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comment Density of polynomials with a prescribed number field extension
It presumably goes to $0$. Consider the case $n = 2$ for concreteness.
Apr
23
revised How seriously should a graduate student take teaching evaluations?
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Apr
23
comment p-adic analogue of the Strong Law of Large Numbers
The $p$-adic Haar measure does not work well for the integration developed in his book, since it is not a bounded distribution. I've never heard of a robust $p$-adic version of the LLN.
Apr
23
comment p-adic analogue of the Strong Law of Large Numbers
Jesse, be more precise about what you mean by a $p$-adic random variable and integration wrt the $p$-adic-valued Haar "measure." Such integration is not well-defined in general, as a limit of Riemann sums can depend on the choice of sample points. See p. 39 of Koblitz's $p$-adics book (2nd ed.) for a problem with the simple example $f(x) = x$. If you want to define this integral using sample points $0,1,\ldots,p^n-1$ at "level $p^n$" then the integral exists for a restricted class of continuous functions. See Robert's $p$-adics book for "Volkenborn integral."
Apr
21
comment Is this theorem on $L$-functions known?
In your theorem, the two conditions do not directly mention $L_1$ and $L_2$. It would be better to write those conditions using "$L_i$ for $i = 1$ and $2$" in each and writing $L_i$ instead of $L$ in the equations.
Apr
20
awarded  Nice Answer
Apr
20
comment Surjectivity of trace map
This already has counterexamples in the simplest example of quadratic fields. Let $R = \mathbf Z$, $F = \mathbf Q$, and $K = \mathbf Q(\sqrt{d})$ where $d$ is a squarefree integer. If $d \not\equiv 1 \bmod 4$ then $A = \mathbf Z[\sqrt{d}]$ and the image of the trace map from $A$ to $R$ is $2\mathbf Z$. A concrete example is $A = \mathbf Z[i]$. In these examples $2$ is wildly ramified.
Apr
16
comment When has the Borel-Cantelli heuristic been wrong?
@DavidSpeyer, for the convenience of anyone reading this let me point out in that example in $\mathbf F_2[u]$ that there is an obstacle, but it's not local: if $f(0) \not= 0$ then $\mu(f^8+u^3) = 1$, so $f^8 + u^3$ can't be irreducible (it's easier to see this if $f(0) = 0$). The calculation of the M\"obius function of $f^8+u^3$ is, as you say, not obvious.
Apr
14
comment distribution of $\sqrt{-1} \mod p$
@JohannesHahn, don't think $p$-adically. Just use the mod $p$ reduction lying between $0$ and $p-1$ in the usual sense, and divide that by $p$. Of course there is a sign ambiguity on $\sqrt{-1}$, but just fix a definite interval $[a,b]$ inside $[0,1]$ and ask how often some square root of $-1 \bmod p$ has a normalized ratio in there, i.e., count for large $x$ how many $p \leq x$ with $p \equiv 1 \bmod 4$ have an $r_p$ from $0$ to $p-1$ such that $r_p^2 \equiv -1 \bmod p$ and $a \leq r_p/p \leq b$. Then let $x \rightarrow \infty$ and look at asymptotics. Isn't that a sensible formulation?
Apr
14
comment What are some mathematical concepts that were (pretty much) created from scratch and do not owe a debt to previous work?
I found this information in "Mathematics of Frobenius in Context" by Hawkins, pp. 288-289.
Apr
14
comment What are some mathematical concepts that were (pretty much) created from scratch and do not owe a debt to previous work?
@Kimball, some special cases of composition of quadratic forms had been worked out before Gauss, not just "trivial" (in sense of well-known and in the sense of trivial class group) cases like $x^2+y^2$ and $x^2+2y^2$ by Fermat and Euler long before Gauss, but also Lagrange had shown in 1783, hence about 20 years before Gauss, that $(2x^2+2xy+3y^2)(2x'^2+2x'y'+3y'^2)$ can be written as $X^2 + 5Y^2$ where $X$ and $Y$ are each bilinear in $(x,y,x',y')$ with integer coefficients. Gauss saw the analogy between his composition law on classes of quadratic forms and multiplication in unit groups.
Apr
14
comment What are some mathematical concepts that were (pretty much) created from scratch and do not owe a debt to previous work?
Certainly Dirichlet was inspired by Fourier. In the book From Fermat to Minkowski the chapter on Dirichlet has a quote from Jacobi to Humboldt in which Jacobi writes that Dirichlet "has introduced a new branch of mathematics, which uses the infinite series, introduced by Fourier for the study of heat, to discover properties of prime numbers." Also, Dirichlet himself very clearly acknowledges in his paper that he was trying to adapt Euler's proof of infinitude of the primes via divergence of $\sum 1/p$.
Apr
10
comment Is there a generalization of the “characteristic polynomial” to other split/quasi-split algebraic groups?
Check the title of Springer's book in the last paragraph.
Apr
8
comment An application of Maschke's theorem
@DavidHill, the ring $\mathbf Z[\sqrt{2}]$ has a division algorithm, so it is a unique factorization domain: all integral domains with a division algorithm have unique factorization. Don't try to find more counterexamples to unique factorization in $\mathbf Z[\sqrt{2}]$; they do not exist and if you think you found one then you're making an error.
Apr
8
comment An application of Maschke's theorem
@DavidHill, that is not a counterexample! The numbers $7$ and $5\sqrt{2}\pm 1$ are not prime. What you wrote is like saying $\mathbf Z$ lacks unique factorization because $4\cdot 9 = 6 \cdot 6$. Each factorization can be broken down further and rearranged to get the other one: $4 \cdot 9 = 2 \cdot 2 \cdot 3 \cdot 3 = 6 \cdot 6$. In ${\mathbf Z}[\sqrt{2}]$, $7 = (3+\sqrt{2})(3-\sqrt{2})$ and $5\sqrt{2}-1 = (3-\sqrt{2})^2(\sqrt{2}+1)$ while $5\sqrt{2}+1 = (3+\sqrt{2})^2(\sqrt{2}-1)$. The factors $3\pm \sqrt{2}$ are prime while the factors $\sqrt{2}\pm 1$ are units (their product is $1$).
Apr
8
comment Interesting mathematical documentaries
The page you linked to has an error on it and the film is not available there.
Apr
8
comment An application of Maschke's theorem
You write that you "ignored the fact that $\mathbf Z[\sqrt{2}]$ is not a unique factorization domain," but in fact it is a unique factorization domain!
Apr
7
awarded  Nice Answer
Apr
6
awarded  Good Answer
Apr
6
comment Some Non-Trivial Algebraic(Rational) Number
@VladimirDotsenko, I think it's pretty clear what the question is about. In elementary situations, like with nested square roots, it's easy to figure out by hand a (nonzero) polynomial with rational coefficients having the number as a root. I don't think the question is seeking examples where the algebraicity is surprising, as with the unexpected appearance of Conway's constant, but rather is seeking examples where algebraicity is not easily checked by someone knowing nothing besides the definition.