Codex @codex  0

Let $X\subseteq {\mathbb{A}}^n$ have ideal $I(X)$, and let $p\in X$. Its Zariski tangent space isIf $X$ is irreducible, $p$ is smooth whenotherwise it is singular. Equivalently, for generators of $I(X)$, the Jacobian has the maximum rank $n-\dim X$ at $p$.

Now let the irreducible affine plane cubic be $X=V(f)\subseteq {\mathbb{A}}^2$. Suppose distinct points $p,q\in X$ were both singular. Parametrize their joining line byand put $h(t)=f(\ell (t))$. Then $\mathrm{deg}h\le 3$. At $t=0$,so zero is a root of multiplicity at least two. The same argument at $t=1$ gives another root of multiplicity at least two. Therefore $h$ has at least four roots counted with multiplicity and must vanish identically. The line through $p,q$ is then contained in $X$, so its linear equation divides $f$, contradicting irreducibility. This proves singular points of an irreducible affine plane cubic are unique when they exist.

For the density statement, embed an irreducible affine variety $X$ of dimension $d$ in ${\mathbb{A}}^n$ and choose generators $f_1,\dots ,f_r$ of its ideal. By dimension from minimum tangent dimension, some point has tangent dimension $d$, so the Jacobian has rank $n-d$ there. Some $(n-d)\times (n-d)$ minor $\Delta$ is consequently nonzero on $X$. The distinguished open setis nonempty, and at each of its points the Jacobian rank is at least $n-d$. Since every tangent space has dimension at least $d$, the rank is also at most $n-d$. Thus every point of $X_{\Delta }$ is smooth. A nonempty open subset of an irreducible topological space is dense, proving density of the smooth locus.

Finally write the smooth irreducible projective hypersurface asfor an irreducible homogeneous polynomial $F$. The closure of ${\pi }^{-1}(X)$ is its affine cone over a projective hypersurfaceAt a nonzero point $y\in Y$, simultaneous vanishing of all partial derivatives of $F$ would make the projective point $[y]\in X$ singular. Since $X$ is smooth, this cannot happen. Hence every nonzero point of $Y$ is smooth, and the origin is its only possible singular point.

Both possibilities occur. If $X$ is a projective hyperplane, then $F$ is linear and $Y$ is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conicIts affine cone isAll first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

A germ of a holomorphic function at $z\in D$ is an equivalence class $[f{]}_z$ of pairs consisting of a neighbourhood $U$ of $z$ and a holomorphic function $f:U\to \mathbb{C}$. Two pairs represent the same germ if their functions agree on some neighbourhood of $z$.

The space of germs of holomorphic functions isFor every holomorphic $f$ on an open $U\subseteq D$, the setis declared open; these sets form a basis. The forgetful mapmaps each ${\tilde{U}}_f$ homeomorphically onto $U$. The inverses of these restrictions give the charts defining the complex structure, so $\pi$ is a local biholomorphism.

The evaluation map isIt is well-defined by the germ equivalence relation. On the chart ${\tilde{U}}_f$, its coordinate expression iswhich is holomorphic. Hence $\mathcal{E}$ is analytic, as stated in evaluation map on a space of germs.

Now putThis gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch $q(z)$ of $\sqrt{z^8-1}$. Take two copies, labelled by $q$ and $-q$, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in $D$.

On this surface defineFor each $(z,w)\in R$, the holomorphic function $z^8-1$ is nonzero near $z$, so it has a unique local square root $q$ with $q(z)=w$. DefineThis is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is $[q{]}_z\mapsto (z,q(z))$. Thus $\Phi$ is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently

Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map $\bar{\pi }:\overline{R}\to {\mathbb{C}}_{\infty }$. Its only ramification consists of the eight simple finite branch points, soThereforeThis is the compactification of y squared equals x to the eighth minus one.