most recent 30 from http://mathoverflow.net 2013-05-25T22:58:41Z http://mathoverflow.net/feeds/user/17761 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81342/elementary-results-with-p-adic-numbers/81470#81470 guy 2011-11-21T00:53:53Z 2011-11-21T00:53:53Z

<p>You could mention Fermat's last theorem: Kummer's proof (in the regular case) uses properties of Bernoulli numbers that are close to the existence of the p-adic zeta function, and Wiles's proof uses p-adic numbers in at least 3 ways.</p>

http://mathoverflow.net/questions/33911/why-linear-algebra-is-funor/80928#80928 guy 2011-11-14T20:14:23Z 2011-11-14T20:14:23Z

<p>One nice application of linear algebra (mainly dimension theory) is the impossibility of the duplication of the cube (problem that dates back to the greeks and was solved only in 1837 by Wantzel).</p>

http://mathoverflow.net/questions/77195/how-has-modern-algebraic-geometry-affected-other-areas-of-math/77318#77318 guy 2011-10-06T01:08:13Z 2011-10-06T01:08:13Z

<p>The size of Fourier coefficients of modular forms can only be studied (so far) via the use of very sophisticated tools from Algebraic Geometry. Of course, one could argue that modular forms are part of Number Theory, but this is not how they arised and they appear in many othe branches of mathematics (combinatorics or theoretical physics for example).</p>

http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/75790#75790 guy 2011-09-18T20:52:10Z 2011-09-18T20:52:10Z

<p>Take two commutative rings $A$ and $B$ such that the polynomial rings $A[X]$ and $B[X]$ are isomorphic. Does this imply that $A$ and $B$ are isomorphic? (I think this is still open.)</p>

http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/75267#75267 guy 2011-09-13T01:18:05Z 2011-09-13T01:18:05Z

<p>Here is a riddle which proves to be extremely hard: Imagine a finite assembly in which some people happen to be friends (friendship is a symmetric relation but not transitive and you are not your own friend). Now it happens that anytime two persons have the same number of friends, they do not have any common friend. The conclusion to be proved is that there is at least one person that has one and only one friend.</p> <p>A proper generalization of the conclusion makes the riddle almost trivial.</p>