Kevin Ventullo
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Registered User
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Grad student at UCLA.
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1d |
answered | Fields whose embeddings into the complex numbers are invariant under complex conjugation |
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2d |
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Finite rank free modules over PIDs This will always be the case. Since $N$ is free, $M/\ker\phi$ is torsion-free, hence free, so there is a section $M/\ker\phi\rightarrow M$; you can take $C$ to be the image of this section. By the way, in the future you should use math.stackexchange.com |
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May 7 |
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Galois group of constructible numbers It might also be easier to fix a finite set of primes $S$ containing $2$, and try to understand the maximal pro-2 quotient of $G_{\mathbb{Q},S}$. For one thing, this group is finitely generated. |
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Apr 30 |
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Generators of class groups A sharp answer for part $b$ is that $|M|\leq \log_2(|G|)$. |
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Apr 21 |
awarded | ● Yearling |
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Apr 4 |
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American put option pricing by “binomial trees” Perhaps it's a rounding issue. |
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Apr 4 |
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American put option pricing by “binomial trees” A rough explanation for the dependence on parity is that if $n$ is odd, the possible share prices at time $t$ are $\lbrace su,sd,su^3,sd^3,\ldots\rbrace$, whereas if $n$ is even the possible prices are $\lbrace s,su^2,sd^2,su^4,sd^4,\ldots\rbrace$. |
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Apr 4 |
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Is there an analog of determinant for linear operators in infinite dimensions as that of finite dimensions? @darij: The condition you described only lets you define $\det (1-Tf)$. This will be a polynomial whose roots are the reciprocals of the nonzero eigenvalues of $f$. |
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Mar 25 |
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Self-containing structures Pseudo-example: the modular curve $X_0(11)$, say over $\mathbb{C}$, parametrizes elliptic curves together with a subgroup of order 11. It has genus one, so is itself an elliptic curve after choosing some base point. Thus it ''contains'' 12 copies of itself. |
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Mar 1 |
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Market-clearing price vector in an “aggregate demand system” Are there additional assumptions? It seems to me if each $D_i$ is constant, equal to say $1/n$, then all your conditions are satisfied, but you cannot obtain any demand vector by varying the prices. |
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Feb 16 |
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What is the spectrum of a ring of holomorphic rational power series? For example, take $p = x^2 -2$. Let $a_0, a_1, a_2, \ldots$ be a sequence of rationals which sum to $\sqrt{2}$, and which tends to zero as fast as you want. Let $f(x) = x + (a_0 + \frac{a_1}{2}x^2 + \frac{a_2}{4}x^4 + \frac{a_3}{8}x^6 +\ldots) \in R_\rho$. Then $f(-\sqrt{2})=0$ and $f(\sqrt{2})=2\sqrt{2}$. One can modify this example to get any pair of real numbers. Thus $R_\rho/(p) \cong \mathbb{Q}(\sqrt{2})\otimes \mathbb{R} \cong \mathbb{R} \times \mathbb{R}$. |
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Feb 16 |
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What is the spectrum of a ring of holomorphic rational power series? So if $p\in \mathbb{Q}[x]$, it seems to me that the ideal generated by $p$ in $R_\rho$ is just the set of functions which vanish at the zeroes of $p$ to at least the same order as $p$. This means $R_\rho/(p)\subset \mathbb{Q}[x]/(p) \otimes_\mathbb{Q} \mathbb{R}$. My guess is this is always an isomorphism. |
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Feb 15 |
answered | Using schemes to prove things about rings |
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Jan 17 |
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Base Change for Eigenvarieties Dear David, that's great! Sure, my email is (mylastname) at math.ucla.edu. Can you remove the condition that $E/F$ is cyclic when $G=GL_2$, using the recent results of Dieulefait? |
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Dec 8 |
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The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field? When I was much younger, this is what I thought Class Field Theory was all about. |
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Dec 3 |
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when is a sum of idempotents idempotent in commutative ring theory? Actually, in my response you should replace "connected component" or "set of connected components" with "clopen subset". Basically, the condition is satisfied if and only if every direct summand of $R$ is characteristic zero. |
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Dec 2 |
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when is a sum of idempotents idempotent in commutative ring theory? If some connected component has characteristic $p$, then taking the sum of the corresponding idempotent with itself $p+1$ times gives back the idempotent, so the statement is not quite true in this case. |
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Dec 2 |
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when is a sum of idempotents idempotent in commutative ring theory? An idempotent in a commutative ring $R$ is the same thing as a set of connected components of Spec$(R)$; the idempotent is identically one on this set, and identically zero on its complement. Thus, if each connected component has characteristic zero (e.g. if $R$ is a $\mathbb{Q}$-algebra), then a sum of idempotents can only be an idempotent if their corresponding sets of components are disjoint, i.e. the pairwise products are all zero. |
