally constant sheaf). So, even if I'm not sure I understood precisely your wishes, I think it is possible that ...

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Proof that a periodic function is bounded and uniformly continuous.
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7 votes

Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function. Now, let $c_0$ be a point where $f_{|[0,1]}$ has a ...

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Are the local rings at smooth points of an irreducible variety isomorphic?
7 votes

It is not true that Aut(X) acts transitively. It's for example false for all Riemann surfaces of genus > 1, as the automorphism group is then finite.

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Did Zariski really define the Zariski topology on the prime spectrum of a ring?
7 votes

Let's listen to the master: Mais en 1945, Jacobson observe [172 c] que le procédé de définition d'une topologie, imaginé par Stone, peut en fait s'appliquer à tout anneau A (commutatif ou non) pourvu ...

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Finding my mistake in a "proof" that there is no non-trivial knot
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6 votes

Proposition 3 is false, or at least phrased in a misleading way: the complement of a knotted solid torus is S³ is certainly not a solid torus: this happens if and only if the solid torus is unknotted, ...

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Can the long line be embedded in euclidean space?
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6 votes

No, because of the Poincaré-Volterra theorem. Here's the statement you can find in Bourbaki's General Topology (I.11.7, corollary 3). Theorem. Let $Y$ be a locally compact, locally connected space ...

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Why is $\mathrm{Spec}(\mathbb{Z})$ a terminal object in the category of affine schemes?
6 votes

The category of affine schemes is nothing but the opposite category of commutative rings. So $\mathrm{Spec}\, \mathbb Z$ is a terminal object in the category of affine schemes because $\mathbb Z$ is ...

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Smooth isometric embeddings of Riemannian manifolds
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6 votes

You should have a look at Berger's monumental A Panoramic View of Riemannian Geometry (specifically section 3.4 p. 131-142 and section 4.6, p. 216-218). Two striking theorems are the following: ...

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