User balazs strenner - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:10:52Z http://mathoverflow.net/feeds/user/11214 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/47846#47846 Answer by Balazs Strenner for Atiyah-MacDonald, exercise 2.11 Balazs Strenner 2010-12-01T00:44:07Z 2010-12-01T00:44:07Z <p>Here is another solution using only the Cayley-Hamilton Theorem for finitely generated modules (Proposition 2.4. in Atiyah-Macdonald) which, even though looks quite innocent, is a very powerful statement.</p> <p>Assume by contradiction that there is an injective map $\phi: A^m \to A^n$ with $m>n$. The first idea is that we regard $A^n$ as a submodule of $A^m$, say the submodule generated by the first $n$ coordinates. Then, by the Cayley-Hamilton Theorem, $\phi$ satisfies some polynomial equation $$\phi^k + a_{k-1} \phi^{k-1} + \cdots + a_1 \phi + a_0 = 0.$$ Using the injectivity of $\phi$ it is easy to see that if this polynomial has the minimal possible degree, then $a_0 \ne 0$. But then, applying this polynomial of $\phi$ to $(0,\ldots,0, 1)$, the last coordinate will be $a_0$ which is a contradiction as it should be zero. </p> http://mathoverflow.net/questions/8521/nice-proof-of-the-jordan-curve-theorem/8522#8522 Comment by Balazs Strenner Balazs Strenner 2011-05-31T04:47:46Z 2011-05-31T04:47:46Z In my opinion the Tietze-trick is the most beautiful part of the proof. I am also one of the many people grown up having told that the Jordan curve theorem is something quite difficult to prove. But after reading this proof I will sleep very well.