User pierre-yves gaillard - MathOverflow

What is your favorite proof of Tychonoff's Theorem?

2010-05-30T04:44:19Z

Here is mine. It's taken from page 11 of "An Introduction To Abstract Harmonic Analysis", 1953, by Loomis:

http://www.archive.org/details/introductiontoab031610mbp

http://ia331316.us.archive.org/3/items/introductiontoab031610mbp/introductiontoab031610mbp.pdf

(By the way, I don't know why this book is not more famous.)

To prove that a product $K=\prod K_i$ of compact spaces $K_i$ is compact, let $\mathcal A$ be a set of closed subsets of $K$ having the finite intersection property (FIP) --- viz. the intersection of finitely many members of $\mathcal A$ is nonempty ---, and show $\bigcap\mathcal A\not=\varnothing$ as follows.

By Zorn's Theorem, $\mathcal A$ is contained into some maximal set $\mathcal B$ of (not necessarily closed) subsets of $K$ having the FIP.

The $\pi_i(B)$, $B\in\mathcal B$, having the FIP and $K_i$ being compact, there is, for each $i$, a point $b_i$ belonging to the closure of $\pi_i(B)$ for all $B$ in $\mathcal B$, where $\pi_i$ is the $i$-th canonical projection. It suffices to check that $\mathcal B$ contains the neighborhoods of $b:=(b_i)$. Indeed, this will imply that the neighborhoods of $b$ intersect all $B$ in $\mathcal B$, hence that $b$ is in the closure of $B$ for all $B$ in $\mathcal B$, and thus in $A$ for all $A$ in $\mathcal A$.

For each $i$ pick a neighborhood $N_i$ of $b_i$ in such a way that $N_i=K_i$ for almost all $i$. In particular the product $N$ of the $N_i$ is a neighborhood of $b$, and it is enough to verify that $N$ is in $\mathcal B$. As $N$ is the intersection of finitely many $\pi_i^{-1}(N_i)$, it even suffices, by maximality of $\mathcal B$, to prove that $\pi_i^{-1}(N_i)$ is in $\mathcal B$.

We have $N_i\cap\pi_i(B)\not=\varnothing$ for all $B$ in $\mathcal B$ (because $b_i$ is in the closure of $\pi_i(B)$), hence $\pi_i^{-1}(N_i)\cap B\not=\varnothing$ for all $B$ in $\mathcal B$, and thus $\pi_i^{-1}(N_i)\in\mathcal B$ (by maximality of $\mathcal B$).

A pdf version is available at http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Tycho/ .

Many people credit the general statement of Tychonoff's Theorem to Cech. But, as pointed out below by KP Hart, Tychonoff's Theorem seems to be entirely due to ... Tychonoff. This observation was already made on page 636 of

Chandler, Richard E.; Faulkner, Gary D. Hausdorff compactifications: a retrospective. Handbook of the history of general topology, Vol. 2 (San Antonio, TX, 1993), 631--667, Hist. Topol., 2, Kluwer Acad. Publ., Dordrecht, 1998

http://books.google.com/books?id=O2Hwaj2SqigC&lpg=PA636&ots=xjvA9nwlO5&dq=772%20tychonoff&pg=PA636#v=onepage&q&f=false

The statement is made by Tychonoff on p. 272 of "Ein Fixpunktsatz"

http://www.springerlink.com/content/n61706447r886l58/?p=328f0106a2634abfb53531fca0ca5a90&pi=0

where he says that the proof is the same as the one he gave for a product of intervals in "Über die topologische Erweiterung von Räumen"

http://www.springerlink.com/content/l656352441w67612/?p=328f0106a2634abfb53531fca0ca5a90&pi=1

Generators of a certain ideal

2012-02-03T13:06:21Z

In view of Mariano Suárez-Alvarez's answer I see how badly phrased my question was, and decided to rewrite it. The drawback is that some comments of Martin Brandenburg are now incomprehensible, but I thought it would suffice to say here that Martin made some legitimate constructive criticisms to the original wording of the question. By the way, thank you also to Vladimir Dotsenko for his comments.

Let $K$ be a commutative ring, and let $X_1,\dots,X_n$ be indeterminates. Here $n$ is an integer $\ge3$. For $1\le i < j\le n$ put $$ x_{ij}:=\frac{1}{X_i-X_j} $$ and let $Y_{ij}$ be an indeterminate. Let $I$ be the kernel of the $K$-algebra morphism $$ \varepsilon:K[(Y_{ij})]\to K[(x_{ij})],\quad Y_{ij}\mapsto x_{ij}. $$

Is $I$ finitely generated? If it is, can one give an explicit finite set of generators?

Note that the identity $$ \frac{1}{a-b}\ \frac{1}{a-c}+\frac{1}{b-a}\ \frac{1}{b-c}+\frac{1}{c-a}\ \frac{1}{c-b}=0. $$ shows that $I$ is nonzero.

(I put the homological algebra tag because the ultimate goal is to know whether there is a functorial free resolution of $K[(x_{ij})]$, viewed as a $K[(Y_{ij})]$-module, and, if it exists, what can be said about it.)

The question had been posted before on Mathematics Stack Exchange (link).

What if Current Foundations of Mathematics are Inconsistent?

2010-10-03T10:17:11Z

The title of the question is also the title of a talk by Vladimir Voevodsky, available here.

Had this kind of opinion been expressed before?

EDIT. Thanks to all answerers, commentators, voters, and viewers! --- Here are three more links:

Question arising from Voevodsky's talk on inconsistency by John Stillwell,

Nelson's program to show inconsistency of ZF, by Andreas Thom,

Pierre Colmez, La logique c’est pas logique !

EDIT. Here the link to the FOM list discussing these themes.

Answer by Pierre-Yves Gaillard for Galois theory timeline

2010-06-02T13:32:26Z

EDIT. Here is the part of the answer that has been rewritten:

We give below a short proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below).

The derivation of the FTGT from (a) and (b) takes about four lines, but I haven't been able to find these four lines in the literature, and all the proofs of the FTGT I have seen so far are much more complicated. So, if you find either a mistake in these four lines, or a trace of them the literature, please let me know.

The argument is essentially taken from Chapter II (link) of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin's proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin's argument) was written with Keith Conrad's help.

Theorem. Let $E/F$ be an extension of fields, let $a_1,\dots,a_n$ be distinct generators of $E/F$ such that the product of the $X-a_i$ is in $F[X]$. Then

the group $G$ of automorphisms of $E/F$ is finite,
there is a bijective correspondence between the sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have $$ S\leftrightarrow H\iff H=\text{Aut}(E/S)\iff S=E^H $$ $$ \implies[E:S]=|H|, $$ where $E^H$ is the fixed subfield of $H$, where $[E:S]$ is the degree (that is the dimension) of $E$ over $S$, and where $|H|$ is the order of $H$.

PROOF

We claim:

(a) If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|\text{Aut}(E/S)|$.

(b) If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.

Proof that (a) and (b) imply the theorem. Let $S/F$ be a sub-extension of $E/F$ and put $H:=\text{Aut}(E/S)$. Then we have trivially $S\subset E^H$, and (a) and (b) imply $$ [E:S]=[E:E^H]. $$ Conversely let $H$ be a subgroup of $G$ and set $\overline H:=\text{Aut}(E/E^H)$. Then we have trivially $H\subset\overline H$, and (a) and (b) imply $|H|=|\overline H|$.

Proof of (a). Let $1\le i\le n$. Put $K:=S(a_1,\dots,a_{i-1})$ and $L:=K(a_i)$. It suffices to check that any $F$-embedding $\phi$ of $K$ in $E$ has exactly $[L:K]$ extensions to an $F$-embedding $\Phi$ of $L$ in $E$; or, equivalently, that the polynomial $p\in\phi(K)[X]$ which is the image under $\phi$ of the minimal polynomial of $a_i$ over $K$ has $[L:K]$ distinct roots in $E$. But this is clear since $p$ divides the product of the $X-a_j$.

Proof of (b). In view of (a) it is enough to check $|H|\ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a $$ b=(b_1,\dots,b_k)\in E^k. $$ We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^\perp\cap(E^H)^k$ is nonzero, where $\bullet^\perp$ denotes the vectors orthogonal to $\bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $b^\perp\cap (E^H)^k$ is necessarily orthogonal to $hb$ for any $h\in H$, so $$ b^\perp\cap(E^H)^k=(Hb)^\perp\cap(E^H)^k, $$ where $Hb$ is the $H$-orbit of $b$. We will show $(Hb)^\perp\cap(E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^\perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^\perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^\perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j=1$ for some $j$. Since the subspace $(Hb)^\perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $hx\in(Hb)^\perp$, so $hx-x\in(Hb)^\perp$. Since $x_j=1$, the $j$-th coordinate of $hx-x$ is $0$, so $hx-x=0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.

[A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II, available here.

PDF version: http://www.iecn.u-nancy.fr/~gaillapy/DIVERS/Fundamental.theorem.of.galois.theory/

Here is the part of the answer that has not been rewritten:

Although I'm very interested in the history of Galois Theory, I know almost nothing about it. Here are a few things I believe. Thank you for correcting me if I'm wrong. My main source is

http://www-history.mcs.st-and.ac.uk/history/Projects/Brunk/Chapters/Ch3.html

Artin was the first mathematician to formulate Galois Theory in terms of a lattice anti-isomorphism.

The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.

The first publications of this formulation by Artin himself were "Foundations of Galois Theory" (1938) and "Galois Theory" (1942).

Artin himself doesn't seem to have ever explicitly claimed this discovery.

Assuming all this is true, my (probably naive) question is:

Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?

I also hope this is not completely unrelated to the question.

Answer by Pierre-Yves Gaillard for "Sums-compact" objects = f.g. objects in categories of modules?

2011-11-19T11:32:40Z

It seems to me the references in this Mathematics - Stack Exchange answer contain the requested information.

EDIT 1. Here is an excerpt from Hyman Bass's book Algebraic K-Theory, W. A. Benjamin (1968), p. 54:

Exercise.

(a) Show that a module $P$ is finitely generated if and only if the union of a totally ordered family of proper submodules of $P$ is a proper submodule.

(b) Show that $\text{Hom}_A(P,\bullet)$ preserves coproducts if and only if the union of every (countable) chain of proper submodules is a proper submodule.

(c) Show that the conditions in (a) and (b) are not equivalent. (Examples are not easy to find.)

EDIT 2. Here is a solution to Exercise (a) above. Let $R$ be an associative ring with $1$, and $A$ an $R$-module. If $A$ is finitely generated, then the union of a totally ordered set of proper submodules is clearly a proper submodule. Let's prove the converse:

Assume that $A$ is not finitely generated. Let $Z$ be the set of those submodules $B$ of $A$ such that $A/B$ is is not finitely generated. The poset $Z$ is nonempty and has no maximal element. By Zorn's Lemma, there is a nonempty totally ordered subset $T$ of $Z$ which has no upper bound. Letting $U$ be the union of $T$, we see that $A/U$ is finitely generated. There is thus a finitely generated submodule $F$ of $A$ which generates $A$ modulo $U$. Then the $B+F$, where $B$ runs over $T$, form a totally ordered set of proper submodules whose union is $A$. QED

I'd be most grateful to whoever would post a solution to the other exercises in Bass's list. (I haven't been able to do them.) The following references might help, but I haven't been able to find them online:

R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933. [Update: see Edit 3 below.]
P.C. Eklof, K.R. Goodearl and J. Trlifaj, Dually slender modules and steady rings, Forum Math. 9 (1997), 61-74.

This paper is available online, but I don't understand it:

Jan Zemlicka, Classes of dually slender modules, Proc. Algebra Symposium Cluj 2005, 129-137.

EDIT 3.

$\bullet$ Rentschler's paper

R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933

is available here in one click, and there in a few clicks. [I'm also giving the second option because it's a trick worth knowing.] Thanks to Stéphanie Jourdan for having found this link!

$\bullet$ Exercise (b) in Bass's list is in fact the easiest. [Sorry for not having realized that earlier.] Here is a solution. --- Let $R$ be an associative ring with $1$, let $A$ be an $R$-module, and let "map" mean "$R$-linear map".

If $A_0\subset A_1\subset\cdots$ is a sequence of proper submodules of $A$ whose union is $A$, then the natural map from $A$ to the direct product of the $A/A_n$ induces a map from $A$ to the direct sum of the $A/A_n$ whose components are all nonzero.

Conversely, let $f$ be a map from $A$ to a direct sum $\oplus_{i\in I}B_i$ of $R$-modules such that the set $S$ of those $i$ in $I$ satisfying $f_i\neq0$ [obvious notation] is is infinite. By choosing a countable subset of $S$ we get a map $g$ from $A$ to a direct sum $\oplus_{n\in \mathbb N}C_n$ of $R$-modules such that $g_n\neq0$ for all $n$. It is easy to check that the $$ A_n:=\bigcap_{k > n}\ \ker(g_k), $$ form an increasing sequence of proper submodules of $A$ whose union is $A$.

EDIT 4. [Version of Nov. 26, 2011, UTC.] The following result is implicit in Rentschler's paper, and solves Bass's Exercise (c):

Theorem. Let $T$ be a nonempty ordered set $ ( * ) $ with no maximum. Then there is a domain $A$ which has the following property. If $P$ denotes the poset of proper sub-$A$-modules of the field of fractions of $A$, then there is an increasing $ ( * ) $ map $f:T\to P$ such that $f(T)$ is cofinal in $P$.

$ ( * ) $ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular:

ordered set = ensemble totalement ordonné,

poset = ensemble ordonné,

increasing = strictement croisssant.

Proof. Let $T_0$ be the ordered set opposite to $T$, let $\mathbb Z^{(T_0)}$ be the free $\mathbb Z$-module over $T_0$ equipped with the lexicographic order. Then $\mathbb Z^{(T_0)}$ is an abelian ordered group (groupe abélien totalement ordonné). By Example 6 in Section V.3.4 of Bourbaki's Algèbre commutative, there is a field $K$ and a surjective valuation $$ v:K\to\mathbb Z^{(T_0)}\cup { \infty }. $$ Say that a subset $F$ of $\mathbb Z^{(T_0)}$ is a final segment if $$F\ni x < y\in\mathbb Z^{(T_0)} $$ implies $y\in F$. Attach to each such $F$ the subset $$ S(F):=v^{-1}(F)\cup { 0 } $$ of $K$. Then $A:=S(F_0)$, where $F_0$ is the set of nonnegative elements of $\mathbb Z^{(T_0)}$, is a subring of $K$. Moreover, by Proposition 7 in Section V.3.5 of the book quoted above, $F\mapsto S(F)$ is an increasing bijection from the final segments of $\mathbb Z^{(T_0)}$ to the sub-$A$-modules of $K$.

Write $e_{t_0}$ for the basis element of $\mathbb Z^{(T_0)}$ corresponding to $t_0\in T_0$. Then the intervals $$ I_{t_0}:=[-e_{t_0},\infty) $$ are cofinal in the set of all proper final segments of $\mathbb Z^{(T_0)}$, and we have $I_{t_0}\subset I_{u_0}$ if and only if $t\le u$. [We denote an element $t$ of $T$ by $t_0$ when we view it as an element of $T_0$.]

Answer by Pierre-Yves Gaillard for When Animals Attack

2011-11-22T08:04:49Z

(1) Let $A$ be a commutative ring, and let $M$ and $N$ be $A$-modules. If the natural morphism from $A$ to $\text{End}_A(M\oplus N)$ is surjective, then the annihilators of $M$ and $N$ are comaximal.

Indeed, this comaximality is the condition for the projectors attached the given direct sum decomposition to be in the image.

Assume now that $A$ is a principal ideal domain. Let $(G,+)$ be the Grothendieck group of the category $C$ of finitely generated torsion $A$-modules, and let $(H,\cdot)$ be the group of (nonzero) fractional ideals of $A$.

There is a (clearly unique) morphism from $G$ to $H$ which maps $A/\mathfrak a$ to $\mathfrak a$.

It is easy to see that the fractional ideal attached to $M\in C$ is in fact integral. Call it the characteristic ideal of $M$. Moreover we have in view of (1):

(2) The natural morphism from $A$ to $\text{End}_A(M)$ is surjective, if and only if the characteristic ideal of $M$ coincides with the annihilator of $M$.

Assume now that $A=K[X]$, where $K$ is a field and $X$ an indeterminate, and that $V$ is a finite dimensional $K$-vector space equipped with an endomorphism $a$. Then the characteristic ideal of $V$ is generated by the characteristic polynomial of $a$, and (1) and (2) imply:

The characteristic polynomial of $a$ coincides with its minimal polynomial if and only if any endomorphism of $V$ commuting with $a$ is in $K[a]$.

Preservation of direct sums and finite generation

2011-11-19T06:21:03Z

I asked this question on Mathematics - Stack Exchange (MSE). Having figured out out how to handle the problem in an extremely particular case, I also posted it as an answer (in the technical sense of the term) to my own question. Getting no other answer, I thought I could post the question on MathOverflow. For the reader's convenience, here is a copy and paste of the question.

This is a follow up on this MSE question, asked by Evariste.

Let $R$ be an associative ring with one. The word "module" shall mean left $R$-module. Say that a module $A$ preserves direct sums if the functor $\hom_R(A,?)$ does.

The main question is

Does the condition that $A$ preserves direct sums imply that $A$ is finitely generated?

The converse is clear: see this MSE answer.

As observed by Mariano Suárez-Alvarez in a comment to this MSE answer, if $A$ can be written as the union of an increasing sequence $(A_n)_{n\in\mathbb N}$ of submodules, then $A$ does not preserve direct sums. [The argument is described in the answer.]

Say that $A$ is countably cofinal if it can be written as such a union. If $A$ is neither finitely generated nor countably cofinal, say that $A$ is uncountably cofinal.

[Here is the motivation for this terminology. A group which can be written as the union of an increasing sequence of subgroups is called countably cofinal, and a group which is neither finitely generated nor countably cofinal, is called uncountably cofinal. Uncountably cofinal groups have been studied by Serre, Tits, MacPherson, Bergman, and many others: see this Google Search. In particular, uncountably cofinal groups do exist.]

The second question is:

Do uncountably cofinal modules exist?

Answer by Pierre-Yves Gaillard for Errata for Atiyah-Macdonald

2010-10-24T08:27:49Z

EDIT OF JUNE 9, 2011

Page 102, penultimate paragraph:

"... $f$ induces a homomorphism $\widehat{f}:\widehat{G}\to\widehat{H}$, which is continuous."

No topology has been defined on $\widehat{G}$ and $\widehat{H}$.

[July 7, 2011, GMT. The topology on $\widehat{G}$ can be described as follows. For any subset $S$ of $G$, let $\widehat{S}\subset\widehat{G}$ be the set of equivalence classes of Cauchy sequences in $S$, and say that a subset $V$ of $\widehat{G}$ is a neighborhood of $0$ if there is a neighborhood $W$ of $0$ in $G$ such that $\widehat{W}\subset V$.]

By the way, there is (I think) a somewhat similar "mistake" in the article Atiyah wrote with Wall in "Algebraic Number Theory" Ed. Cassels and Froehlich (see http://mathoverflow.net/questions/11437/erratum-for-cassels-froehlich). Atiyah and Wall forgot to mention the crucial compatibility between change of groups and connecting morphisms. (See p. 99.)

END OF EDIT OF JUNE 9, 2011

Page 25, first line of the proof of (2.13): change (2.11) to (2.12).

Page 29, about two third of the page: change (2.14) to (2.13).

EDIT. Page 39, last line: change $m$ to $m_i$ (three times).

EDIT OF NOV. 22, 2010. Page 63, proof of Lemma 5.14. The current text reads

Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=\sum a_i\,x_i$ for some $n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_n]\ \dots$

It would be better (I think) to write something like

Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=a_1\,x_1+\cdots+a_m\,x_m$ for some $m,n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_m]\ \dots$

[July 8, 2011, GMT. Page 90. It seems to me that the second part of the proof of Theorem 8.7 can be simplified. We must check the uniqueness of the decomposition of an Artin ring $A$ as a finite product of Artin local rings $A_i$. To do this it suffices to observe that, for each minimal primary ideal $\mathfrak q$ of $A$, there is a unique $i$ such that $\mathfrak q$ is the kernel of the canonical projection onto $A_i$.]

[July 7, 2011, GMT. Page 107, lines 4-5. Instead of $A^*=A[x_1,\dots,x_r]$ read $A^*=A[y_1,\dots,y_r]$ where $y_i=(0,x_i,0,\dots)$.]

[July 7, 2011, GMT. Page 112, proof of Proposition (10.24). Instead of $\mathfrak{a}^{k+n(i)}$ read $\mathfrak{a}^{\max(0,k-n(i))}$.]

[July 9, 2011, GMT. Page 122, proof of Proposition 11.20. There is a minor typo in the third line of the proof: read $\mathfrak{q}^2$ instead of $\mathfrak{q}$. Another problem is that the notation $d(A)$ is used with two different meanings: the one given on p. 117 for graded modules, and the one given on p. 119 for noetherian local rings. I'll use the notation $D(A)$ for the first meaning. The definition of $D(M)$ for a graded module $M$ makes sense only if the Poincaré series $P(M,t)$ is nonzero. In particular Proposition 11.3 makes sense and holds (with the proof given in the book) only if

$P(M,t)\not=0\not=P(M/xM,t)$,
$D(M)\ge1$,
$D(M/xM)\ge0$ if $D(M)=1$,

and we must take this into account when using Proposition 11.3 to prove Proposition 11.20. The simplest way to do that is (in my opinion) to treat separately the case $s=0$. Indeed, when $s\ge1$, Proposition 11.3 (as amended above) applies.]

Answer by Pierre-Yves Gaillard for Can a vector space over an infinite field be a finite union of proper subspaces?

2011-07-01T03:44:57Z

Anton Geraschenko's comment prompted me to write a new version of this short answer. I'm leaving the old version to make Anton's comment clearer (and also to increase the probability of having at least one correct answer).

NEW VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each (affine) line. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.

Indeed, as pointed out by Anton, the $K$-valued functions on $A$ which are polynomial on each line form obviously a ring $R$. This ring is a domain, because if $f$ and $g$ are nonzero elements of $R$, then there is a line on which none of them is zero, and their product is nonzero on this line.

OLD VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each finite dimensional affine subspace. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.

Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.

Answer by Pierre-Yves Gaillard for Expressing adj(A) as a polynomial in A?

2010-07-20T09:35:01Z

EDIT OF AUG. 31, 2010. The proof of the Cayley-Hamilton Theorem I like best (among the ones I know) is on page 21 (proof of Proposition 2.4) of Introduction to Commutative Algebra by Atiyah and MacDonald. The argument can be phrased as follows.

Let $K$ be a commutative ring; let $n$ be a positive integer; let $A=(a_{ij})\in M_n(K)$ be an $n$ by $n$ matrix with entries in $K$; let $\chi$ be its characteristic polynomial; define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}\,A-a_{ij}$; let $(e_i)$ be the canonical basis of $K^n$; observe $$\sum_i\ \ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Applying (a trivial case of) Fubini's Theorem to the double sum $\sum_{i,j}\ c_{jk}\ b_{ij}\ e_i$, we get $\chi(A)=0$.

Thank you very much to darij grinberg! [I'm leaving the previous edits "for the record".] END OF EDIT OF AUG. 31, 2010.

EDIT OF DEC. 11, 2010. For a nice application of the Cayley-Hamilton Theorem, see this answer by Balazs Strenner.

PREVIOUS EDITS:

Here is a proof of the Cayley-Hamilton Theorem.

Let $K$ be a commutative ring, let $n$ be a positive integer, let $X$ be an indeterminate, let $A\in M_n(K)$ be an $n$ by $n$ matrix with coefficients in $K$, and let $\chi:=\det(X-A)$ be the characteristic polynomial. Equip $K^n$ with the $K[X]$-module structure induced by $A$. We must check $\chi K^n=0$. Form the right $M_n(K[X])$-module $$H:=\mathrm{Hom}_{K[X]}(K[X]^n,K^n).$$ Let $e\in H$ be the evaluation at $A$ (note $K[X]^n=K^n[X]$). As $e$ is surjective, it suffices to show $e\chi=0$. As $X-A$ divides $\chi$ on the left, it suffices to show $e(X-A)=0$. But this is obvious.

EDIT OF AUG. 1, 2010. Here is a diagrammatic rewriting of the argument.

EDIT OF AUG. 30, 2010. Here is a coordinate version of the above argument. [Compare with the proof of Propositon 3 page 81 of Weil's Basic Number Theory, and with the proof of Propositon 2.4 page 21 of Introduction to Commutative Algebra by Atiyah and MacDonald].

Weil's formulation. Put $$B(X)=(b_{ij}(X)):=X-A\in M_n(K[X]),$$ and let $C(X)=(c_{ij}(X))$ be the adjugate of $B(X)$. We have $$\sum_j\ c_{jk}(X)\ b_{ij}(X)=\delta_{ik}\ \chi(X)\in K[X].$$ Replacing $X$ with $A$, evaluating on $e_i$ (the $i$-th vector of the canonical basis of $K^n$), and summing over $i$ gives $$\sum_j\ c_{jk}(A)\ \sum_i\ b_{ij}(A)\ e_i=\chi(A)\ e_k\in K^n.$$ But the second sum is 0 by definition of $b_{ij}(X)$.

Atiyah-MacDonald's formulation. Put $A=(a_{ij})$ and define $B=(b_{ij})\in M_n(K[A])$ by $b_{ij}:=\delta_{ij}A-a_{ij}$; observe $$\sum_i\ b_{ij}\ e_i=0,\quad\det B=\chi(A);$$ and write $(c_{ij})$ for the adjugate of $B$. Computing $\sum_{i,j}\,c_{jk}\,b_{ij}\,e_i$ in two ways we get $\chi(A)=0$.

Answer by Pierre-Yves Gaillard for Closedness of finite-dimensional subspaces

2010-09-08T04:21:06Z

This holds indeed for complete fields: see Theorem 2, Section I.2.3, of Bourbaki's "Espaces Vectoriels Topologiques".

Here is the argument.

Let $K$ be a (not necessarily commutative) field equipped with a complete nontrivial absolute value $x\mapsto|x|$, let $n$ be a positive integer, let $\tau$ be a Hausdorff vector space topology on $K^n$, and let $\pi$ be the product topology on $K^n$.

THEOREM $\tau=\pi$.

REMINDER A topological group $G$ is Hausdorff iff {1} is closed. [Proof: {1} is closed $\Rightarrow$ the diagonal of $G\times G$ is closed (because it's the inverse image of {1} under $(x,y)\mapsto xy^{-1}$) $\Rightarrow$ $G$ is Hausdorff.]

LEMMA The Theorem holds for $n=1$.

The Lemma implies the Theorem. We argue by induction on $n$. The continuity of the identity from $K^n_\pi$ to $K^n_\tau$ (obvious notation) is clear (and doesn't use the Lemma). To prove the continuity of the identity from $K^n_\tau$ to $K^n_\pi$, it suffices to prove the continuity of an arbitrary nonzero linear form $f$ from $K^n_\tau$ to $K_\pi$. By induction hypothesis, the kernel of $f$ is closed, and the Theorem follows from the Reminder and the Lemma.

Proof of the Lemma. We'll use several times the fact that $K^\times$ contains elements of arbitrary large and arbitrary small absolute value. As already observed, we have $\tau\subset\pi$. If $x$ is in $K^\times$, write $B_x$ for the open ball of radius $|x|$ and center 0 in $K$. Let $a$ be in $K^\times$, and let $\tau_0$ be the set of those $U$ such that $0\in U\in\tau$.

It suffices to check that $B_a$ contains some $U$ in $\tau_0$.

We can find a $b$ in $K^\times$ and a $V$ in $\tau_0$ such that $a$ is not in $B_bV$, and then a $c$ in $K$ with $|c|>1$ and a $W$ in $\tau_0$ such that $a$ is not in $B_cW$. Then $U:=c^{-1}W$ does the job.

Answer by Pierre-Yves Gaillard for Examples of common false beliefs in mathematics.

2010-05-12T14:48:30Z

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

Answer by Pierre-Yves Gaillard for Unbounded operator bounded in a dense subset

2010-06-18T04:44:53Z

This is just to make Nate Eldridge's answer selfcontained.

For any normed vector space $V$ and any $r > 0$, write $V_r$ for the open ball of radius $r$ and center $0$ in $V$.

Let $X$ be a normed vector space, $Y$ a closed space, $Z$ the quotient, $\pi$ the canonical projection, $\tau$ the quotient topology, $\nu$ the topology on $Z$ induced by the quotient norm $$|\pi(x)|:=\inf_{y\in Y}|x+y|.$$

(It's easy to see that this is a norm.)

We claim $\tau=\nu$.

Both topologies are translation invariant.

The set $\{\pi(X_r)\ |\ r > 0\}$ is a basis for the $\tau$-neighborhoods of $0$ in $Z$.

The set $\{(Z_r)\ |\ r > 0\}$ is a basis for the $\nu$-neighborhoods of $0$ in $Z$.

As $\pi(X_r)=Z_r$, we're done.

Answer by Pierre-Yves Gaillard for What are examples of mathematical concepts named after the wrong people? (Stigler's law)

2010-05-21T04:42:10Z

I think the Kazhdan-Lusztig Conjectures are due to Vogan.

EDIT.

True or false, the claim is mainly based on the very first two paragraphs of

[II] Irreducible characters of semisimple Lie groups II. The Kazhdan-Lusztig conjectures. David A. Vogan, Jr. Duke Math. J. Volume 46, Number 4 (1979), 805-859. --- The link

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077313724

gives a universal access to the first page, which contains the two paragraphs in question. In case you don't have access to the full paper, here is a scan of the references (to completely understand the two paragraphs):

http://www.iecn.u-nancy.fr/~gaillard/vogan_ref.pdf

Here are two more references:

[I] Irreducible characters of semisimple Lie groups I, David A. Vogan, Jr., Duke Math. J. Volume 46, Number 1 (1979), 61-108.

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077313255

[KL] David Kazhdan and George Lusztig, Representations of Coxeter groups and Hecke algebras, Inventiones Mathematicae, Volume 53, Number 2, 165-184.

http://gdz.sub.uni-goettingen.de/en/dms/load/img/?PPN=PPN356556735_0053&DMDID=dmdlog14

I would summarize things as follows.

Step 1. In [I] Vogan made a certain conjecture.

Step 2. [II] and [KL] were written simultaneously. Each paper cites the other. In [KL] Kazhdan and Lusztig also made a certain conjecture. When he learned this, Vogan immediately (or at least very fast) proved that the "Step 1 conjecture" implies that of Kazhdan and Lusztig. (He even showed that the "Step 1 conjecture" generalizes that of Kazhdan and Lusztig.)

But, again, the best is to read carefully the first two paragraphs of [II]. Vogan explains this much more clearly than I, and it's always better to hear things from the horse's mouth.

Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?

2010-08-02T06:03:49Z

The posting of this question was suggested by Yemon Choi: see http://mathoverflow.net/questions/30092/discrete-cyclic-subgroup/34005#34005. The question is not mine; it's just a rephrasing of http://mathoverflow.net/questions/30092/discrete-cyclic-subgroup

EDIT 4. This post claims that the answer is No in general. I hope it's correct. END OF EDIT 4.

By page 110 of Weil's book L'intégration dans les groupes topologiques et ses applications, the answer is No in the abelian case.

I know almost nothing about locally compact groups. The question might be very easy for experts, and perhaps even for laymen. In the unlikely event the question is difficult, here is a particular case:

Let G be a non-compact connected Lie group. Does G admit a discrete infinite cyclic subgroup?

EDIT 1. I think that, by known results about lattices, the answer is No for semisimple Lie groups. Thanks for correcting me if I'm wrong, or (even better) for providing precise statements and references. Again, plenty of MathOverflowers know this stuff much better than I. I'm making it a Community Wiki. END OF EDIT 1.

EDIT 2. I scanned a few pages of Weil's L'intégration dans les groupes topologiques et ses applications and of Raghunathan's Discrete subgroups of Lie groups, and highlighted some statements. The highlighted statements from Raghunathan's book imply that the answer to the main question is No for semisimple Lie groups. (Of course, there might be more elementary arguments.)

[On page 100 of Raghunathan's book (one of the scanned pages) one reads "As will be seen later ... any lattice in a connected Lie group is finitely generated". Unfortunately, I haven't been able to find where, in the sequel of the book, this is proved. If somebody could indicate the appropriated page (and even scan it), it would be great!] END OF EDIT 2.

EDIT 3. Keivan Karai's answer convinced me that there are elementary arguments showing the negativity of the answer to the main question for semisimple Lie groups. END OF EDIT 3.

Answer by Pierre-Yves Gaillard for Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?

2010-08-03T06:04:47Z

The answer to the title question is No, that is

A locally compact group has a compact open subgroup or a discrete infinite cyclic subgroup.

[Keivan Karai helped me a lot, without being responsible for the possible mistakes in this post.]

Step 1. The case of a connected noncompact Lie group.

We claim that a connected noncompact Lie group $G$ has a discrete infinite cyclic subgroup.

Arguing by contradiction, assume $G$ is a counterexample of smallest dimension, and let $Z$ be its center.

Suppose $Ad(G)$ is compact. Then $Z$ is noncompact. If $\dim Z=0$, then $G$ is a noncompact connected covering of the connected compact group $Ad(G)$, which is impossible. Thus, $\dim Z\ge1$. Let $Z_1$ be the connected component of $Z$. If $Z_1$ were compact, $G/Z_1$ would be a counterexample of smaller dimension. If $Z_1$ were noncompact, it would contain a discrete infinite cyclic subgroup, again a contradiction.

Thus, $Ad(G)$ is noncompact. Replacing $G$ with $Ad(G)$, we can suppose $G\subset GL_n(\mathbb R)$.

Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $exp(\mathbb Z x)$ of $GL(V)$ is not an infinite discrete subgroup.

Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ is nonzero and satisfies (C), the same holds for $ad(x)$, implying $Killing(x,x) < 0$. If it were so for all nonzero $x$ in $\mathfrak g$, then $\mathfrak g$, and thus $G$, would be compact.

Step 2. The general case.

Let $G$ be our locally compact group, $C$ its connected component, and recall the following facts (see [1], [2], and references therein):

(1) If $G$ is totally disconnected, then $G$ contains a compact open subgroup.

(2) $C$ is a normal subgroup of $G$, and $G/C$ is totally disconnected.

(3) If $G$ is connected, then every neighborhood of 1 contains a compact normal subgroup $K$ such that $G/K$ is a connected Lie group.

Assume $G$ has no compact open subgroups.

We must show that $G$ has a discrete infinite cyclic subgroup.

As $C$ is noncompact by (1) and (2), we can assume $G=C$, that is $G$ is connected. Then (3) implies that $G$ contains a compact normal subgroup $K$ such that $G/K$ is a connected noncompact Lie group, and we can assume $K=1$, that is $G$ is a connected noncompact Lie group, and the conclusion follows from Step 1.

[1] Willis, G. The structure of totally disconnected, locally compact groups. Math. Ann. 300 (1994), no. 2, 341-363.

http://gdz.sub.uni-goettingen.de/en/dms/load/img/?IDDOC=167209

[2] Willis, G. Totally disconnected, nilpotent, locally compact groups. Bull. Austral. Math. Soc. 55 (1997), no. 1, 143-146.

http://journals.cambridge.org/action/displayFulltext?type=1&fid=4856560&jid=BAZ&volumeId=55&issueId=01&aid=4856552

Answer by Pierre-Yves Gaillard for Discrete cyclic subgroup.

2010-07-31T11:33:53Z

This is a comment. I'm putting it as an answer just to draw attention to the question. (I hope it's ok.) It might be very easy. I'm very curious to know the answer. I'd state the question as follows:

Are there locally compact groups which have no compact open subgroups and no discrete infinite cyclic subgroups?

EDIT OF AUG. 1, 2010

By page 110 of Weil's book [1], the answer is No in the abelian case.

A particular case of the question in the nonabelian case is this. Let G be a non-compact connected Lie group. Does G admit a discrete infinite cyclic subgroup? Again, I suspect that this subquestion is very easy (at least for experts).

[1] Weil, André, L'intégration dans les groupes topologiques et ses applications, Actual. Sci. Ind., no. 869. Hermann et Cie., Paris, 1940. 158 pp.

Answer by Pierre-Yves Gaillard for Applications of the Chinese remainder theorem

2010-07-30T08:39:10Z

The Chinese Remainder Theorem gives a way to compute matrix exponentials.

Indeed, let $A$ be a complex square matrix, put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $S$ be the set of eigenvalues of $A$, $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$.

The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective.

Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$.

Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=\frac{\mu}{(X-s)^{m(s)}}\ g\ \mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

Answer by Pierre-Yves Gaillard for How would one even begin to try to prove that a simple number-theoretic statement is undecidable?

2010-06-12T09:36:26Z

EDIT

Here is my problem. To prove that statement S is undecidable is to

(1) prove that one cannot prove S.

I think I understand the meaning of the second "prove". (It depends of course on the context.) But I don't understand the meaning of the first "prove".

A "solution" would be to replace (1) by

(2) give an informal proof showing that one cannot prove S,

(3) give an acceptable argument showing that one cannot prove S.

This seems to be in tune with the following quotation from Cohen's "Set theory and the Continuum Hypothesis" (p. 41):

We have now arrived at a rather peculiar situation. On the one hand $\sim A$ is not provable in $Z_1$ and yet we have just given an informal proof that $\sim A$ is true. (There is no contradiction here since we have merely shown that the proofs in $Z_1$ do not exhaust the set of all acceptable arguments.)

I think it would be an enormous progress to replace (1) by (2) or (3).

In other words, instead of talking about "proving" that some statements are undecidable, it would be wiser, I believe, to talk about giving "informal proofs", or "acceptable arguments", or "convincing evidence", ... that these statements are undecidable.

END OF EDIT

Here is, for what it's worth, my personal conviction on this.

If (like me) you don't believe that "you can't get something for nothing", then you don't believe in Gödel and Cohen's results.

The claim to "get something for nothing" is very openly expressed (in my opinion) by Cohen on page 39 of "Set theory and the Continuum Hypothesis":

The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions.

Cohen really says: "The theorems of the previous section are proved without invoking any axiom, that is, they are gotten for nothing". Or am I putting words in his mouth?

I think the key is to understand the respective STATUS of the various statements involved. In particular, a clear distinction should be made between mathematical and metamathematical statements.

I also think we should all make an effort to talk unemotionally about such questions.

Answer by Pierre-Yves Gaillard for Solutions to the Continuum Hypothesis

2010-05-20T05:59:46Z

I believe Gödel and Cohen's arguments are not acceptable. My reasoning is a common sense one:

A mathematical proof is a text whose spelling correctness can be checked by a machine. The text in itself has no meaning. Only an interpretation of it can have any meaning. So, at best, Gödel and Cohen proved some mathematical statements which can be interpreted as implying that the continuum hypothesis is undecidable (in a given axiom system).

It's surprising that Bourbaki, in his Set Theory book, takes Gödel and Cohen's arguments for granted. I think that Bourbaki's approach, coherently followed, leads to a rejection of these arguments.

Here is a short text about this:

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Continuum_Hypothesis/

I think many people share this opinion. (For some reason they don't express it.)

In another answer I gave three quotations from Cohen's book "Set theory and the Continuum Hypothesis". For the reader's convenience I'll paste them below. I think we'll all find them interesting.

Quotation 1 (pages 26-27)

It should be emphasized that these functions are "real" mathematical objects and not objects of any formal system ...

Question. What is a "real mathematical object"? (And what is an "unreal" or "irreal" mathematical object?)

Quotation 2 (page 39)

The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions.

Questions. What is an "absolute statement"? Is there an "axiom free" mathematics?

Quotation 3 (page 41)

We have now arrived at a rather peculiar situation. On the one hand $\sim A$ is not provable in $Z_1$ and yet we have just given an informal proof that $\sim A$ is true. (There is no contradiction here since we have merely shown that the proofs in $Z_1$ do not exhaust the set of all acceptable arguments.)

Question. What is an "acceptable argument"? It seems that "true" is defined as "following from an acceptable argument".

Here is another quotation (page 13):

If a set of statements S has a model then it is consistent.

If the underlying set theory is inconsistent, then any set of statements has a model.

I suggest the following three books to the interested reader.

(1) P. Cohen, Set theory and the Continuum Hypothesis.

(2) S.C. Kleene, Introduction to Metamathematics.

(3) N. Bourbaki, Theory of Sets. French original: Théorie des ensembles.

(You may have an institutional access to some Bourbaki's books.)

Let me ask explicitly the question which is implicit in this post:

Is the Continuum Hypothesis true, false or undecidable in Bourbaki's set theory?

There is no mathematical questions whose answer I more ardently wish I knew.

Thank you to all the people in charge of MathOverflow for everything you're doing!

Answer by Pierre-Yves Gaillard for Cardinality of the permutations of an infinite set

2010-06-12T04:51:02Z

As already mentioned, we have $$2^k\le k^k\le(2^k)^k=2^{kk}=2^k,$$ and thus $2^k=k^k$. The inequality $k!\le k^k$ is obvious. To check $2^k\le k!$, note that $2^k$ subsets of $X$ are the set of fixed points of some permutation. Conclusion: $k!=2^k$.

(I don't understand Robin Chapman's argument.)

Answer by Pierre-Yves Gaillard for Examples of common false beliefs in mathematics.

2010-05-19T16:03:12Z

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Hodgegaillard/

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

Answer by Pierre-Yves Gaillard for Knuth's intuition that Goldbach might be unprovable

2010-06-11T06:26:11Z

I don't understand why people don't follow Bourbaki's approach, as expounded in his set theory book, on this kind of questions. I find this book incredibly simple and clear. From Bourbaki's viewpoint, there is no reason to expect Goldbach's conjecture to be decidable.

Answer by Pierre-Yves Gaillard for Shortest/Most elegant proof for $L(1,\chi)\neq 0$

2010-05-25T05:56:31Z

I know it's not an answer to the question, but rather another (probably naive) question suggested by the original question.

It seems to me that proving the prime number theorem for arithmetic progressions isn't harder than proving the arithmetic progression theorem alone. In other words, you can get then prime number theorem for nothing.

I know that many people have already done that (see this text for references), but I'd expect almost everybody to do it, and I don't understand why they don't.

Answer by Pierre-Yves Gaillard for Examples of common false beliefs in mathematics.

2010-05-19T16:02:12Z

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

Answer by Pierre-Yves Gaillard for Quick proofs of hard theorems

2010-05-18T04:40:58Z

Cantor's proof of the existence of transcendental numbers. With a (now) obvious one-line argument he showed that there are uncountably many of them --- when Liouville, Hermite and others had to take (putative) transcendental numbers one at a time ...

Newman's argument (especially Korevaar's and Zagier's version of it) turned the Prime Number Theorem, which took a century to be proved, into something that can be explained in a few minutes to any graduate student.

Answer by Pierre-Yves Gaillard for Math paper authors' order

2010-05-18T03:46:03Z

It seems to me nobody mentioned the Zariski-Samuel and Grothendieck-Dieudonné cases.

Answer by Pierre-Yves Gaillard for Suggestions for good books on class field theory

2010-05-15T11:04:16Z

Among the few books on class field theory I tried to read, Weil's Basic Number Theory is the one I found most accessible. By far.

Answer by Pierre-Yves Gaillard for Subset of the plane that intersects every line exactly twice

2010-05-14T05:59:32Z

Here is a minor variation of the proof.

We show that there is a subset $Z$ of $\mathbb R^2$ which intersects each line exactly twice.

We define the ordinals in such a way that each ordinal $a$ is the set of those ordinals $< a$. For any set $S$ we write $|S|$ for the least ordinal equipotent to $S$, and call it the cardinality of $S$. Let $c$ (for continuum) be the cardinality of $\mathbb R$. For any subset $S$ of $\mathbb R^2$ we denote by $\langle S\rangle$ the set of lines generated by $S$ (a line being generated by $S$ if it has at least two points in common with $S$). Let $d\mapsto L_d$ be a bijection from $c$ onto the set of lines in $\mathbb R^2$.

For each $d\in c$ we define $Z_d\subset\mathbb R^2$, $f(d)\in c$, and $z_d\in\mathbb R^2$, as follows. Let $z_0$ be any point of $L_0$, put $f(0)=0$, and let $Z_0$ be the empty set. Now assume $0< d< c$, and $f(e), z_e$ already defined for $e < d$. Put $Z_d:=$ {$z_e\ |\ e< d$}. As $|\langle Z_d\rangle|< c$ (because $|Z_d|< c$), there is a least $f(d)$ in $c$ such that $L_{f(d)}\notin\langle Z_d\rangle$. Let $z_d$ be any point of the set

$$L_{f(d)}-(Z_d\cup(\cup\langle Z_d\rangle)),$$

which is easily seen to be nonempty.

Let $Z$ be the union of the $Z_d$ and $L$ any line in $\mathbb R^2$. We claim $|L \cap Z|=2$, that is, $Z$ is the sought-for subset of $\mathbb R^2$. To prove $|L\cap Z|\le2\ (*)$ we assume by contradiction that there are $g< h< i$ in $c$ such that $z_g,z_h,z_i\in L\cap Z$. We have $z_g,z_h\in Z_i$ by definition of $Z_i$, and thus $L\in\langle Z_i\rangle$, contradicting the definition of $z_i$. To prove $|L\cap Z|\ge2$ we assume by contradiction $|L\cap Z|< 2$. Put $L=L_d$ and let $g$ be in $c$. The inequality $|L_d\cap Z_g|<2$ implies $L_d\notin\langle Z_g\rangle$, and thus $f(g)\le d$ by minimality of $f(g)$. This shows that $Z$ is contained into the union of the $L_e$ such that $e\le d$. As $|L_e\cap Z|\le2$ by $(*)$, we get $|Z|\le2|d|+2< c$, contradicting the obvious equality $|Z|=c$.

A pdf version is available here.

Answer by Pierre-Yves Gaillard for What does it mean for a mathematical statement to be true?

2010-05-12T09:38:30Z

In my humble opinion, the best reference for this kind of questions is Bourbaki's "Set Theory" ... Actually, I would recommend Bourbaki's book to people who, like me, have trouble to understand other texts on the same subject.

Comment by Pierre-Yves Gaillard

2012-04-01T08:36:14Z

You want the basis $\frac{\partial}{\partial\overline z}\ $, $\frac{\partial}{\partial z}$ to be dual to the basis $dz,d\overline z$.

Comment by Pierre-Yves Gaillard

2012-03-26T13:11:26Z

Dear @Robert: I suggest that you post an answer.

Comment by Pierre-Yves Gaillard

2012-03-21T14:41:57Z

Link to "Is the dream solution to the continuum hypothesis attainable?" by Joel David Hamkins: arxiv.org/abs/1203.4026

Comment by Pierre-Yves Gaillard

2012-03-18T09:48:52Z

Typo on the first line (Guassian). (I don't have edit privileges.)

Comment by Pierre-Yves Gaillard

2012-03-11T10:27:50Z

Dear Jacques: A detail: In Bourbaki's theory you don't have $2\in3$. (You do have other junk theorems.) I'm mentioning that because some of the mathematicians using (at least implicitly) Bourbaki's theory are usually regarded as important ones, like Serre or Grothendieck.

Comment by Pierre-Yves Gaillard

2012-02-26T11:09:37Z

This article is freely and legally available at ams.org/journals/proc/1970-025-03/…

Comment by Pierre-Yves Gaillard

2012-02-14T17:51:53Z

Possibly related: Section 18 of Hecke's Lectures on the Theory of Algebraic Numbers.

Comment by Pierre-Yves Gaillard

2012-02-13T16:19:17Z

Dear Pham Hung Quy: Thanks for having answered my comment. I'm sorry, I still don't see the point of your edit, because I think it is a standard fact that the radical of any primary ideal is irreducible.

Comment by Pierre-Yves Gaillard

2012-02-13T08:57:00Z

Dear Pham Hung Quy: About your edit: It seems to me that if 0 was primary, its radical would be prime, and thus irreducible.

Comment by Pierre-Yves Gaillard

2012-02-03T18:44:53Z

Dear Mariano: Thanks a lot! Your answer is very concise, but I'm sure it contains a lot of maths. I'll try to digest and assimilate it.

Comment by Pierre-Yves Gaillard

2012-02-03T15:12:36Z

Dear @Martin: Thanks for your comments. I edited the question.

Comment by Pierre-Yves Gaillard

2012-02-03T11:39:34Z

It seems to me that the paper "The diamond lemma for ring theory", Advances in Mathematics 29 (1978) 178-218, by George M. Bergman has not been quoted in this thread. Related link: math.berkeley.edu/~gbergman/papers/updates/…

Comment by Pierre-Yves Gaillard

2012-01-14T19:51:25Z

The "S" comes from the Latin "sinister", which means "left" indeed. Bourbaki uses systematically the letter "s" to mean "left". For example $A_s$ mean "$A$ viewed as a left $A$-module".

Comment by Pierre-Yves Gaillard

2011-12-14T10:31:29Z

Dear @Martin: Thanks! I agree. In the previous versions of the answer, what is now Part 1 was Part 2. Old Part 2 was much shorter and added as a loosely related complement. --- I'm kind of obsessed by the question of knowing if my argument is correct. If it is, it's seems hard to believe that it had not been written somewhere before. It's true that I'm taking advantage of the post to ask this question.

Comment by Pierre-Yves Gaillard

2011-12-10T10:21:28Z

David Roberts's Google+ post: plus.google.com/u/0/103404025783539237119/posts/…