User steven landsburg - MathOverflow

Answer by Steven Landsburg for objects which can't be defined without making choices but which end up independent of the choice

2013-05-20T22:31:54Z

You can define $Ext^n$ as the set of isomorphism classes of $n$-step extensions, equipped with the Baer sum. This eliminates choices from the definition of $Ext$ in exactly the same spirit as your original post eliminates choices from the definition of the trace.

Answer by Steven Landsburg for objects which can't be defined without making choices but which end up independent of the choice

2013-05-20T19:11:14Z

I doubt that you can define addition and multiplication on the real numbers (defined as equivalence classes of Cauchy sequences) without making choices.

[Edited to add: As Toink points out in comments, it's actually quite easy to define addition and multiplication without making choices. So let me modify the above by replacing "addition and multiplication on the real numbers" with "the square root function on the positive real numbers".]

In general, when an infinite set (or, say, an infinitely generated group, etc) is defined via an equivalence relation, as often happens in mathematics, there's no avoiding the fact that functions on that set have to be defined by choosing representatives (unless, of course, you first prove that your definition is equivalent to some other definition --- e.g. the reals can also be defined by Dedekind cuts --- but that only pushes the making of choices back a step, into the proof that the definitions are equivalent).

(One class of exceptions: If you're defining a function into an ordered set, you can avoid making choices by defining the value of your function to be its maximum over all possible choices --- this works for, say, the dimension of a vector space. But if the function's codomain has no extra structure, it seems clear that choices are unavoidable. Edited to add: Again, per Toink's comment, another class of exceptions occurs when the structure on the quotient is inherited from above.)

Answer by Steven Landsburg for What's the definition of continuous of set-valued functions?

2013-05-20T11:44:12Z

$\phi$ is upper semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\subset W\rbrace $ is open in $X$.

$\phi$ is lower semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\cap W\neq \emptyset\rbrace$ is open in $X$.

$\phi$ is continuous if it is both upper semincontinuous and lower semicontinuous.

Answer by Steven Landsburg for When does "second annihilator" of a (principal) ideal equal the ideal itself , ie $Ann_R(Ann_R(r))=Rr$?

2013-05-07T12:29:28Z

For what it's worth, it suffices for $(r)$ to be an interesection of minimal primes; in fact, more generally if $J$ is any intersection of minimal primes then $J=Annih(Annih(J))$. (This requires only that $R$ be commutative and noetherian; you don't need local.)

This is Lemma 2.17 in an old paper of mine called "Patching Modules of Finite Projective Dimension", where the stated hypothesis is "Let $R$ be any (commutative) ring", but it now seems to me that the proof requires $R$ to be noetherian.

Answer by Steven Landsburg for Category with a "metric" for arrow composition

2013-05-05T22:58:53Z

1) In the category of finite sets (or finite groups or finite topological spaces....) let $d(f)$ be the cardinality of the image of $f$. This satisfies the strong triangle condition $$d(x\rightarrow z)\le\text{min }(d(x\rightarrow y),d(y\rightarrow z))$$

2) In any category, for each object $x$, let $\xi(x)$ be an (arbitrarily assigned) positive real number and define $$d(f)=\text{min }\lbrace{\xi(c)|f \hbox{ factors through } c}\rbrace$$ This also satisfies $$d(x\rightarrow z)\le \text{min }(d(x\rightarrow y),d(y\rightarrow z))$$

3) Fix a formal language for describing arrows in your category, let $l(f)$ be the length of the shortest description of $f$, and let $d(f)=l(f)+5$. The triangle inequality follows because $g\circ h$ always has a formal description just slightly longer than the sum of the shortest formal descriptions of $g$ and $h$ (say by putting each of these descriptions between parentheses and inserting a $\circ$ between them, which adds five characters).

In case 3), you have to allow $d$ to take the value infinity, or restrict to categories in which everything has a finite description.

Edited to add: 4) For the category of topological spaces, you can fix a non-negative integer $r$ and let $d(f)= \hbox{rank} (H^r(f,{\mathbb Q}))$ . This requires either allowing $d$ to take the value infinity or restricting to some subcategory where the homology groups are finite dimensional.

Answer by Steven Landsburg for Dense Affine Subvarieties of Algebraic Varieties

2013-05-01T19:33:30Z

If $X$ is irreducible, then any open subset is dense, and since $X$ can be covered by open affines, any one of these will do. If $X$ isn't irreducible, take an affine dense subvariety of each component, and take their union.

Edited to add: As commenters have noted, you'll want to be sure the affine dense subvarieties you choose are disjoint from one another (in order to insure that their union is affine). But this is easy: For each of these subvarities, just throw away all its intersections with the other components, then take an affine open subvariety of what remains.

Answer by Steven Landsburg for History of the high-dimensional volume paradox

2013-04-27T01:32:52Z

A related (and to me, when I first saw it, much more suprising) Fun Fact: Divide the n-dimensional cube in half in each of $n$ dimensions, to create $2^n$ smaller cubes of edge length 1/2. Inscribe a ball in each of these subcubes, and then construct the smallest ball tangent to each of those (and centered at the center of the original cube) like so:

What happens to the diameter of the central ball as $n$ gets large?

This question received much attention at an algebraic K-theory conference in Boulder in the early 1980s, where each new arrival was presented with a multiple choice problem: Without stopping to compute, is the limit $-1$, $0$, $1/2$, $1$, $10$ or $\infty$? You were allowed to choose any three answers out of six, and place a bet on whether the right answer was among them. I can report that an overwhelming majority of algebraic K-theorists reason thusly: the answer can't be negative and can't be greater than 1 (the ball, after all, is obviously contained inside a box of side 1!); therefore it's safe to bet on the set $\lbrace 0,1/2,1 \rbrace $. Feel free to make money off this.

Answer by Steven Landsburg for What is the Q-construction, metaphysically?

2013-04-25T23:31:30Z

1) One wants the $Q$-construction to have the property that $\pi_1(QP)=K_0(P)$.

2) To get this, one wants, for any covering space of $BQP$, that $K_0(P)$ acts naturally on the fiber over $0$.

3) To get this, one wants to associate to any monomorphism $i$ in $P$ a morphism $i_!$ in $QP$ and to any epimorphism $j$ in $P$ a morphism $j^!$ in $QP$ in a way that satisfies certain simple properties; the statement of the properties, and the proof that they suffice to get this result, is in Quillen's Algebraic K-Theory I (Theorem I).

4) Quillen's $Q$-construction is the Universal construction yielding such $i_!$ and $j^!$. (The proof is in QUillen's paper, immediately preceding Theorem 1.)

5) Therefore, there's a sense in which Quillen's $Q$-construction is the natural first guess for what should work. (Of course the "naturality" of this guess appears only in hindsight; a lot of other people failed to find this construction.)

PS. After you work through the constructions, you see that this is another way to see the same thing: For any object $A$ in your category $P$, you want to associate the $K_0$-class $[A]$ to some loop in $BQP$. The simplest thing to hope for is two canonically defined maps from $0$ to $A$ in $QP$, which together give you your loop. Quillen's construction provides those two maps (recognizing $0$ as a quotient of both $0\subset A$ and $A\subset A$) in the simplest possible way.

Answer by Steven Landsburg for ring with prescribed K group

2013-04-23T14:23:10Z

Every abelian group $G$ is the class group of some Dedekind domain $R$ (theorem of Luther Claborn), so we have $K_0^{red}(R)= G$.

Answer by Steven Landsburg for K-theory of monoidal categories

2013-04-22T13:40:54Z

First, see Thomason's paper "Beware the Phony Multiplication on Quillen's $S^{-1}S$" (Proceedings of the AMS, 1990) for why the most obvious construction is wrong. In the bibliography, you'll find references to papers of Loday, Waldhausen, Segal/Wolfson and May with constructions that are more complicated, but correct.

Answer by Steven Landsburg for Localization sequence for K^0(X)

2013-04-15T02:38:23Z

An element of $K_0(U)$ is represented by a perfect complex $F^.$ in the derived category of $U$-modules. As long as $X$ and $U$ are quasi-compact and quasi-separated, the class $[F^.]$ lifts to $K_0(X)$ if and only if $F^.$ is the restriction (in the derived category) of a perfect complex on $X$. (This is the "key proposition" of Thomason and Trobaugh's paper on "Higher Algebraic K-Theory of Schemes and of Derived Categories".) So any perfect complex that doesn't lift gives a counterexample to the surjectivity on $K_0$.

Answer by Steven Landsburg for Correspondence between submodules and quotient modules

2013-04-12T17:24:35Z

If $M=R/(p^n)$ with $p$ prime, the result is clear. Since an arbitrary $M$ is a direct sum of such modules, the result is still clear.

Answer by Steven Landsburg for Tensoring with descending chain of modules

2013-04-08T01:20:52Z

Let $A=k[X]$, let $M_i$ be the $A$-ideal generated by $X^i$, and let $B=k(X)$. Then $\cap M_i=0$ is certainly finite and free, but

$$0=(\cap M_i)\otimes B\neq \cap(M_i\otimes B)=B$$

which is a counterexample to what you're looking for (as clarified in your response to Eric Wofsey's comment), and it's hard to imagine rings nicer than $A$ and $B$.

Answer by Steven Landsburg for Self-containing structures

2013-03-24T02:26:58Z

In the same vein as your Gromov-Hausdorff example, the set of all isomorphism classes of finitely generated monoids is a monoid under direct sum. And this recent MO question concerns the skeleton-of-the-category of all skeletons-of-categories.

Answer by Steven Landsburg for Six operations for (quasi)-coherent sheaves

2013-03-23T04:31:49Z

Well, you could read SGA. But my two favorite sources for this material are here for an abstract treatment that doesn't (as far as I remember) talk specifically about quasi-coherent sheaves, and here for a considerably longer but readable treatement that does.

(More precisely, the first reference has a section that basically takes various properties of quasi-coherent sheaves as axioms and proceeds from there. If you're willing to accept these axioms without working through all the geometry, that's probably the reference you're looking for.)

Answer by Steven Landsburg for Is the primitive element theorem a cohomological statement?

2013-03-21T03:01:47Z

The vanishing of the cohomology group $H^1(Spec(R),GL_n)$ doesn't actually say that all projectives of rank $n$ are free; it says only that all projectives of rank $n$ are isomorphic. Combining this with the observation that at least one such projective is free, we get that they're all free.

But in the case of field extensions, it is not true that all finite separable extensions are isomorphic, even though they're all generated by primitive elements. Therefore, I think the analogy you're seeing is largely illusory.

Answer by Steven Landsburg for Group G hasn't all conditions of Lie group

2013-03-16T17:00:49Z

Bryant requires in the definition of a Lie group only that the multiplication map be smooth, and then proves that the inversion map must be smooth also. (Proposition 1, page 14.)

Answer by Steven Landsburg for Let $M$ be a $R$-Bimodule that happens to be projective, is its associated left $R \otimes R^{op}$-module projective too?

2013-03-12T06:05:43Z

I'm not sure what hypothesis you intend, but I don't think there's any reasonable interpretation under which it implies your conclusion.

Let $R=M={\mathbb Z}[X]$, acting on itself by multiplication from both the left and the right. Then $R\otimes R^{op}\approx {\mathbb Z}[X,Y]$ is a domain, and $(X-Y)$ annihilates $M$, whence $M$ cannot be free.

Answer by Steven Landsburg for Defining Transfers Algebraically

2013-03-12T03:38:38Z

Let $M$ be a $G$-module. Then there's an obvious injection $$M\rightarrow {\mathbb Z}G\otimes_{{\mathbb Z}H}M$$
Taking homology gives $$H_\bullet(G,M)\rightarrow H_\bullet(G,{\mathbb Z}G\otimes_{{\mathbb Z}H}M)$$
To compute the group on the right, start with a projective resolution $P$ of ${\mathbb Z}$ over ${\mathbb Z}G$, then take the homology of
$$P\otimes_{{\mathbb Z}G}{\mathbb Z}G{\otimes_{\mathbb ZH}}M=P\otimes_{{\mathbb Z}H}M$$
Notice that $P$ is also a projective resolution of ${\mathbb Z}$ over ${\mathbb Z}H$, so the homology you've just computed is in fact equal to $H_\bullet(H,M)$.
Thus the map in 2) is actually a map $H_\bullet(G,M) \rightarrow H_\bullet(H,M)$ and is in fact the transfer map.
To make the same argument work for a ring inclusion $A\rightarrow B$ you'd need to know that a projective $B$-resolution is also a projective $A$-resolution. This works, for example, if $B$ is $A$-free.

Answer by Steven Landsburg for Putting objects into boxes so that each box gets <=2 objects

2013-03-10T23:05:23Z

Let $S_n$ be the symmetric group on $n$ letters, so that $S_k\times S_n$ acts on the set of allowable placements in the obvious way.

For a placement with $A$ boxes having 2 balls, $B$ boxes having 1 ball and $C$ boxes having 0 balls, the order of the isotropy subgroup is easily seen to be $$A!B!C!2^A$$ so the orbit size is $n!k!$ divided by this expression.

Adding up the sizes of all the orbits, and accounting for the facts that $k=2A+B$ and $n=A+B+C$, we get the expression $$\sum_{A=0}^{ceiling(k/2)}{n!k!\over A!(k-2A)!(n-k+A)!2^A}$$

Answer by Steven Landsburg for A question on generic point and a question on Hartshorne

2013-03-10T16:02:44Z

If $X=Spec(R)$ is an affine scheme (as it is in the example you refer to) and $Y$ is the subscheme defined by a prime ideal $P$, then the generic point of $Y$ is the point $[P]\in Spec(R)$. The local ring at that point is the localization $R_P$.

In the Hartshorne example, $R={\mathbb C}[x,y,z]/(xy-z^2)$, and $P$ is the ideal $(y,z)$. (You really should have mentioned this in your question). In the ring $R_P$, the maximal ideal is $(y,z)$, which is the same as $(z)$ (because $x$ is a unit). Because $y$ is a unit times $z^2$, the divisor of $y$ is twice the divisor of $z$.

Answer by Steven Landsburg for Product and quotient of ideals

2013-03-09T04:49:03Z

Assume we're in an integral domain. (I realize your example is actually a polynomial ring over ${\mathbb C}$, but let's work in a more general domain for now.) Let's also suppose our domain to be an algebra over a field of characteristic $\neq 2$.

Let's look for ideals $J$ and $K$, and an element $x$ such that $xK\subset JK$ but $x\notin J$. (This would give counterexamples to both (1) and (2).)

This is surely impossible if $K$ is principal, so let's investigate the case where $K$ is generated by 2 elements.

Then I claim the following are equivalent:

1) The ring $S$ contains a counterexample to your (1) and/or (2) with $K$ two-generated.

2) The ring $S$ contains elements $A,B,C,D,F$ with $(A-D-F)(A-D+F)=4BC$ and $F\notin (A,B,C,D)$.

${\bf Proof:}$ Let $\alpha, \beta$ generate $K$. Then given a counterexample, we can write $$x\pmatrix{\alpha\cr\beta\cr}=\pmatrix{A&B\cr C&D\cr}\pmatrix{\alpha\cr\beta\cr}$$ for some $A,B,C,D\in J$, which we might as well assume generate $J$. Thus $x$ is an eigenvalue of the displayed two-by-two matrix and so satisfies its characteristic equation, whence there exists $F$ with $x=A-D-F$ and $(A-D-F)(A-D+F)=4BC$. Also, $(\alpha,\beta)$ must be the transpose of an eignvector, which we can take to be $(A+D-F,-C)$.

This will be a counterexample iff $x\notin J$, hence iff $F\notin J$. QED.

Thus, for algebras over a field $k$ of characteristic $\neq 2$, the universal counterexample is given by $$R=k[A,B,C,D,F]/((A-D-F)(A-D+F)-4BC)$$ $$x=A-D-F$$ $$J=(A,B,C,D)$$ $$K=(A+D-F,-C)$$

Your ring $S$ will contain a counterexample (with $K$ two-generated) iff it contains a homomorphic image of $R$ in which $F\notin (A,B,C,D)$. When $S$ is a polynomial ring, I'm not sure whether this is the case but it might not be too hard to settle.

Answer by Steven Landsburg for Product and quotient of ideals

2013-03-07T03:28:43Z

There's a good chance you already know this, but $IK\subset JK$ at least implies $I\subset\sqrt{J}$.

Proof: Let $k_1,\ldots,k_n$ generate $K$. Then for any $x\in I$, we have $xk_\alpha=\sum j_{\alpha\beta}k_\beta$ for some $j_{\alpha\beta}\in J$. Putting these together gives a matrix equation $$(x\cdot 1-M)k=0$$

where $1$ is the identity matrix, $M$ has all its entries in $J$, and $k$ is the column vector consisting of the $k_\alpha$.

This implies that $(x\cdot 1-M)$ has determinant zero, but clearly this determinant is of the form $x^n-j$ with $j\in J$. So $x\in \sqrt{J}$.

I realize this is unlikely to be all you need, since you went out of your way to say that $J$ might be nonreduced.

Examples where it's useful to know that a mathematical object belongs to some family of objects

2011-09-01T01:25:23Z

For an expository piece I'm writing, it would be useful to have good examples of the following phenomenon:

(1) ${\cal X}$ is a parameterized family of somethings. (Varieties, schemes, manifolds, rings, groups, whatever.)

(2) ${\cal X}_0$ is an member of this family.

(3) ${\cal X}_0$ has some interesting property $P$ that, on the surface, appears to have nothing to do with the inclusion of ${\cal X}_0$ in the family ${\cal X}$.

(4) Nevertheless, the only (or ``best'', or simplest, or most natural) way to prove that ${\cal X}_0$ has property $P$ is to invoke the existence of the family ${\cal X}$.

So, to put this in the form of a question, what are some good examples of this phenomenon? I'm especially (but not exclusively) interested in examples that could be explained to an undergraduate.

(It was hard to find the right tags for this question; feel free to change them.)

Answer by Steven Landsburg for Is every projective $\mathbf{Z}[x]$-module free?

2013-03-03T00:39:44Z

While it's certainly true (per Fernando's comment) that this is a special case of the Quillen-Suslin theorem, it was certainly known long before Quillen and Suslin came along.

There's a paper of Murthy from the mid-1960s which shows that every projective $R[x]$-module is extended whenever $R$ is a regular ring of dimension at most 2. ("Extended" here means "of the form $P[x]$ where $P$ is a projective $R$-module". Since all projective ${\mathbb Z}$-modules are free, extended is equivalent to free in this case.)

But there's an even earlier paper of Bass which covers the case where $R$ is regular of dimension 1, which is all you need. The paper is called "Torsion Free and Projective Modules".

Edited to add: And the case of a PID predates even Bass; I think it's due to Seshadri in the 1950s.

Answer by Steven Landsburg for Formula for finite sum

2013-02-20T17:46:25Z

The generating function is:

$${x^2(3-4x)\over 2(2x-1)^2}$$

Edited to add: This is wrong for two reasons, both related to my not having carefully read the question. First, I treated the "less than or equal" signs as equal signs. Second, and more important, I failed to see that $k$ was fixed, and thought the sum ran over all possible values of $k$. Therefore this is still the right answer, but to an entirely different question.

Answer by Steven Landsburg for Set theory question

2013-02-21T01:22:49Z

There's a little bit of ambiguity in your question. Cohen's 1963 paper gives a model of $$ZF+\hbox{(not AC)} + CH$$ so this theory is consistent.

But when you say that you want to assume $ZF$ is consistent, I'm not sure whether you're asking about the above theory or the theory $$ZF+\hbox{Con(ZF)}+\hbox{not AC}+ CH$$

If the latter, then I don't know the answer to your question.

Answer by Steven Landsburg for many expected streaks imply high probability for a streak

2013-02-19T17:14:25Z

It seems to me that you're right; in principle there could be a few sequences with many streaks and many sequences with no streaks, yielding a high expected value and yet a low probability for a streak.

On the other hand, fix $k$ and let $f(n)$ be the number of sequences of length $n$ that contain a streak of heads having length $k$. Then clearly $f(1)=f(2)=...=f(k-1)=0$ and $f(k)=1$. And (unless I screwed up) it's not hard to get the following recursion: $$f(n)=2^{n-k}+(n-k)2^{n-k-1}-\sum_{j=k}^{n-k-1}f(j)2^{n-k-1-j}$$

The probability $g(n)$ that a randomly chosen sequence of length $n$ contains a string of heads of length $k$ is $f(n)/2^n$, which gives us the following recursion:

$$g(1)=g(2)=...=g(k-1)=0$$ $$g(k)=1/2^k$$ $$g(n)= {1\over 2^{k+1}}\left(2+n-k-\sum_{j=k}^{n-k-1}g(j)\right)$$

The generating function for $g$ is then

$${x^k\over(1-x)(2^k-2^{k-1}x-2^{k-2}x^2-...-x^k)}$$

The question, then, is what the $n$th power series coefficient looks like when $n$ is approximately $e^{2k}$. I don't have a theorem for you, but numerical tests suggest that this is very close to 1. In particular, even for $k=3$, we're looking at roughly the 403'd coefficient, which is approximately .9999999999999975 --- and this increases monotonically with $k$.

In other words, yes, at least one such streak is very likely to occur.

Answer by Steven Landsburg for Question on resolutions for arbitrary chain complexes.

2013-02-17T00:26:28Z

Start with a (possibly bounded) sequence of maps satisfying $dd=0$. Per Fernando's comment, you can always add an infinite number of zeroes on the left and/or right to create a ${\mathbb Z}$-graded complex. You can then build a Cartan-Eilenberg resolution of that ${\mathbb Z}$- graded complex. Your question (I think) is whether this Cartan-Eilenberg resolution can be truncated to give a Cartan-Eilenberg-like resolution of your original sequence (i.e. a resolution whose coboundaries and cohomology are resolutions of your original sequence's coboundaries and cohomology). The answer is yes, because (thinking of your original sequence as a row) the C-E construction puts a column of zeros wherever your row has a zero --- and throwing away columns of zeros can't change the coboundaries and cohomologies of the rows.

(This assumes that you define "cohomology" in the obvious way at the beginning and end of your sequence --- you haven't actually told us what your definition is.)

Answer by Steven Landsburg for Are all variables in a set of random variables independent if all pairs are independent?

2013-02-08T13:21:11Z

The simplest of the many standard counterexamples is when $(X_1,X_2,X_3)$ takes the values $(1,1,1)$, $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ all equiprobably.

Comment by Steven Landsburg

2013-05-24T21:08:20Z

It would be difficult to imagine anything more irrelevant to this question than the Axiom of Choice.

Comment by Steven Landsburg

2013-05-24T16:41:19Z

Rhett: There's no need to repeat what you've repeatedly said; we can let a jury of statistics professors decide which procedure is appropriate to the question as stated. Please send your contact info so we can confirm our bet.

Comment by Steven Landsburg

2013-05-24T16:30:23Z

A quantum particle is always in exactly one state.

Comment by Steven Landsburg

2013-05-24T16:16:53Z

Rhett: And I stand by the remainder of the bet as well: If you and I disagree about the proper procedure, we submit the question to 5 professors of statistics randomly chosen from the top ten departments. So if we disagree about the procedure, we can still sign our contract, leaving it to the professors to decide which procedure to follow. Do send me you're info; I'm eager to get this signed.

Comment by Steven Landsburg

2013-05-24T16:04:32Z

PS---note that this is exactly as in the original blog post ( thebigquestions.com/2010/12/27/… )which you've said repeatedly is wrong. So feel free to try to win my money.

Comment by Steven Landsburg

2013-05-24T16:03:30Z

Rhett: The conditions of the contract are exactly what I stated in my blog post --- i.e. the terms on which you've asserted repeatedly that you're sure to win. We throw the dice till we observe the 4th odd number. Then we compute the ratio of odds and evens. To guard against statistical flukes, we;ll repeat this experiment 3000 times (though if we're throwing physical dice, we might want to make it, say, 100 times instead) and take the average of the outcomes. I win your money if the answer is less than 47%; you win mine if it's over 47%.

Comment by Steven Landsburg

2013-05-24T15:03:21Z

Rhett: Sure! I'll be happy to substitute dice rolls for computer simulations. Email me your contact info so we can draw up a binding contract.

Comment by Steven Landsburg

2013-05-24T14:17:30Z

Rhett: so take my bet.

Comment by Steven Landsburg

2013-05-24T13:09:13Z

Rhett: Your position does not match Lubos's. Lubos claimed (incorrectly) that the expected value of the random variable $G/(G+B)$ is 1/2. You are claiming (downright ludicrously) that the random variable $G/(G+B)$ is identically equal to 1/2. It's nearly impossible you actually believe this. But if you do, please accept the bet I offered on my blog. If I win, I'll donate all my winnings to MathOverflow. If you're so very sure you're right, why not take the bet?

Comment by Steven Landsburg

2013-05-22T23:44:03Z

I don't follow the sentence beginning "Moreover, since $B$ is arbitrary.....". I thought $B$ was the set defined by the wff $x\notin x$.

Comment by Steven Landsburg

2013-05-22T20:50:44Z

This is not a math question.

Comment by Steven Landsburg

2013-05-22T16:52:53Z

I hope that people who might be tempted to answer this will restrain themselves; this is the third off-topic post by this user in the last three hours. Please don't encourage this.

Comment by Steven Landsburg

2013-05-22T02:44:50Z

Adding to Kevin's comment, this would still be true over a Dedekind ring, since $M/ker\phi$ is a submodule of a free module and hence projective.

Comment by Steven Landsburg

2013-05-20T22:44:15Z

Toink: I'm pretty sure there's not.

Comment by Steven Landsburg

2013-05-20T22:43:09Z

Adeel: But how do you define maps in the derived category without making choices?