Al Don Gate

Hi, Let [math]L/K[/math] be a simple extension : [math]L/K[/math] is a a field extension and it exits [math]x\in L[/math] such that L=K(x). I consider the endomorphism of L [math]m_x[/math] define by [math]m_x(y)=xy[/math]. My problem : I find ( and i can explain if somebody want...) that the characteristic polynomial of m_x is 0 !!! It involves that the degree of the extension is 0??? How is it possible?

Hi, I like this one : [math] \sum_{i=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6} [/math] That's a wonderful thing that [math]\pi[/math] appears in the result of this sum. And i find the way you can prove that e.g. by Fourier series is quite elegant.

Hi, You know, it's not because an object is fractal his volume tends to [math]\infty[/math]. I think particularly to the sponge of Menger : [url=http://en.wikipedia.org/wiki/Menger_sponge][/url]http://en.wikipedia.org/wiki/Menger_sponge But I don't answer to your question...

Hello, I hope i have understood your question... Let be the multiplicative group G={-1,1} acting on itself by translation : [math](g,x)\mapsto gx[/math] If i look at the sub group G (which is well a sub group of G), G can't be written as th stabilizater of a element of G because : Stab(1)={1} Stab(-1)={1} But maybe you think at a stric subgroup of G... Hi, Another example more significant (i hope) : Let be [math]G=\mathbb{Z}/4\mathbb{Z}(=\{\overline{0},\overline{1},\overline{2},\overline{3}\})[/math] the additive subgroup of congruence modulo 4. We see that [math]H=\{\overline{0},\overline{2}\}[/math] is a subgroup of G. Suppose G act on itself by translation. Then there is no element of G which have H for stabilizer. And here H is a strict subgroup.

Hi, i'm Al, i have a problem... i'm French... and i love mathematics...