= Solution
Inside $\mathbb A^{n^2}$,
$$
GL_n(k)=D(\det)=\{A:\det A\ne0\},
$$
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 map
$$
A\longmapsto(A,(\det A)^{-1})
$$
identifies $GL_n$ with the closed hypersurface
$$
\{(A,t)\in\mathbb A^{n^2+1}:t\det A-1=0\}.
$$
In 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 $n\geq2$, 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.
Back to article page
