User zev chonoles - MathOverflow

Answer by Zev Chonoles for Fitting algebraic expression to a number [algorithm]

2013-04-05T23:29:02Z

The Inverse Symbolic Calculator sounds like what you want (it's not clear from your question what form you have the number in to start with).

Answer by Zev Chonoles for Jokes in the sense of Littlewood: examples?

2010-09-16T01:39:37Z

Let "$\int$" denote $\int_0^x$. We want to find the solution to

$$\int f = f-1.$$

We simply "factor out" $f$, getting $1=\left(1-\int\right)f$. Thus, $f=(1-\int)^{-1}1$.

Using the geometric series,

$$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots$$

Thus,

$$f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots=e^x,$$

as expected. (This was told to me by Steve Miller)

Are the field norm and trace the unique "nice" maps between fields?

2009-12-03T00:48:23Z

This seems like an obvious fact, but I'm not sure what the necessary meaning of "nice" is to get a result like this. I'm wondering if there is a theorem of the form:

For any <1> field extension $K/F$, a map from $\phi:K\rightarrow F$ that satisfies <2> is the field norm (or trace).

where <1> could be something like finite, algebraic, etc., and <2> could be anything (obviously there would be different <2>'s for norm and trace).

Generalizations of "standard" calculus

2009-12-23T05:11:24Z

We have the usual analogy between infinitesimal calculus (integrals and derivatives) and finite calculus (sums and forward differences), and also the generalization of infinitesimal calculus to fractional calculus (which allows for real and even complex powers of the differential operator). Have people worked on a "fractional finite" calculus, where instead of a differintegral we had some sort of "differsum"? I don't know much about it, but I was thinking maybe the answer might come from umbral calculus?

To give a motivating example/special case for this question: the Wikipedia article on fractional calculus uses the example of the $\frac{1}{2}$th derivative, which when applied twice gives the standard derivative. What is the operator $D$ on sequences such that, when applied twice, it gives the forward difference of the original sequence?

Also, I have perhaps a related question: The solution to $\frac{d}{dx}f=f$ is $f=e^x$, while the solution to $\Delta f = f$ is $f=2^x$. Is the fact that $e$ is close to 2 a coincidence, or is there something connecting these results? Is there more generally some sort of spectrum of calculi lying between "finite" and "infinitesimal" each with its own "$e$"?

EDIT: After looking around some more I found time scales, which are pretty much what I was thinking of in the second part of my question (though many of the answers people have provided are along the same general lines). I'm surprised I don't hear more about this in analysis - unifying discrete and continuous should make it a pretty fundamental concept!

Fractional powers of Dirichlet series?

2010-01-14T04:57:41Z

Let $R$ be the ring of Dirichlet series with integer coefficients. I'd often wondered about whether $R$ was a UFD; this post cleared that up, because it turns out that $R\simeq\mathbb{Z}[[x_1,x_2,\cdots]]$ (the $x_i$ correspond to primes, apparently, but I'm not sure what the explicit isomorphism is).

My first (slightly mundane) question is: what is the group of units $U(R)$? I know that $f\in R$ is a unit iff $f(0)$ is a unit (which in this case means $f(0)=\pm1$); similarly, $f\in\mathbb{Z}[[x_1,x_2,\cdots]]$ is a unit iff $f$'s constant term is $\pm1$, and part D of this link would seem to help a bit (using $\mathbb{Z}[[x_1,x_2,\cdots]]\simeq \mathbb{Z}[[x_2,\cdots]][[x_1]]$), but I couldn't get very far figuring out what $U(R)$ actually is.

Now, my main question: Can we take arbitrary $n$th roots (and hence, arbitrary fractional powers) of Dirichlet series which are units in $R$? I believe this is equivalent to asking whether the group of units is divisible, but I'm not sure.

A motivating example / special case of my question

$\mu$ and 1 (where $\mu$ is the Mobius function, $1$ is the constant 1 function) are units in $R$. In fact, $\mu\cdot 1=\epsilon$ (where $\epsilon(0)=1$, $\epsilon(n)=0$ for $n>0$ is the identity). Expressing this using the actual series,

$\displaystyle\left(\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{1}{n^s}\right)=\sum_{n=1}^\infty \frac{\epsilon(n)}{n^s} = 1$

and hence $\displaystyle\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$. Indeed, we can find the Dirichlet series for $\zeta(s)^k$ for any $k\in \mathbb{Z}$ by looking at the corresponding element $1^k\in R$ (note that $1^{-k}=\mu^k$). However, I would like to know what Dirichlet sequence corresponds to $\displaystyle\zeta(s)^{\frac{a}{b}}$ for $\frac{a}{b}\in\mathbb{Q}$.

Consequences of not requiring ring homomorphisms to be unital?

2010-08-03T04:12:03Z

As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1_R)=1_S$. But there has been a minority who do not require this, one prominent example being Herstein in Topics in Algebra.

What are some of the most striking consequences of not requiring ring homomorphisms to be unital? For example, what aspects of algebraic geometry would need to be reworked if we no longer required it? What interesting theorems or techniques arise in the not-necessarily-unital theory which do not apply (or are degenerate) for unital homomorphisms?

Special map from a manifold to GL_n(R)?

2009-12-15T03:49:22Z

I've just finished my first course in differential geometry, so forgive me if this is maybe a silly or well-known question, but given any, say, diffeomorphism of $n$-manifolds $\phi:M\rightarrow N$, I was wondering whether the map $f:M\rightarrow GL_n(\mathbb{R})$ defined by $f(p)=d\phi_p$ has any important or nice properties? One guess I had was that further sending $p$ to det$(d\phi_p)$ could say something about orientation? I had been thinking about the Gauss map and how it's a map from any surface $S$ to the sphere $\mathbb{S}^2$, and I was trying to come up with other canonical ways of creating maps of surfaces and/or manifolds to "special" ones.

If K/k is a finite normal extension of fields, is there always an intermediate field F such that F/k is purely inseparable and K/F is separable?

2010-02-21T20:39:25Z

I was feeling a bit rusty on my field theory, and I was reviewing out of McCarthy's excellent book, Algebraic Extensions of Fields. Out of Chapter 1, I was able to work out everything "left to the reader" or omitted except for one corollary, stated without proof (see here for the page in the book):

Let $K/k$ be a finite normal extension. Then $K$ can be obtained by a purely inseparable extension, followed by a separable extension.

The text immediately preceding this implies that the intermediate field that's going to make this happen is $F=\{a\in K:\sigma(a)=a$ for all $\sigma\in Gal(K/k)\}$, and I understand his argument as to why $F/k$ is purely inseparable (in fact, that's the theorem, Theorem 21, which this is a corollary to). What I don't understand is why $K/F$ is separable; I don't see how we've ruled out it being non-purely inseparable.

Note that I will be making a distinction between non-purely inseparable (inseparable, but not purely inseparable) and not purely inseparable (either separable or non-purely inseparable).

Here are some observations / my general approach:

One big thing that seemed promising was Theorem 11 (at the bottom of this page), which is basically the reverse of the corollary I'm having trouble with:

Let $K$ be an arbitrary algebraic extension of $k$. Then $K$ can be obtained by separable extension followed by a purely inseparable extension.

(the separable extension referred to is of course the separable closure of $k$ in $K$). It seems like we want to use Theorem 11 on $K/F$, and argue that there can't be "any more" pure inseparability, but I couldn't figure out a way of doing this.

Theorem 21 is actually an "if and only if" (that is, $a\in K$ is purely inseparable over $k$ iff $\sigma(a)=a$ for all $\sigma\in Gal(K/k)$). Because this implies that any $a\in K$ with $a\notin F$ is not purely inseparable over $k$, we have that $F$ is the maximum (not just maximal) purely inseparable extension of $k$ in $K$.
If any $a\in K$ were purely inseparable over $F$, by Theorem 8 (see here), there is some $e$ for which $a^{p^e}\in F$. But by the same theorem, since $F/k$ is purely inseparable, there is some $b$ for which $(a^{p^e})^{p^b}=a^{p^{e+b}}\in k$. Thus $a$ would be purely inseparable over $k$ by the converse (Corollary 1 to Theorem 9, see here), and hence be in $F$. Thus, $K$ (and any field between $K$ and $F$, besides $F$ itself) is not purely inseparable over $F$.

So, that's why I don't see how we've ruled out $K/F$ being non-purely inseparable. Sorry about making lots of references to the book - I'm just not sure what previously established results McCarthy intended to be used, and I wanted to point out what I saw as the important ones for people not familiar with the book. I'm sure I'm missing something obvious here. Does anyone see the last bit of the argument?

MaxSpec, Spec, ... "RadSpec"? Or, why not look at all radical ideals?

2010-02-05T21:04:41Z

I was reading this question on why algebraic geometry looks at prime ideals instead of only maximal ideals, and I understand Anton's answer, but I'm a little confused as to how this fits with Hilbert's Nullstellensatz - affine algebraic sets are in bijection with radical ideals, not prime ideals, and it seems like we'd want the extra information we'd get by looking at "RadSpec(R)" (my own imagined notation). Also, the preimage of a radical ideal is radical, so there isn't the same objection as to maximal ideals - "RadSpec" would also be a contravariant functor.

So, why not radical ideals instead of only prime ideals, and what kinds of things could we say about RadSpec(R) even if they aren't very interesting?

Answer by Zev Chonoles for Which book would you like to see "texified"?

2011-05-13T18:07:39Z

Atiyah + Macdonald, Introduction to Commutative Algebra.

Kahler differentials of a hypersurface over a non-algebraically closed field

2011-04-17T04:36:08Z

The following was recently on my algebraic geometry homework:

Let $k$ be an algebraically closed field, $f\in B=k[x_1,\ldots,x_n]$, and $A=B/(f)$. Show that $\Omega_{A/k}$ is locally free of rank $n-1$ $\iff$ $\nexists\, p\in k^n$ such that $f(p)=0$ and all $\frac{\partial f}{\partial x_i}(p)=0$.

Here, $\Omega_{A/k}$ is just the module of differentials, not the sheaf of differentials on the corresponding variety (so locally free is meant in the sense of modules). My solution (at least seems to) crucially depend on the Nullstellensatz, so my question is, are there any non-algebraically closed fields $k$ for which this result is still true? If so, is there an argument that treats them simultaneously? Or, if not, is there a good intuition for why algebraically closed is necessary?

Answer by Zev Chonoles for Sets of equations

2011-04-17T07:07:57Z

Here is a second attempt (see edit history for previous version).

For each $t\in\mathbb{N}$, let $$P_{i,j,k,t}=\{1_{i,j,k,t},\ldots,n_{i,j,k,t},\ldots,\gamma(i,j,k)_{i,j,k,t}\}$$ (so that for each choice of $i\in I$, $j\in J$, $k\in K$, and $t\in\mathbb{N}$, we have a disjoint set of size $\gamma(i,j,k)$).

For each $t\in\mathbb{N}$, let $$Q_t=\{a_{k,t}\mid k\in K\}$$ (so for each $t\in\mathbb{N}$, this is just a copy of $K$, up to relabeling).

Let $$X=\coprod_{t\in\mathbb{N}}\left(Q_t\coprod_{\substack{i\in I,j\in J\\k\in K}}P_{i,j,k,t}\right).$$ Define $$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k,1}\subset X,$$ and $f_i:X\rightarrow X$ by $$f_{i_0}(n_{i,j,k,t})=\begin{cases}a_{k,1}\text{ if }i=i_0,t=1\\ n_{i,j,k,t+1}\text{ otherwise}\end{cases}$$ $$f_i(a_{k,t})=a_{k,t+1}$$

Thus $$f_{i}^{-1}(n_{i,j,k,t})=\begin{cases}\emptyset\text{ if }t=1,2\\ \{n_{i,j,k,t-1}\}\text{ if }t>2\end{cases}$$ $$f_i^{-1}(a_{k,t})=\begin{cases}\coprod_{j\in J}P_{i,j,k,1}\text{ if }t=1\\ \{a_{k,t-1}\}\text{ if }t>1\end{cases}$$ We choose $p_k=a_{k,1}$.

Thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k,1}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.

Unfortunately this still doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable, because if $J$ is uncountable then $f_i^{-1}(a_{k,1})$ is uncountable (I added the whole mess with the $t$'s to make the preimages of all the other elements countable). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.

Can we relate Cech cohomology and derived functor cohomology even when the cover we choose isn't nice?

2011-04-07T05:39:51Z

In my algebraic geometry class this semester, we've learned about Leray's Theorem, which states that for a sheaf $\mathcal{F}$ on a topological space $X$, and $\mathcal{U}$ a countable cover of $X$, if $\mathcal{F}$ is acyclic on every finite intersection of elements of $\mathcal{U}$ then the Cech cohomology $\check{H}^p(\mathcal{U},\mathcal{F})$ and derived functor cohomology $H^p(X,\mathcal{F})$ agree.

The potential for disagreement between them is covered well in these two MO questions. However, what neither of them seem to address is whether we can salvage any information about $H^p(X,\mathcal{F})$ from $\check{H}^p(\mathcal{U},\mathcal{F})$ even when $\mathcal{U}$ does not have the property that $\mathcal{F}$ is acyclic on all finite intersections, which is what I'd like to find out about here. I'm aware of Hartshorne Lemma 3.4.4, which says that there is a natural map $\check{H}^p(\mathcal{U},\mathcal{F})\rightarrow H^p(X,\mathcal{F})$ which is functorial in $\mathcal{F}$, but this is gotten by abstract nonsense - my feeling is that the existence of this map is not conveying much useful information. For all we know (?), all these maps could be the trivial homomorphism.

What I'm imagining is that perhaps the higher cohomology of $\mathcal{F}$ on the finite intersections of $\mathcal{U}$ can be related to the "difference" between $\check{H}^p(\mathcal{U},\mathcal{F})$ and $H^p(X,\mathcal{F})$, and that when the higher cohomology vanishes (i.e. $\mathcal{F}$ is acyclic), we get back the original theorem (that Cech and derived functor agree).

So, is there a useful relationship betwen Cech and derived functor cohomology even when $\mathcal{U}$ is not a nice open cover with respect to $\mathcal{F}$? Am I mistaken in assuming that the map $\check{H}^p(\mathcal{U},\mathcal{F})\rightarrow H^p(X,\mathcal{F})$ is not (particularly) useful?

Also I would like to avoid if possible the operation of taking the limit over all covers of $X$. I want to relate the specific Cech cohomology with respect to the cover $\mathcal{U}$, whatever its failings may be, with the derived functor cohomology.

Answer by Zev Chonoles for What are the units of $\mathbb{Z}/4\mathbb{Z}[x]$?

2011-03-24T09:06:58Z

Problem 1.2i in Atiyah-Macdonald states that for any ring $A$ and $f=a_0+\cdots+a_nx^n\in A[x]$, $f$ is a unit $\iff$ $a_0$ is a unit in $A$, and the $a_i$ are nilpotent.

Note that $A=\mathbb{Z}/n\mathbb{Z}$ has nontrivial nilpotents $\iff$ $n$ is not squarefree.

In your specific case, $\mathbb{Z}/4\mathbb{Z}$ has 0 and 2 as nilpotents, and 1 and 3 as units, hence the units of $(\mathbb{Z}/4\mathbb{Z})[x]$ are, as you predicted, exactly the polynomials of the form $\pm1+2xp$ for $p\in(\mathbb{Z}/4\mathbb{Z})[x]$.

The natural inclusion of an infinite abelian group $G$ into $\widehat{\widehat{G}}$

2011-03-01T01:24:09Z

I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into $$\widehat{\widehat{G}}=\text{Hom}_{\mathsf{Ab}}(\text{Hom}_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$ is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete).

Obviously, $G$ has to be infinite. However, the only example I could think of (much less actually prove worked) was $G=\mathbb{Z}$, in which case $\widehat{G}=\text{Hom}_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that $$\widehat{\widehat{G}}\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$ which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$, and therefore not isomorphic to $\mathbb{Z}$. However, (my understanding is that) the existence of the anything more than the two obvious automorphisms of $\mathbb{C}$ requires the axiom of choice. So my questions are,

Does the claim that $\mathbb{Z}\not\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$ require AC?

and, if the answer to the above is yes,

Does the claim that there exists some abelian group $G$ such that $G\not\cong\widehat{\widehat{G}}$ require AC? Is it equivalent?

I initially asked this on math.SE but I suspect it's harder than I initially thought.

Problems where we can't make a canonical choice, solved by looking at all choices at once

2010-12-21T04:04:38Z

It's a common theme in mathematics that, if there's no canonical choice (of basis, for example), then we shouldn't make a choice at all. This helps us focus on the heart of the matter without giving ourselves arbitrary stuff to drag around.

However, in this question, I'm looking for examples of problems solved by a specific type of "not making a choice" - namely, making all available choices, and looking at all the end results together as a whole. We can't necessarily discern any individual piece, but the average behavior, or some other information about the big picture, provides (or at least points towards) a solution.

I really wish I had an example of this phenomenon to provide, but even one escapes me at the moment, which is what spurred me to ask this question. I imagine combinatorics is full of examples; unfortunately I haven't really studied that field in any depth yet.

Something close to what I'm after is Burnside's (a.k.a. not-Burnside's) Lemma. There's no good way of directly counting orbits, i.e. choosing to look at a particular orbit one at a time, so we just look at the average number of fixed points of elements of $G$ (I'm reluctant to call this an example of the kind of result I'm looking for because I'm not entirely clear on why the fixed points of an element should be thought of as substitutes for orbits. Perhaps that's a separate question).

This question is in a similar vein.

Extending arithmetic functions to groups

2011-02-09T09:05:50Z

Thinking along the lines of Tom Leinster's fascinating recent question, I'm wondering more generally about how to extend questions about natural numbers to groups, with the cyclic groups representing the natural numbers, and normal subgroups representing divisors. For example, let $\mathbb{G}$ be the set of all isomorphism classes of finite groups. Then an "arithmetic function" could be defined to be simply a function $f:\mathbb{G}\rightarrow\mathbb{C}$. Here are a few analogs I wrote down rather quickly (so perhaps there are better proposals for generalizations than these): $$\text{id}(G)=|G|\quad\quad\quad \epsilon(G)=\begin{cases}1\text{ if $G$ is trivial}\\ 0\text{ otherwise}\end{cases}\quad\quad\quad z(G)=0$$ $$\sigma_k(G)=\sum_{N\,\triangleleft \,G} |N|^k \quad\quad\quad\quad\phi(G)=|G|\prod_{\substack{N\,\triangleleft \,G \\ |N|\text{ prime}}}\left(1-\frac{1}{|N|}\right)$$ Tom Leinster's question is whether there is a solution to $\sigma_1(G)=2|G|$.

Question 1: What is known, if anything, about these functions? What is a good proposal for an analog of the Mobius function (I couldn't think of one offhand)? Can anyone demonstrate that they satisfy formulas that are analogs of their natural-number counterparts, or perhaps instead give examples that would indicate that these functions act weird and aren't nice generalizations to make?

I suspect that if these functions act badly, the most likely fix would be to redefine $\mathbb{G}$ to be isomorphism classes of finite abelian groups. All subgroups are normal, and the structure theorem makes things much more controlled. I would guess that defining (for example) what it means for two groups to be "coprime", and hence what it means for an "arithmetic function" to be "multiplicative", would go over much more smoothly with abelian groups.

Going further: given two "arithmetic functions" $f,g:\mathbb{G}\rightarrow\mathbb{C}$, we can define a "Dirichlet convolution" by $$(f\ast g)(G)=\sum_{N\,\triangleleft \,G}f(N)g(G/N)$$ I've got to say, my jaw dropped a bit when I wrote that down. But one immediate difference I can see, somewhat discouraging, is that $\ast$ would not be abelian, since we aren't guaranteed that there are any normal subgroups $M\triangleleft G$ isomorphic to $G/N$ (see this MO question), much less that if $M\cong\! G/N$ then $G/M\cong\!\! N$. However, the function $\epsilon$ is still a left and right identity for $\ast$, and $z$ is still a zero.

Question 2: Can anyone prove or disprove that $\ast$ is associative? If it is, we at least get a non-commutative ring under $\ast$ and pointwise addition (that $\ast$ distributes over pointwise sums is obvious).

Now I suspect I am really getting "greedy" with my generalizing. We could further define "Dirichlet series", such as the zeta function (note that this is not the same thing as the zeta function of a group): $$\zeta_{\mathbb{G}}(s)=\sum_{G\in\mathbb{G}}\frac{1}{|G|^s}=\lim_{n\rightarrow\infty}\sum_{\substack{G\in\mathbb{G}\\ |G|\leq n}}\frac{1}{|G|^s}$$ I seriously doubt there is any hope for an Euler product-like expression. But perhaps, if we restricted ourselves to finite abelian groups...

Also, I'm familiar with the result that the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3+O(n^{8/3})}$, and I imagine the growth rate is at least as bad for not-necessarily $p$-groups.

Question 3: Is there any $s>0$ for which $\zeta_{\mathbb{G}}(s)$ converges?

I wrote this question rather fast, and I welcome any feedback on how to improve it, or make it more appropriate for MO. Should I break this up into multiple questions? Is it too open-ended?

Can Eisenstein's lattice point proof of quadratic reciprocity be generalized?

2011-02-09T06:50:15Z

I'm referring to this proof. The key formula ("Eisenstein's Lemma") is $$\left(\frac{q}{p}\right)=(-1)^{\sum_{u}\lfloor\frac{qu}{p}\rfloor},\text{ where $u=2,4,\ldots,p-1$}$$
The sum in the exponent is easily seen to count the number of lattice points in this rectangle that are below the diagonal and having even $x$-coordinate,

where $p$ and $q$ are our primes in question, and via some clever flipping, quadratic reciprocity pops out.

I seem to recall in my first summer at PROMYS, some counselors tried to work out a version for quartic reciprocity, using (if I remember correctly) a similarly-defined lattice in $\mathbb{Z}[i]\times\mathbb{Z}[i]$. However, I don't remember the details, or if they were successful. I'd be interested to see if a "low-tech" proof using lattice point counting can work for cubic or quartic reciprocity laws (or even more generally, but I suspect that would be overly optimistic).

Answer by Zev Chonoles for Errata for Atiyah-Macdonald

2011-02-05T03:10:16Z

On page 91, the second line in the second Example should refer to Proposition 8.8, not Theorem 8.7.

Answer by Zev Chonoles for Are some numbers more irrational than others?

2011-01-29T16:18:24Z

Yes, there is such a thing as the irrationality measure of a real number (I'm not sure if it can be / has already been extended to complex numbers). It is based on the idea that all algebraic numbers (including the golden ratio) are hard to approximate well by rationals, relative to the size of the denominator of the rational used, while it is sometimes possible for a transcendental number to be approximated better. In particular, if a number $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ has the property that there are infinitely many rational approximations $\frac{p}{q}\in\mathbb{Q}$ with $|\,\alpha-\frac{p}{q}|< q^{-t}$, then $t$ is a lower bound for the irrationality measure of $\alpha$; the larger $t$ is, i.e. the better your approximations are relative to the denominator, the "more irrational" you are, at least from a Diophantine approximation point of view.

From Wikipedia: The irrationality measure of a rational number is 1; the very deep theorem of Thue, Siegel, and Roth shows that any algebraic number that isn't rational has irrationality measure 2; and transcendental numbers will have an irrationality measure $\geq2$. However, as Douglas Zare has pointed out in the comments, the set of transcendental numbers of irrationality measure $>2$ has measure 0, so that in most cases it's unfortunately not useful as a comparison.

It appears that the irrationality measure of $\pi$ is not currently known, but that there are upper bounds; the most recent one I could find is this, which would appear to show that $\mu(\pi)\leq7.6063$. The Wikipedia article claims that $\mu(e)=2$, so whether or not $\pi$ is "more irrational" than $e$ looks like an open question.

Answer by Zev Chonoles for trace zero elements in algebraic number fields

2011-01-26T05:30:13Z

The answer is no. Let $\alpha\in\mathbb{C}$ be a root of $x^3+tx+1$ for $t\in\mathbb{Q}\setminus\mathbb{Z}$, and let $L=\mathbb{Q}(\alpha)$, so that $N_\mathbb{Q}^L(\alpha)=1$ and $tr_{\mathbb{Q}}^L(\alpha)=0$. The element $\alpha$ cannot be integral over $\mathbb{Z}$, because its minimal polynomial over $\mathbb{Q}$ is not integral.

Answer by Zev Chonoles for Proving theorems by using functions with fixed points.

2011-01-23T15:17:45Z

There is a very slick proof (discussed here on MO) that every prime $p=4k+1$ is a sum of two squares, which looks at the set $S= \{(x,y,z) \in N^3: x^2+4yz=p \}$ and shows that a particular involution of $S$ has exactly one fixed point.

Intuitive explanation of Burnside's Lemma

2010-12-21T05:42:22Z

Burnside's Lemma states that, given a set $X$ acted on by a group $G$,

$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|$$

where $|X/G|$ is the number of orbits of the action, and $|X^g|$ is the number of fixed points of $g$. In other words, the number of orbits is equal to the average number of fixed points of an element of $G$.

Is there any way in which the fixed points of an element $g$ can be thought of as orbits? I had wondered aloud on my recent question here how (or if) Burnside's Lemma can be interpreted as having the same kind of object on both sides, so as to be a "true" average theorem, e.g.

"number of orbits = average over $g\in G$ of (number of orbits satisfying (something to do with $g$))"

"number of orbits = average over $g\in G$ of (number of orbits of some new action which depends on $g$)"

Since Qiaochu stated the comments to my question that he suspects Burnside's Lemma can be categorified, and that this may be related, I have also added that tag.

Factoring a field extension into one which adds no roots of unity, followed by one which adds only roots of unity

2010-12-19T22:40:57Z

I am asking my question here, since it's been voted up a fair bit on math.SE, but without answers, so it may be harder than I assumed it was.

Can we always break an arbitrary field extension $L/K$ into an extension $F/K$ in which the only roots of unity of $F$ are those in $K$, i.e. $\mu_F=\mu_K$, followed by an extension $L/F$ which is of the form $L=F(\{\omega_i\})$, where the $\omega_i$ are roots of unity? If not, are there reasonable hypotheses (e.g. separable, finite) on $L/K$ that would make this true?

My motivation was simply that the other order, i.e. breaking an arbitrary extension $L/K$ into one where $F=K(\{\omega_i\})$ for some roots of unity $\omega_i$, followed by $L/F$ where $\mu_L=\mu_K$, is obvious - specifically, set $F=K(\mu_L)$.

Now, my first (naive) approach was to try to construct the "maximum" intermediate field which does not add roots of unity by taking the compositum of all such intermediate extensions. However, this doesn't exist even for number fields, e.g. setting $K=\mathbb{Q}$, $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$, $E_1=\mathbb{Q}(\sqrt[3]{2})$, and $E_2=\mathbb{Q}(\zeta_3\sqrt[3]{2})$, we have $\mu_{E_1}=\mu_{E_2}=\mu_K$, but $\mu_{E_1E_2}=\mu_{L}\supsetneq\mu_K$.

Note that $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ isn't a counterexample to the actual problem - for example, $F=E_1$ works, because $\mu_{E_1}=\mu_{\mathbb{Q}}$, and $L=E_1(\zeta_3)$.

So, to prove the claim / construct a counterexample, it seems to me that we want to look at intermediate fields $E$ which are maximal among those such that $\mu_E=\mu_K$, and determine whether or not there always exists at least one such $E$ such that $L=E(\text{some roots of unity})$.

Here is Arturo Magidin's comment on the original question:

not a proof/counterexample, but an observation: suppose $L$ is Galois over $K$; we can let $M$ be the extension of $K$ obtained by adding all roots of unity in $L$; this is Galois over $K$, so corresponds to a normal subgroup $H$ of $G=\text{Gal}(L/K)$. If we can break up the extension as you mention, then $L$ is Galois over $F$, and $\text{Gal}(L/F)=\text{Gal}(M/K)=G/H$. So $G$ would necessarily have normal subgroup $H$ and a subgroup isomorphic to $G/H.$

Good lattice theory books?

2010-02-06T01:29:08Z

A recent answer motivated me to post about this. I've always had a vague, unpleasant feeling that somehow lattice theory has been completely robbed of the important place it deserves in mathematics - lattices seem to show up everywhere, the author or teacher says "observe that these ____ form a complete lattice" or something similar, and then moves on, never to speak of what that might imply. But, not currently knowing anything about them, I can't be sure. What would be a good place to learn about lattice theory, especially its implications for "naturally occurring" lattices (subgroups, ideals, etc.)?

Answer by Zev Chonoles for What is the p-adic valuation of a number?

2010-11-07T09:24:23Z

The conflict is just that some people use the words valuation and absolute value interchangeably. The term "p-adic valuation", used correctly, refers to $\nu$, though perhaps in some areas of math the prevailing choice is the other way around.

How badly can Krull's Hauptidealsatz fail for non-Noetherian rings?

2010-10-17T15:15:04Z

Krull's Hauptidealsatz (principal ideal theorem) says that for a Noetherian ring $R$ and any $r\in R$ which is not a unit or zero-divisor, all primes minimal over $(r)$ are of height 1. How badly can this fail if $R$ is a non-Noetherian ring? For example, if $R$ is non-Noetherian, is it possible for there to be a minimal prime over $(r)$ of infinite height?

Answer by Zev Chonoles for Errata for Atiyah-Macdonald

2010-10-16T22:53:15Z

On page 31, the first line refers to Proposition 2.11, when it should be 2.12.

Answer by Zev Chonoles for Errata for Atiyah-Macdonald

2010-10-16T22:48:02Z

On page 29, the example at the top has two typos: it says "$(x)=2x$", when it should be "$f(x)=2x$", and the exact sequence at the end of that same line says "$0\rightarrow\mathbb{Z}\otimes \stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$", when it should be

"$0\rightarrow\mathbb{Z}\otimes N\stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$".

Answer by Zev Chonoles for Errata for Atiyah-Macdonald

2010-10-16T22:28:20Z

On page 8, the proof of part ii of Proposition 1.11 begins "Suppose $\mathfrak{p}\not\subseteq\mathfrak{a}_i$ for all $i$." It should be $\not\supseteq$.

Comment by Zev Chonoles

2013-06-04T13:43:38Z

Originally posted on math.SE: math.stackexchange.com/q/402350/264

Comment by Zev Chonoles

2013-05-30T08:00:53Z

@Jiang: I've edited the MathJax code. There are some issues with how it works here, but they should go away when MathOverflow moves to the SE 2.0 network sometime soon: meta.mathoverflow.net/discussion/1416/1/…

Comment by Zev Chonoles

2013-05-26T22:37:44Z

There is at least one entire book on this subject: amazon.com/dp/088385578X which makes me feel that this question is too broad.

Comment by Zev Chonoles

2013-05-26T20:46:20Z

Crossposted to math.SE: math.stackexchange.com/q/403236/264

Comment by Zev Chonoles

2013-05-26T06:35:29Z

This question is off-topic here. It would be much better suited to mathematica.stackexchange.com

Comment by Zev Chonoles

2013-05-23T04:17:21Z

Crossposted from math.SE: math.stackexchange.com/q/399698/264

Comment by Zev Chonoles

2013-05-23T04:16:29Z

Crossposted from math.SE: math.stackexchange.com/q/399296/264

Comment by Zev Chonoles

2013-05-18T01:18:28Z

This was crossposted from math.SE: math.stackexchange.com/q/395000/264. In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere.

Comment by Zev Chonoles

2013-04-19T06:56:09Z

This site is for questions of interest to research mathematicians (as is indicated in the FAQ), so your question is off-topic here. Try math.stackexchange.com or one of the other sites listed in the FAQ, and please be sure to explain what you've tried on the problem when you post.

Comment by Zev Chonoles

2013-04-12T21:30:52Z

Crossposted from math.SE: math.stackexchange.com/q/359883/264 In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere.

Comment by Zev Chonoles

2013-04-09T08:32:04Z

This is far better suited to mathematica.stackexchange.com, so I've voted to close here.

Comment by Zev Chonoles

2013-04-08T12:12:47Z

It can also be fixed by enclosing a few of the pieces of MathJax with backticks, e.g. `(dollar sign)...(dollar sign)`, but since this hack will not work on the Stack Exchange network, I'm hesitant to add to what is already a daunting problem (given how commonly this has been used in the past on MO). See here for an example of how math inside backticks appears on the SE network: math.stackexchange.com/questions/354677/…

Comment by Zev Chonoles

2013-04-08T00:52:43Z

Any particular reason you put [closed] at the end of your own question? I'm guessing you thought people wouldn't look at it then, and it'd stay open? (See meta.mathoverflow.net/discussion/1543)

Comment by Zev Chonoles

2013-04-02T21:35:16Z

Crossposted from math.SE: math.stackexchange.com/q/349491/264

Comment by Zev Chonoles

2013-03-24T18:41:21Z

MathOverflow is for questions of interest to research mathematicians (this is explained in the FAQ). Therefore your question will be closed. The FAQ provides a list of some more appropriate sites. Moreover, wherever you decide to post this question, it is quite helpful - and polite - to actually explain what you've tried so far, not just issue commands to people.