Qiaochu Yuan
Reputation
109/100 score
 1h comment How much information about $R-\mathrm{Mod}$ can be extracted from $\underline{R-\mathrm{Mod}}$ and $K_0(R)$? Fields, or more generally commutative rings, are Morita equivalent iff they are isomorphic; the center of a ring is a Morita invariant. 20h comment Why are algebraic cycles rational? Use Poincare duality over $\mathbb{Z}$. 23h awarded Nice Answer 1d awarded Nice Answer 1d awarded Good Answer 2d comment Where does the proof of $\sqrt 2$ is irrational break down when trying to prove the same for $\sqrt 4$? @Asaf: In a direct proof, I can check the statements I'm making against mathematical reality; in this case, I can look at examples of natural numbers, and I can verify that everything I've just said holds for those examples. In the standard proof by contradiction that $\sqrt{2}$ is irrational I can't check anything I'm saying against examples of positive integers $n, m$ such that $n^2 = 2m^2$, because there aren't any! 2d answered Where does the proof of $\sqrt 2$ is irrational break down when trying to prove the same for $\sqrt 4$? Apr24 revised computing Lefschetz number added 225 characters in body Apr24 revised computing Lefschetz number deleted 16 characters in body Apr24 answered computing Lefschetz number Apr24 comment Why does $\bigwedge^p(L_1\oplus\cdots\oplus L_p)\cong L_1\otimes\cdots\otimes L_p$? Prove a more general formula for the exterior powers of a direct sum. A clean way of stating it is that $\Lambda^{\bullet}(-)$ converts direct sums into (graded) tensor products. Apr24 comment Is $\mathbb{Z}(\sqrt[3]{5})$ a PID? Factorisation of the ideal $(2)$ Incidentally, solving Diophantine equations algorithmically is an incredibly hard problem, so it's not exactly surprising that WA is bad at it. The general problem is just undecidable, full stop (en.wikipedia.org/wiki/Diophantine_set#Matiyasevich.27s_theorem), and even the special case of Diophantine equations in two variables is open, at least as far as I know. Apr24 comment Is $\mathbb{Z}(\sqrt[3]{5})$ a PID? Factorisation of the ideal $(2)$ I don't know how WolframAlpha attempts to solve Diophantine equations, but checking a few examples it seems to know how to solve linear equations and Pell equations and that's about it. It told me that $y^2 = x^3 + 9$ has no integer solutions, but of course $(3, 6)$ is a solution. Apr24 answered Is $\mathbb{Z}(\sqrt[3]{5})$ a PID? Factorisation of the ideal $(2)$ Apr24 comment Are there terminologies distinguishing these two ranks? I find at least that in practice the second notion of rank never arises. Apr24 comment What are some good invariants for low dimensional Lie algebras? The dimension of the center, the signature of the Killing form... there's lots of stuff to do depending on what kind of computations you're willing to do and how much structure theory of Lie algebras you know. Also, I don't know what you mean by "orthonormal" here. Apr23 comment Prove that there is no smallest positive real number Because I wrote down true statements and true implications between them instead of pretend statements and untrustworthy implications between them, I have learned something true and valuable about the rationals beyond what I set out to prove, which is that the idea of $p$-adic valuation is useful. And indeed the use of $p$-adic valuations is a fundamental tactic in number theory. Apr23 comment Prove that there is no smallest positive real number Here is an example where the distinction is maybe clearer. It is standard to prove that $\sqrt{2}$ is irrational by contradiction: you assume that $\sqrt{2} = \frac{p}{q}$ and fiddle around a bit. But instead you can just prove the statement directly. $\sqrt{2}$ being irrational means that if $\frac{p}{q}$ is a rational number, then it doesn't square to $2$. Here is a direct proof: the $2$-adic valuation of the square of a number is even, and the $2$-adic valuation of $2$ is $1$. Again, the distinction between this proof and the proof by contradiction is that everything I've just said is true. Apr23 comment Prove that there is no smallest positive real number The problem with proofs by contradiction is that they are easy to mess up. In a long proof by contradiction, if you reach a contradiction at any point you can't tell whether you've finished your proof or whether you've made a mistake that introduced a new contradiction. When you start making statements using hypotheses you believe to be contradictory, you can't check your work based on what you believe to be true. This is an easy way to end up believing you've proven the Riemann hypothesis or whatever. Apr23 comment Prove that there is no smallest positive real number @Marc: again, there is absolutely no need to use proof by contradiction. The hypothesis that $r$ is the least positive real number is not used at all in the proof, which shows no more and no less than what I've shown above: that $r$ is not the least positive real number. The distinction between what I wrote and what the OP wrote is that at every step in my argument, every statement I've written down is a true statement about the real numbers. (I think there are other distinctions one can make about whether one argument is intuitionistically valid or not but I think this is less important.)