User lubin - MathOverflow

Answer by Lubin for Writing papers in pre-LaTeX era?

2012-07-13T20:40:15Z

I typed my thesis in late Winter 1963, on an IBM Executive in the Bowdoin College Math Dept office after hours. This was before the Bouncing Ball, but it had two or three removable type-bars at the side, and we had a couple of dozen special bars, each with its own character. If you wanted to type “$\alpha\beta$”, you’d have to remove the alpha-stick and attach the beta-stick. I think that there were relatively few characters I had to put in by hand: $\mathfrak{p}$ maybe, and certainly the inclusion symbols. It took me 45 minutes or so per page, and according to the rules at Harvard, there could be no corrections on any page (not even white-out).

Higher-dimensional algebraic subgroups of the proalgebraic Nottingham group?

2012-01-28T06:36:03Z

Let $R$ be a commutative ring, and, for $n\ge0$, ${\mathcal{A}}_n={\mathcal{A}}_n(R)$ the group of series $u(x)=\sum_0^\infty a_jx^{j+1}\in R[[x]]$ for which $a_0\in R^\times$ and $u(x)\equiv x\pmod{x^{n+1}}$. The group operation is composition of series. A series in ${\mathcal{A}}_n$ that’s not in ${\mathcal{A}}_{n+1}$ is said to have depth $n$. When $\kappa$ is a field with $p^s$ elements, ${\mathcal{A}}_1(\kappa)$ is often called the Nottingham group, but it seems to have been designated by different names in different times and places.

One sees that ${\mathcal{A}}_n$ is the projective limit of the groups ${\mathcal{A}}_n/{\mathcal{A}}_{n+r}$ , and for $n>0$ these are unipotent algebraic groups of dimension $r$. So it's the proalgebraic group ${\mathcal{A}}_1$ , defined over ${\mathbb{F}}_p$ , that I'm asking about, and the existence of (finite-dimensional) algebraic subgroups in it.

There's one that's obvious, namely the image of ${\mathbf{G}}_{\mathrm a}\to {\mathcal{A}}_1$ by the homomorphism $$ t\mapsto \frac{x}{1-tx} =x+tx^2+t^2x^3+\cdots=u_t(x) $$ fractional-linear transformation leaving the origin fixed, expanded as series.

Almost as obvious is to apply the transformation $$ u(x)=x(1+g(x))\mapsto (u(x^m))^{1/m} =x(1+g(x^m))^{1/m} $$ clearly a group homomorphism, well defined when $m$ is prime to $p$, because you're raising a principal unit to an exponent that's in ${\mathbb{Z}}_p$. This takes $u_t$, a subgroup of Nottingham of depth one to a group of depth $m$. In particular, the Nottingham proalgebraic group has lots of one-dimensional algebraic subgroups.

There are no commutative connected algebraic subgroups of dimension greater than $1$: I've found an argument that depends on higher ramification theory and uses Hasse-Arf, but breaks down completely in any noncommutative case. I would like to settle the question of whether there are any algebraic subgroups of Nottingham of dimension greater than $1$, necessarily noncommutative: it will be enough to dispose of the two-dimensional case. So it occurs to me that this may already be known, or maybe someone out there can suggest an approach to me.

Answer by Lubin for Finite connected groups over a perfect field of characteristic p

2011-12-12T17:29:44Z

Others can do this much better than I, but here's what's happening: to describe a group scheme of any kind, you need to talk about not only the underlying space, but also the law of composition on the group. In this case, the kernel of $[p]$ in the muliplicative group, you describe the law of composition by writing down the the comultiplication on the affine ring $k[X]/(X^p)$. This is simply $X\mapsto 1 \otimes X + X \otimes 1 + X \otimes X$.

Answer by Lubin for squares in dyadic local fields

2011-11-11T02:22:41Z

You expect that when you try to take square root of $1+4u$, you're led to an unramified extension, just as ${\mathbb{Q}}_2(5^{1/2})$ is unramified over $ {\mathbb Q}_2$. Indeed, from $x^2 - (1+4u)$ you are led (by an appropriate change of variables) to $X^2 + X - u$, clearly either irreducible with roots in an unramified extension of your $2$-adic ground field, or reducible, depending on whether the corresponding polynomial in characteristic $2$ doesn't or does have roots in the residue field. That drops out of Hensel's Lemma.

The upshot is that you can ``always get a unit $u$ such that $1+4u$ is not a square'' if and only if the residue field has quadratic extensions. Always, in particular, if the residue field is finite.

Answer by Lubin for When are roots of power series algebraic?

2011-10-23T21:09:47Z

I feel considerable trepidation in muddying the waters that Keith has so nicely clarified, but it seems to me, in view of the discreteness of the valuation on $K$, that first, the hypothesis that there are only finitely many roots is unnecessary, and second, it's unnecessary to go by way of the closed unit disk.

As an example of a series with infinitely many roots, hold in mind the logarithmic series, $\sum_n (-1)^{n+1}x^n/n$, convergent on the open unit disk. Its roots are the numbers $\zeta-1$, $\zeta$ running through the $p$-power roots of unity. For a series to be convergent on the open disk of ${\bf C}_p$, it's necessary and sufficient that the limit slope of the Newton polygon be nonnegative. You can define this as $\lim_n v(a_n)/n$ if you like. In case the series has positive limit slope, then there's a segment with positive slope, and you can use Weierstrass Prep directly to show that there are only finitely many roots of the series in the open disk, and these are roots of an integral monic $K$-polynomial. If, as in the case of the log, your series has limit slope zero, then you need a suitably jazzed-up version of W-Prep saying every vertex $V$ of the polygon gives you a monic polynomial factor, still with $K$-integers for coefficients. Or, since in this case we're only worried about the roots being algebraic, you can just replace $f(x)$ by $f(p^\lambda x)$, where $\lambda$ is a positive rational between the negatives of the slopes on either side of $V$. And then apply the first case, 'cause now the limit slope is $\lambda$. In any case, and by whatever method, the roots of $f$ are all algebraic.

Answer by Lubin for Closed but not rational points of a real cubic

2011-08-23T22:19:37Z

Starting always from the knowledge that a closed point over $\mathbb{R}$ is either an $\mathbb{R}$-rational point or a pair of conjugate $\mathbb{C}$-rational points, let's think of a complex point $P=(z,w)$ together with its conjugate $\overline P$, if need be, and call $z=a+bi$, $w=c+di$. Then your maximal ideal corresponding to $P$ is $(x^2-2ax+a^2+b^2,y^2-2cy+c^2+d^2)$, with special forms in case either $z$ or $w$ is real, for example $(x-a,y^2-2cy+c^2+d^2)$ in case $z$ is real but not $w$. Of course the condition that $w^2=z^3-z$ is expressed by a pair of equations in the real variables $a,b,c,d$, namely $a + c^2 - d^2 - a^3 + 3ab^2=0$ and $b + 2cd - 3a^2b + b^3=0$, which I haven't gotten much help from, even though I've stared at them long and hard hoping to verify your very interesting insight about the bordered region $S=\lbrace (x,y):y^2 \ge x^3-x\rbrace$. If you want a surface in $\mathbb{R}^3$ to look at, you can take the points $(a,c,b^2+d^2)$, subject to the two conditions above. Its intersection with the plane $(*,*,0)$ is just the locus of real-rational points, but its projection onto that plane is not your region $S$. Notice that conjugate points have the same image under this mapping, and nonconjugate points have different images.

The edition of Mumford's Red Book that I'm looking at does not support your guess about $S$, in my opinion, even though as a topological space with border, $S$ is exactly right. Perhaps there's a transcendental argument justifying your insight, using the $\wp$-function for the appropriate lattice.

Answer by Lubin for Algorithm for Weierstrass Preparation Theorem for Formal Power Series

2011-08-18T05:19:45Z

Here's an algorithm that I use. Let's call $S$ the degree-$n$ shift operation, sending $\sum c_kz^k$ to $\sum c_{n+k}z^k$, in other words the quotient when you divide a power series by $z^n$. Step 0: divide $f$ by $Sf$, giving you a power series $f_1$ such that $Sf_1\equiv 1$ modulo $M$. Step $i$, for $i > 0$: repeat. At each stage, you get a power series $f_i$ for which $Sf_i\equiv 1 $ modulo $M^i$. For a quicker variant of Step $i$ (for $i > 0$), instead multiply by $2-Sf_i$. It works because you've constructed a convergent infinite product.

Answer by Lubin for k-th powers in the field of p-adics

2011-07-06T20:03:51Z

There are many different ways of attacking this, and though I have nothing against series, I always prefer a polygon argument, when one exists. First, by dividing by a suitable $n_0$, you can assume that $v(x-1)\ge1$; second, since Silvain has already taken care of the question when $k$ is prime to $p$, you can assume that $k=p^s$ for some $s\ge1$.

Now consider the polynomial $f(t)=(1+t)^p-1$, its Newton polygon, and especially the polygon of $f(t)-\alpha$, for $\alpha\in\mathbb{Z}_p$. If $v(\alpha)\le1$, then there's a single segment of (negative) slope $1/p$ or zero, while if $v(\alpha)\ge2$, then there's a segment of slope $v(\alpha)-1$ and width $1$, and a segment of slope $1/(p-1)$ and width $p-1$. Consequence is that $f$ maps $p\mathbb{Z}_p$ onto $p^2\mathbb{Z}_p$ in such a way that $v(f(z))=v(z)+1$, and more, that $f^{\circ s}$ maps $p\mathbb{Z}_p$ onto $p^ {s+1}\mathbb{Z}_p$. (Argument needs slight modification when $p=2$) So for $x/n$ to be a $p^s$-th root of a $p$-adic integer, it's enough for the condition $v(x-1)\gt s$ to hold, which can be done because the reciprocals of integers are dense in the $p$-adic units.

Answer by Lubin for Integral interpolation by polynomial

2011-05-03T02:58:19Z

It's a topic I liked to cover when I was still teaching junior-level Algebra, even if it didn't fit in well with the other topics. You start with a function defined on the set of integers from $0$ to $n$ inclusive, and end with a polynomial of degree $\le n$ that agrees at those $n+1$ points, and if the values you started with are integers, your function always sends integers to integers. You take successive differences, as indicated by Robert Israel above, and then you list $f(0)$, $\Delta f(0)$, $\Delta^2f(0)$, up to $\Delta^n$, and use these as coefficients, which you multiply to $C_0(x)=1$, $C_1(x)=x$, $C_2(x)=x(x-1)/2$, etc., the binomial polynomials. Your assignment is to try it out for a few examples, and then prove that the method works.

Answer by Lubin for About Frobenius of Witt vectors

2011-04-23T05:41:13Z

You may find the following more transparent, since it uses only the fact that the Witt vectors are a complete DVR with residue field $k$. Call the Witt vectors $R$, and let $y$ be a unit for which you want to find $z$ with $z^\sigma=yz$. First do it mod $p$, by solving $\zeta^p=\eta\zeta$ for $\zeta$ in $k$, where $\eta$ is the image of $y$ in $k$. Now you can assume that you have $z\in R$ satisfying $z^\sigma\equiv yz \mod{(p^m)}$, in other words $z^\sigma \equiv yz + p^m\delta \mod{(p^{m+1})}$. Now you want to adjust $z$ to $z'=z+p^m x$ so that $z'$ satisfies your congruence modulo $(p^{m+1})$. This boils down to solving $\xi^p - \xi \eta + \delta = 0$ in $k$, which you can do. So you see that you don't need $k$ to be algebraically closed, just separably closed.

Answer by Lubin for normal domains with algebraically closed quotient field

2011-04-09T21:21:03Z

Try this: Let $B_0$ be the ring of real algebraic integers, and let $B=B_0[1/2]$, so the ring of real algebraic numbers integral except possibly at $2$. But $B[i]$ is equal to the ring of algebraic numbers integral except possibly at $2$, and this is integrally closed. And so we take $A=B[3i]$, not integrally closed, and of course the fraction field is algebraically closed.

Answer by Lubin for When should a supervisor be a co-author?

2011-03-04T20:31:27Z

I guess my viewpoint is rather parochial, but when I saw the question, I wondered where OP got such ideas of politeness. Reading the previous answers, though, I realized that different practices may be current, depending on the national culture, or even the culture of the university in question. Speaking strictly for myself, I would have considered it quite improper to have my name on a student's thesis, in effect robbing the student of the credit that was deserved.

Answer by Lubin for What is the different in the cyclotomic tower over a finite ramified extension of Qp?

2011-02-04T16:41:06Z

Here's a relatively easy counterexample. Take $p=2$, $L=\mathbb{Q}_2(2^{1/3})$. I did some direct computation and saw that $v_2(\mathfrak{D}^{L_2}_L)=2/3$, $v_2(\mathfrak{D}^{L_3}_{L_2})=5/6$. Looks like there's a pattern. But there's a better way.

We're in a situation where not only $L/\mathbb{Q}_2$ is tamely ramified of degree $3$, but also every $L_n/K_n$. Now we use the functoriality of the Hasse-Herbrand transition function: if $k\subset F\subset K$, then $\varphi^K_k=\varphi^F_k\circ \varphi^K_F$. Use the relation $\varphi^{L_n}_{\mathbb{Q}_2}=\varphi^{K_n}_{\mathbb{Q}_2}\circ \varphi^{L_n}_{K_n}= \varphi^L_{\mathbb{Q}_2}\circ \varphi^{L_n}_ L$ and the fact that a tamely ramified extension has all the transition-function action at the origin. That is, the function is $y=x$ for $x\le 0$, but $y=x/e$ for $x\ge 0$, where $e$ is the ramification index. So as real functions, $\varphi^L_{\mathbb{Q}_2}=\varphi^{L_n}_{K_n}$, namely this is just $y=x/3$. Consequently, the transition function of $L_n/L$ is gotten by conjugating that of $K_n/{\mathbb{Q}_2}$ with the tame transition function. The effect is to multiply all coordinates of vertex points by $3$.

But we also know the transition function of $K_n$ over ${\mathbb{Q}_2}$: its vertices are at all $(2^{i-1}-1,i-1)$ for $2\le i\le n$.The new vertices are at $(3,3)$, $(9,6)$, $(21,9)$, etc. This means that the lower breaks of $L_n/L$ are at $3(2^{i-1}-1)$ for $2\le i\le n$, and in particular the unique break of $L_n/L_{n-1}$ is at $3(2^{n-1}-1)$.

Now use the formula \begin{align}{ v_F(\mathfrak{D}^F_k)=\sum_{j\ge 0}\bigl(|G_j|-1\bigr) }\end{align} where the $G$'s are the lower ramification groups, and where in this case all the numbers being added up are $1$ or $0$, to see that $v_{L_n}\bigl(\mathfrak{D}^{L_n} _ {L_{n-1}}\bigr)=3(2^{n-1}-1)+1=3\cdot2^{n-1}-2$. Divide by the ramification number of $L_n$ over $\mathbb{Q}_2$ to get $1-1/(3\cdot 2^{n-2})$, agreeing with my computations for $n=2$ and $n=3$.

It's not an issue of tame versus wild ramification in the extension $L/\mathbb{Q}_2$, either. I used $L=\mathbb{Q}_2(2^{1/4})$ to find that the numbers are $1-3/2^m$; the argument is similar but a bit more delicate, since you have no a priori idea of what the transition function of $L_n/K_n$ might be.

Answer by Lubin for How to get explicit unramified covers of an elliptic curve?

2011-01-15T01:15:24Z

In my perhaps skewed experience, it's much easier, from a computational standpoint, to get isogenies out of an elliptic curve than into one. So I'd say that you're really looking into the more uncomfortable end of the telescope.

Again, computationally, once you identify a finite subgroup $S$ of your elliptic curve, it's not too hard to get formulas for the isogeny with $S$ as kernel: use translation (implemented by chord-and-tangent) by the elements of $S$ to define the elements of a Galois group, and find its fixed field inside $\mathbb{C}(x,y)$.

But the curve you gave is same as $y^2=x^3-x$, and clearly has complex multiplication by $i$: you have $ \lbrack i\rbrack (\xi,\eta) = (-\xi, i\eta)$. This permits you to write down an example of a cyclic $n$-isogeny any time that $n$ is the sum of two squares in $\mathbb{Z}$, so let's do it for $n=2$. Use the endomorphism $1+i$, which has degree two, so that your curve covers itself in a cyclic $2$-cover. You add $(\xi,\eta)$ to $(-\xi, i\eta)$ on your curve and the result is $(-i/2)(\xi^2-1)/\xi$ for the $x$-coordinate, and $-((1+i)/4)\eta(1+1/\xi^2)$ for the $y$-coordinate. (This point really is on the curve.) The two preceding quantities generate a subfield over which $\mathbb{C}(x,y)$ is quadratic, the extension is unramified as desired.

Answer by Lubin for Computing (on a computer) higher ramification groups and/or conductors of representations.

2010-12-16T22:40:54Z

It's rather late in the day, but there's an easy way of getting the whole Hasse-Herbrand function $\varphi^K_{\mathbb{Q}_p}(x)$ if you know the minimal polynomial $F$ of a prime element $\pi$. First, you write down the copolygon (valuation function) of $F(X+\pi)$ using the valuation normalized to have $v(p)=1$, then you stretch it horizontally by a factor of $[K\colon\mathbb{Q}_p]$, then you move it down and to the left by one unit, to get the numberings consistent with Serre's convention. The vertices in KB's case are $(1,1)$, $(3,2)$ and $(5,5/2)$. I couldn't figure out the prime of the Galois closure till I saw the extension as quadratic over $\mathbb{Q}_2(\zeta_8)$. At any rate, the chain of fields corresponding to the ramification filtration is $\mathbb{Q}_2\subset\mathbb{Q}_2(i)\subset\mathbb{Q}_2(\zeta_8) \subset K$. Needless to say, you don't need any kind of powerful package to do this kind of computation.

Comment by Lubin

2013-03-09T16:17:23Z

Even though a formal group law is a power series in two variables, it’s not clear to me that this is the best or most informative way to think about such a thing. Up in characteristic zero, one gets much more useful information out of the logarithm than out of the two-variable series.

Comment by Lubin

2013-03-09T16:12:19Z

Neat pictures!

Comment by Lubin

2013-01-24T15:07:01Z

Overreliance on word-origins can lead one astray. My little dictionary of classical Greek gives “income; profit; gain; gratification” as the meaning of $\lambda\widetilde\eta\mu\mu\alpha$.

Comment by Lubin

2013-01-12T21:23:14Z

One might point out that the above mentioned author is not “Oliver” but Olver.

Comment by Lubin

2013-01-05T14:35:42Z

Welcome, Mike!!

Comment by Lubin

2012-11-29T16:37:47Z

The old-fashioned way I see this, and Felipe can slap me down if I’m wrong, is that the transcendence-degree-one field of functions on $A$ has only one purely inseparable subfield of each possible degree $p^r$.

Comment by Lubin

2012-10-30T02:51:45Z

A bit late to add to the above comment, but just what “Hensel’s Lemma” refers to, is in a state of confusion. For me, H’s L relates not at all to finding the root of a polynomial but rather to lifting a characteristic-$p$ factorization back to characteristic zero (or the appropriate generalization in the equal-characteristic case).

Comment by Lubin

2012-10-20T13:54:58Z

Ordinarily, "\lbrace" and "\rbrace" should work.

Comment by Lubin

2012-08-06T00:56:26Z

Very large degree polynomials $f$ and $g$? I have something home-made that's primitive in comparison to Sage, but designed for just this sort of thing, yet may choke at huge degree. If you don't get better suggestions, contact me by e-mail.

Comment by Lubin

2012-07-26T05:15:11Z

Interesting that the French use the word in several ways, but English has only the one meaning, the geological.

Comment by Lubin

2012-07-06T16:45:33Z

Or: “you can trisect a 90-degree angle, but not a 30-degree angle.”

Comment by Lubin

2012-05-24T20:11:53Z

I’ve never had occasion to do so in a publication, but what I use in my own notes is $A^{(p^{-1})}$.

Comment by Lubin

2012-05-19T04:54:59Z

I notice that the Wikipedia page for Todd is decorated with a photograph of Jack Todd, the widower of Olga Taussky-Todd, who died in 2007.

Comment by Lubin

2012-05-19T04:32:36Z

Sorry, have I misread or seriously misunderstood the question? I don't see that there is such an exact sequence of ${\mathrm{Gal}}(\overline K\colon K)$-modules, because the kernel of reduction becomes isomorphic to the multiplicative (formal) group only after extension of the base to the complete maximal unramified extension of $K$.

Comment by Lubin

2012-05-17T02:27:40Z

Although I always prefer Newton poly., factorization follows also from Hensel’s Lemma, with the one polynomial in char. $p$ being $1+x$ and the other being $1$. HL guarantees a factor of degree $1$.