Timeline for What elementary problems can you solve with schemes?

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Jul 29, 2017 at 23:36	comment	added	Noam D. Elkies		That's a good elementary approach, but you need to use ${\mathbb C}^{n\times n}$, not ${\mathbb R}^{n\times n}$, because the matrices diagonalizable over the reals are not dense in ${\mathbb R}^{n\times n}$ once $n \geq 2$. Deducing the result for arbitrary fields doesn't require the infinite transcendence degree of $\mathbb R$: for each $n$, Cayley-Hamilton is a polynomial identity with integral coefficients, so once it's proved over $\mathbb Q$ it works for all fields (induction on the number of variables).
May 4, 2011 at 14:30	comment	added	Johannes Hahn		You don't even need that: You prove density in $\mathbb{R}^{n\times n}$ w.r.t. the usual topology and derive the Cayley-Hamilton-theorem from that and then use the fact that there are infinitely many algebraicly independent real numbers. This allows you to view $\mathbb{Z}[X_{11},...,X_{nn}]$ as a subring of $\mathbb{R}$ and by specializing $X_{ij}$ to the value $x_{ij}\in R$ in any fixed commutative ring $R$ you can transport the Cayley-Hamilton-equation from $\mathbb{Z}$ to $R$.
May 4, 2011 at 10:16	comment	added	Asaf		You can prove this result without the usage of schemes. One just need to know what's an irreducible variety is.
May 4, 2011 at 4:43	history	answered	Yuhao Huang	CC BY-SA 3.0