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A morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^1$ is represented by homogeneous forms $F,G$ of the same degree with no common projective zero. If that degree were positive, the two plane curves $F=0$ and $G=0$ would intersect, producing a base point. Thus the degree is zero and $\varphi$ is constant.

If a closed subvariety of $({\mathbb{P}}^1{)}^3$ were isomorphic to ${\mathbb{P}}^2$, composing its embedding with each coordinate projection would give three constant maps. Their product would be constant, contradicting that it is an embedding.

The Riemann--Hurwitz theorem statesfor a degree-$d$ nonconstant map of smooth projective curves.

Project $X$ to the second ${\mathbb{P}}^1$. This is a degree-two map. Its quadratic discriminant in the $x$ variables is homogeneous of degree six in $y$, so smoothness gives six branch points counted with multiplicity. Henceand $g_X=2$. A smooth plane curve has genus $(d-1)(d-2)/2$, which is never two for an integer $d$, so $X$ is not isomorphic to a smooth plane curve.

For $K=M\cup N$, the simplicial Mayer--Vietoris sequence is

Construct a simplicial mapping cone of a degree-$k$ map of a circle. Explicitly, let the target circle have vertices $v_0,v_1,v_2$, and let the source circle have vertices $w_0,\dots ,w_{3k-1}$. Map $w_j$ to $v_{j\mathrm{mod}3}$, triangulate each quadrilateral of its mapping cylinder, and cone the source circle to one new vertex. The resulting finite simplicial complex is the mapping cone $K_k$.

Its reduced cellular, or equivalently simplicial, chain complex collapses toin degrees two and one. ConsequentlyThe same computation follows from Mayer--Vietoris applied to the cone and the mapping-cylinder neighborhood.

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

The Möbius function isMöbius inversion says thatNowis the sum of $(-1{)}^{|S|}$ over subsets of the primes whose $k$th powers divide $n$. It is one if there are no such primes and zero otherwise, proving the power-free criterion.

LetGrouping all $n$th roots of unity by their exact order givesMöbius inversion therefore yields $c_n=\mu (n)$.