Al Don Gate

Hi, Let $L/K$ be a simple extension : $L/K$ is a a field extension and it exits $x\in L$ such that L=K(x). I consider the endomorphism of L $m_x$ define by $m_x(y)=xy$. 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 : $\sum_{i=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ That's a wonderful thing that $\pi$ 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 $\infty$. 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 : $(g,x)\mapsto gx$ 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 $G=\mathbb{Z}/4\mathbb{Z}(=\{\overline{0},\overline{1},\overline{2},\overline{3}\})$ the additive subgroup of congruence modulo 4. We see that $H=\{\overline{0},\overline{2}\}$ 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...