 Solution

Inside ${\mathbb{A}}^{n^2}$,a nonempty Zariski-open set. Affine space is smooth and irreducible; smoothness passes to open subsets, and every nonempty open subset of an irreducible space is irreducible.

The mapidentifies $G{L}_n$ with the closed hypersurfaceIn these coordinates multiplication is $(A,s)(B,t)=(AB,st)$, whose entries are polynomial functions, so it is a morphism.

A morphism ${\mathbb{A}}^n\to {\mathbb{A}}^1\setminus \{0\}$ is a polynomial with no zero. Over an algebraically closed field every nonconstant polynomial in $n$ variables has a zero: specialize all but one variable so that a nonconstant coefficient remains, then use algebraic closure. The morphism is therefore constant.

A morphism ${\mathbb{P}}^n\to {\mathbb{P}}^1$ is represented by homogeneous polynomials $F,G$ of the same degree with no common projective zero. If the degree is positive and $n\ge 2$, the Projective dimension theorem forces the two hypersurfaces $V(F)$ and $V(G)$ to meet. Thus the degree is zero and the morphism is constant.