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The induced map on the spectrum of a commutative ring is contraction of an ideal:To see that it is well defined, let $\mathfrak{q}$ be a prime ideal of $S$. Its inverse image is an ideal of $R$, and it is proper because $\phi (1)=1\notin \mathfrak{q}$. If $ab\in {\phi }^{-1}(\mathfrak{q})$, then $\phi (a)\phi (b)\in \mathfrak{q}$, whence $\phi (a)\in \mathfrak{q}$ or $\phi (b)\in \mathfrak{q}$. This proves that the inverse image is a prime ideal.

For any ideal $I\subseteq R$, its extension of an ideal to $S$ is $IS=\phi (I)S$. We haveThus inverse images of Zariski-closed sets are Zariski-closed sets, which proves that $f$ is a continuous map.

We first establish the nilradical as the intersection of prime ideals:Every nilpotent element belongs to every prime ideal. In the other direction, if $a$ is not nilpotent, the multiplicative subset $T=\{1,a,a^2,\dots \}$ avoids zero. By the Zorn lemma, there is an ideal $J$ maximal among those disjoint from $T$. It is a prime ideal: if $uv\in J$ and neither $u$ nor $v$ belongs to $J$, then $J+(u)$ and $J+(v)$ each meet $T$. Multiplying elements in these two intersections gives an element of $T$ lying in $J$, a contradiction. Since $a\notin J$, this proves the identity. The identity also holds for the zero ring, with the intersection over an empty family interpreted as the whole ring.

A Zariski-closed set $V_R(I)$ contains the image of $f$ exactly whenThe last equality follows because $\phi (r)$ is a nilpotent element exactly when $r^n\in \ker \phi$ for some $n\ge 1$. Consequently the smallest Zariski-closed set containing the image, namely its closure, isThis is all of $\mathrm{Spec}R$ precisely when the kernel of a ring homomorphism is contained in every prime ideal of $R$. Hence

Write a polynomial $f\in W$ by its coefficients. Evaluation of $f$ and of each first partial derivative at $p$ is polynomial in these coefficients and in the coordinates of $p$. Consequently the bad incidence locusis Zariski closed, and $U=(W\times k^3)\setminus B$ is Zariski open. Here a hypersurface is regarded as smooth at an ambient point outside it automatically.

In fact $B$ is empty. For homogeneous $f$ of degree $d>0$, the Euler homogeneous function theorem saysIf $(f,p)\in B$, its left side is zero while its right side is $d$, which is nonzero because $\mathrm{char}k=0$. Thus every nonzero level is covered by the smoothness of a nonzero homogeneous level set, andwhich is certainly nonempty and Zariski open.

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 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.

Write a conic as $x^{T}Ax=0$ for a symmetric $3\times 3$ matrix $A$, with off-diagonal entries equal to half the corresponding coefficients. A union of two distinct lines has matrix rank two. Thus $R^{\circ }$ is the rank-two part of the determinant hypersurface $\det A=0$. Its closure also contains rank-one matrices, representing double lines, so it is not Zariski closed. Since the determinant is one nonzero homogeneous equation in ${\mathbb{P}}^5$, its dense rank-two locus has dimension $4$.

The final sentence of the converted question writes $R^c$, but the natural intended set is $R^{\circ }$. The homogeneous ideal of its closure is the principal ideal generated by the cubic polynomial $\det A$: the determinant hypersurface is irreducible, and $R^{\circ }$ is dense in it.

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.

A Zariski-open set is the complement of a Zariski-closed set. A distinguished affine open has the form $D(f)=\{x:f(x)\ne 0\}$.