If \mathbb C[X,Y]/(XY) had a nontrivial idempotent, then there would exist polynomials P(X), Q(Y) with XY \mid (P(X)+Q(Y))^2 - (P(X)+Q(Y)) and P(X)+Q(Y) \neq 0,1. Looking at the highest ...

Denote x=u^{v-1} + v^{u-1}. As you have proved, from FLT, x=u^{v-1} + v^{u-1} \equiv 1 (\mod u) and x=u^{v-1} + v^{u-1} \equiv 1 (\mod v). In other words, you have ...

Reposting my comment for completeness: The first-order Taylor approximation will do the job. You can apply it to \int f(x,y^*)\,dx with a fixed y^*. This gives an estimate for the integral ...

Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general). The join can be thought of as "lines" from X to Y, with some ...

The target space "Y" of these maps is supposed to be the same as the domain (otherwise, it would be quite a coincidence if f(x_0)=x_0, since x_0 is a point of the domain and f(x_0) is a point ...

I think you have the right thing in your mind. However, the function should be y=\frac{d-c}{b-a}(x - a) + c It maps, a to c and b to d.