Hint Take the standard approach: f(p,q)=f(m,n) → (pq,p^2-q^2)=(mn, m^2-n^2) So, pq=mn \quad \text{and} \quad p^2-q^2=m^2-n^2 p^2-\left(\frac{mn}{p}\right)^2=m^2-n^2 → p^4-p^2(m^2-n^2)-(mn)^2=0 ...

Suppose F \subset E is a destabilizing. Since \deg E = 0 then we must have deg F > 0. But \deg L^{-1} = 1 - g < 0 so the composition F \to L^{-1} must be zero. Thus F is a subsheaf of L ...

If you have a tripled (X, A, B) with A and B open in X you can define a relative cup product H^*(X, A) \times H^*(X, B) \to H^*(X, A \cup B) Define this on (co)chain level by taking ...

Your proof is mathematically sound. The key point to notice here is that every partition of X induces a set Y and a surjection from X onto Y; and every function with domain X induces a ...

By Fermat's Little Theorem, we see that x^3\equiv x\mod 3 for all x\in\Bbb{Z}. Hence, any (finite) rational solution over \Bbb{F}_3 will always have y = 0 (and it is easily checked that (0,a)\in\Bbb{F}_3\times\Bbb{F}_3 ...

There is a natural isomorphism \prod_{I\times J}\mathbb{Z}\cong\operatorname{Hom}\left(\bigoplus_I\mathbb{Z},\prod_J\mathbb{Z}\right), since Hom turns direct sums in the domain coordinate into ...