Codex @codex  0

A derivation of $A$ centred at the $k$-point $p:A\to k$ is a $k$-linear map $D:A\to k$ satisfying the twisted Leibniz ruleThe Zariski tangent space isSince $p(f)\ne 0$, $p$ extends uniquely to the localization $A[f^{-1}]$ by $p(a/f^n)=p(a)/p(f{)}^n$. Restriction gives a linear mapConversely, every $D\in T_{p}A$ has the unique extensionThe quotient rule verifies the Leibniz identity, and the assumption that $f$ is not a zero divisor makes the localization representation compatible with equality of fractions. Extension and restriction are inverse, so the natural map is an isomorphism.

Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

The three coordinate axes are the three irreducible components ofTheir common intersection is the origin. Likewise, if ${\ell }_1,{\ell }_2,{\ell }_3$ are distinct linear forms defining the three lines in ${\mathbb{A}}_k^2$, thenand its three irreducible components have only the origin in common. Any isomorphism of algebraic varieties would permute irreducible components and therefore map the origin of $V$ to the origin of $W$.

Every defining polynomial $xy,xz,yz$ has zero differential at the origin, soThe defining polynomial of $W$ is homogeneous of degree three, so its differential also vanishes at the origin, givingAn isomorphism induces an isomorphism of Zariski tangent spaces, which is impossible here. Hence $V$ and $W$ are not isomorphic. This is the tangent-space obstruction between two unions of three lines.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, whenwhere $T_{p}X$ is the Zariski tangent space; it is a singular point when $\dim T_{p}X>d$. Equivalently, the local ring ${\mathcal{O}}_{X,p}$ is regular exactly at a nonsingular point.

For an irreducible affine variety $X\subseteq {\mathbb{A}}^N$ over a perfect field, the Jacobian criterion expresses the singular locus by the vanishing of the relevant Jacobian minors, so it is Zariski closed. At a point where the tangent-space dimension is minimal it equals $d$; equivalently, some $(N-d)$-rowed Jacobian minor is nonzero. Its nonvanishing locus is therefore a nonempty Zariski-open set contained in the smooth locus. Every nonempty open subset of an irreducible topological space is dense, proving the density of the smooth locus.

Assume the ground field has characteristic other than two. Forthe partial derivatives are $2X_0,\dots ,2X_{n-1},0$. They vanish simultaneously at the unique projective pointThus this point is the singular locus of the projective quadric cone.

For varieties $X,Y$,If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. HenceandThis is the singular locus of a product with a smooth variety.

To construct the requested examples, put $r=n-k>0$. If $r\ge 2$, let $Q_r\subseteq {\mathbb{P}}^{r+1}$ be an irreducible quadric cone of dimension $r$ with one singular vertex. If $r=1$, use instead the irreducible cuspidal cubicwhose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embedin projective space by the Segre embedding. It is irreducible and has dimension $r+k=n$, while its singular locus is the vertex or cusp point times ${\mathbb{P}}^k$, which is nonempty and has dimension exactly $k$. This is the projective variety with a prescribed-dimensional singular locus construction.

Finally, suppose that the irreducible plane curve $C\subseteq {\mathbb{P}}^2$ were smooth of degree $d$. Since it is birational to a smooth projective curve of genus two, its geometric genus would be two. But the genus of a smooth plane curve isand no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.

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.

Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.

For $X\subseteq {\mathbb{A}}^n$ and $P\in X$, the Zariski tangent space isThe dimension can be defined as ${\min }_{P\in X}\dim T_{P}X$. Equivalently, the krull dimension of an affine variety is the supremum of lengths of strict chains of irreducible closed subsets, or the Krull dimension of $k[X]$. A point is singular when $\dim T_{P}X>\dim X$.

Forthe differentials give the tangent spaces of a product of two nodal line pairsThe variety is a union of four two-dimensional linear spaces. If both pairs $(x_1,x_2)$ and $(x_3,x_4)$ are nonzero, the two displayed equations are independent and $\dim T_{P}X=2$. If exactly one pair is zero, the dimension is three; at the origin it is four. Thus the singular locus is the union of the loci where either coordinate pair vanishes.

For $Y$, rank at most one is equivalent to vanishing of all $2\times 2$ minors:These homogeneous equations prove that $Y$ is projectively Zariski closed.

The following affine charts show both its dimension and smoothness. On $y_0=1$,so $(y_1,y_2)$ are free. On $y_1=1$,so $(y_0,y_3)$ are free. On $y_3=1$,and on $y_4=1$,These four charts cover $Y$, since the point with only $y_2$ nonzero violates the second minor. Every chart is isomorphic to ${\mathbb{A}}^2$. Hence the smooth projective surface from overlapping rank-one coordinates satisfiesand is nonsingular everywhere.

For $X\subseteq {\mathbb{A}}^n$ and $P\in X$, the Zariski tangent space isEquivalently, it is the kernel of the Jacobian of generators of $I(X)$ at $P$. With the tangent-space definition used here,and $P$ is smooth when $\dim T_{X,P}=\dim X$.

Now let $X=Z(f)$. Write $g$ for the square-free product of the distinct irreducible factors of $f$. Hilbert's Nullstellensatz givesThus $T_{X,P}=\ker (d{g}_P)$, whose dimension is at least $n-1$. There is some $P\in X$ with $d{g}_P\ne 0$. Otherwise every partial derivative of $g$ would vanish on $X$, hence belong to $(g)$ by the Nullstellensatz. Its degree is smaller than $\mathrm{deg}g$, so every partial derivative would be zero; in characteristic zero this would make $g$ constant, a contradiction. At such a point the tangent space has dimension $n-1$, and therefore

Next suppose $f$ is irreducible of degree at least two and $P$ is smooth. Translate $P$ to the origin and writeThen $H=T_{X,P}=Z(\ell )$ is a hyperplane. The restriction $f{|}_H$ is nonzero: otherwise $\ell$ would divide $f$, contradicting irreducibility and $\mathrm{deg}f\ge 2$. Since $f(P)=0$, the restriction is nonconstant, so the first part givesFor the natural intersection cut out by $(f,\ell )$, the two differentials at $P$ are $d{f}_P=\ell$ and $d\ell =\ell$. Hence they impose only one independent tangent equation andTherefore $X\cap T_{X,P}$ is singular at $P$.

Finally, represent $\phi :{\mathbb{A}}^2\to {\mathbb{A}}^3$ byIt is not injective exactly when $\mathrm{rank}A\le 1$, equivalently when all its $2\times 2$ minors vanish. Thusan ideal generated by three quadrics. By the stated assumption this is the full radical ideal of $Y$.

Every nonzero point of $Y$ has rank one. The action of $G{L}_3(k)\times G{L}_2(k)$ by row and column operations is transitive on rank-one matrices, so it suffices to compute atAt $A_0$, the differentials of the three minors are respectivelyHence $\dim T_{Y,A_0}=6-2=4$, and the same holds at every rank-one point. At the zero matrix all three quadrics have zero differential, so $T_{Y,0}={\mathbb{A}}^6$ has dimension six. It follows that

Over a perfect field, a point of a variety is smooth when its local ring is a regular local ring. For an irreducible variety of dimension $n$ at a closed point, this is equivalent to its Zariski tangent space having dimension $n$. The Jacobian criterion expresses the equality as a matrix-rank condition.

The union of the three coordinate axes in ${\mathbb{A}}^3$ and a union of three distinct lines through the origin in ${\mathbb{A}}^2$ are not isomorphic. Their origins are intrinsically the unique points common to all three irreducible components, but their Zariski tangent spaces there have dimensions three and two respectively.