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An integral scheme is a nonempty scheme that is both reduced and irreducible. An affine scheme $\mathrm{Spec}A$ is integral exactly when $A$ is an integral domain.

A nonempty closed subset which, with its induced topology, is an irreducible topological space. In an affine algebraic set these subsets correspond, with reversed inclusion, to prime ideals of the coordinate ring.

An irreducible scheme is one whose underlying topological space is a nonempty irreducible topological space. It can have nilpotent elements in its structure sheaf; irreducibility concerns only the topology. A reduced scheme is an integral scheme precisely when it is irreducible.

An algebraic variety whose underlying topological space with its Zariski topology is an irreducible topological space: it is nonempty and cannot be a union of two proper closed subsets. For a reduced affine variety, this is equivalent to its coordinate ring being an integral domain.

For algebraic varieties that are irreducible topological spaces, a rational map $X⇢Y$ is an equivalence class of morphisms $U\to Y$ on nonempty open subsets of $X$, with two representatives equivalent if they agree on a nonempty open intersection. Such subsets are dense. A birational map has a rational inverse; equivalently it restricts to an isomorphism between dense open subsets. In function field language it induces an isomorphism $k(Y)\cong k(X)$ over the ground field.

The Hilbert basis theorem makes $k[x_1,\dots ,x_n]$ a Noetherian ring. Reversing the inclusion of ideals shows that its Zariski-closed subsets satisfy the descending-chain condition, so every affine variety is a Noetherian topological space.

Suppose some nonempty closed subset had no finite decomposition into irreducible closed subsets. It would be reducible, so it could be writtenwith $V_1,V_2$ proper closed subsets. At least one of these would again admit no finite irreducible decomposition. Repeating that choice would produce an infinite strictly descending chain of closed subsets, contradicting Noetherianity. ThereforeRemoving any member contained in another leaves exactly the irreducible components.

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.

In the classical convention over an algebraically closed field $k$, a prevariety is an irreducible topological space with a sheaf of rings which has a finite open cover by affine varieties, with compatible local identifications. A prevariety need not be separated. A classical algebraic variety is a separated prevariety: its diagonal morphism has closed image. Some authors allow reducible prevarieties and varieties; the separation criterion and product construction remain the same under that convention.