Let $X = \{P\}$ be a single point. The only irreducible closed subset is $\{P\}$ itself, so the longest chain is $Z_0 = \{P\}$, which has length 0. Thus $\dim\{P\} = 0$.

In $\mathbb{A}^1$ with the Zariski topology, the irreducible closed subsets are:

• The whole space $\mathbb{A}^1$ (corresponding to the prime ideal $(0)$).
• Individual points $\{a\}$ for $a \in k$ (corresponding to maximal ideals $(x - a)$).

The longest chain of irreducible closed subsets is:

$\{a\} \subsetneq \mathbb{A}^1,$

which has length 1. Therefore $\dim \mathbb{A}^1 = 1$.

In $\mathbb{A}^2$, the irreducible closed subsets are:

• $\mathbb{A}^2$ itself (prime ideal $(0)$).
• Irreducible curves $V(f)$ for irreducible $f \in k[x,y]$ (prime ideals $(f)$).
• Points $\{(a,b)\}$ (maximal ideals $(x-a, y-b)$).

A maximal chain:

$\{(0,0)\} \subsetneq V(y) \subsetneq \mathbb{A}^2,$

which has length 2. We cannot insert any irreducible closed set strictly between a point and an irreducible curve, or between an irreducible curve and $\mathbb{A}^2$ (this follows from the fact that $k[x,y]$ has Krull dimension 2). Therefore $\dim \mathbb{A}^2 = 2$.

By the same reasoning, the irreducible closed subsets of $\mathbb{A}^n$ are in bijection with prime ideals of $k[x_1,\ldots,x_n]$ (via $V \leftrightarrow I$). A maximal chain of prime ideals in $k[x_1,\ldots,x_n]$ is:

$(0) \subsetneq (x_1) \subsetneq (x_1, x_2) \subsetneq \cdots \subsetneq (x_1, \ldots, x_n),$

which has length $n$. This gives $\dim \mathbb{A}^n = n$. More precisely, the Krull dimension of the polynomial ring $k[x_1,\ldots,x_n]$ is $n$ (a nontrivial theorem in commutative algebra).

Let $Y = V(y - x^2) \subseteq \mathbb{A}^2$. Then $k[Y] = k[x,y]/(y-x^2) \cong k[x]$. The prime ideals of $k[x]$ are $(0)$ and $(x - a)$ for $a \in k$. The longest chain is $(0) \subsetneq (x-a)$, of length 1. So $\dim Y = 1$. This confirms that the parabola is a curve.

Let $Y = V(y^2 - x^3)$. Then $k[Y] = k[x,y]/(y^2 - x^3) \cong k[t^2, t^3]$. This is an integral domain of Krull dimension 1: the only prime ideals are $(0)$ and the maximal ideals (one for each point of $Y$). So $\dim Y = 1$ --- the cuspidal cubic is a curve, despite its singularity.

Let $X = V(xy) \subseteq \mathbb{A}^2$, the union of the coordinate axes. The ring $k[x,y]/(xy)$ has the following prime ideals: $(x)$, $(y)$ (corresponding to the two irreducible components), and $(x, y-a)$, $(x-a, y)$ for $a \in k$ (the points on each axis). A longest chain is:

$(x) \subsetneq (x, y-a),$

which has length 1. So $\dim X = 1$. For a reducible algebraic set, the dimension is the maximum of the dimensions of its irreducible components.

$K(\mathbb{A}^n) = k(x_1, \ldots, x_n)$, the field of rational functions in $n$ variables. The elements $x_1, \ldots, x_n$ are algebraically independent over $k$, and $K(\mathbb{A}^n)$ is generated by them, so

$\operatorname{tr.deg}_k\, k(x_1,\ldots,x_n) = n.$

$Y = V(y - x^2)$. Then $k[Y] \cong k[x]$, so $K(Y) = k(x)$. This is a purely transcendental extension of $k$ of degree 1, so $\operatorname{tr.deg}_k\, K(Y) = 1$.

$Y = V(y^2 - x^3)$. Then $k[Y] = k[t^2, t^3]$ and $K(Y) = k(t)$ (since $t = t^3/t^2 \in K(Y)$). So $\operatorname{tr.deg}_k\, K(Y) = 1$.

Let $Y = V(z - xy) \subseteq \mathbb{A}^3$. Then $k[Y] = k[x,y,z]/(z-xy) \cong k[x,y]$ via $z \mapsto xy$, so $K(Y) = k(x,y)$. We have $\operatorname{tr.deg}_k\, k(x,y) = 2$, confirming $Y$ is a surface.

The twisted cubic $Y = V(y-x^2, z-x^3) \subseteq \mathbb{A}^3$ has coordinate ring $k[Y] \cong k[x]$. The element $x$ is algebraically independent over $k$, and $k[Y]$ is already equal to $k[x]$. So $d = 1$ and the Noether normalization map is the identity: $Y \to \mathbb{A}^1$ given by the coordinate $x$.

This confirms $\dim Y = \operatorname{tr.deg}_k\, k(x) = 1$.

Consider $Y = V(x^2 + y^2 - 1) \subseteq \mathbb{A}^2$ over $\mathbb{C}$. The coordinate ring is $A = k[x,y]/(x^2+y^2-1)$. We need to find an element $t \in A$ such that $A$ is a finite module over $k[t]$.

Take $t = x$. In $A$, we have $y^2 = 1 - x^2$, so $y$ satisfies the monic polynomial $T^2 - (1-t^2) = 0$ over $k[t]$. Thus $A = k[t] + k[t] \cdot y$ is a free $k[t]$-module of rank 2, and $A$ is integral over $k[t]$.

The Noether normalization is the projection $Y \to \mathbb{A}^1$, $(x,y) \mapsto x$, which is a 2-to-1 map (generically). This confirms $\dim Y = 1$.

$\dim \mathbb{P}^n = n$. Indeed, $\mathbb{P}^n$ contains the affine open set $U_0 \cong \mathbb{A}^n$, and $\dim \mathbb{A}^n = n$. The chain of irreducible closed subsets is:

$\{[1:0:\cdots:0]\} \subsetneq V(x_1) \subsetneq V(x_1,x_2) \subsetneq \cdots \subsetneq V(x_1,\ldots,x_{n-1}) \subsetneq \mathbb{P}^n,$

where we abuse notation and write $V(x_1) = V(x_1/x_0)$ restricted to the standard chart.

More precisely, in homogeneous coordinates, a maximal chain is:

$[1:0:\cdots:0] \subsetneq V(X_1) \cap \cdots \cap V(X_n) \subsetneq V(X_1) \cap \cdots \cap V(X_{n-1}) \subsetneq \cdots \subsetneq V(X_1) \subsetneq \mathbb{P}^n.$

The Fermat cubic $V(X^3 + Y^3 - Z^3) \subseteq \mathbb{P}^2$ is an irreducible projective variety. In the affine chart $Z \neq 0$, it becomes $V(x^3 + y^3 - 1) \subseteq \mathbb{A}^2$. The coordinate ring is $k[x,y]/(x^3+y^3-1)$, which has Krull dimension 1 (since $x^3 + y^3 - 1$ is irreducible, it generates a prime ideal of height 1 in $k[x,y]$). So the Fermat cubic is a curve of dimension 1.

In $\mathbb{A}^3$ with coordinates $(x,y,z)$:

• $V(z) \cong \mathbb{A}^2$: a plane, dimension 2. (A hyperplane is always a hypersurface.)
• $V(x^2 + y^2 + z^2 - 1)$: a sphere (over $\mathbb{C}$), dimension 2.
• $V(xy - z^2)$: a quadric cone, dimension 2. This is singular at the origin (the vertex of the cone).
• $V(x^2 + y^3 + z^5)$: an $E_8$ surface singularity, dimension 2.

All hypersurfaces in $\mathbb{A}^3$ (defined by a single irreducible equation) have dimension $3 - 1 = 2$: they are surfaces.

$Q = V(X_0 X_3 - X_1 X_2) \subseteq \mathbb{P}^3$ is an irreducible quadric hypersurface, so $\dim Q = 3 - 1 = 2$. We saw earlier (Segre embedding) that $Q \cong \mathbb{P}^1 \times \mathbb{P}^1$, which is consistent: $\dim(\mathbb{P}^1 \times \mathbb{P}^1) = 1 + 1 = 2$.

In $\mathbb{A}^n$:

| Subvariety | Dimension | Codimension in $\mathbb{A}^n$ | |---|---|---| | $\mathbb{A}^n$ itself | $n$ | $0$ | | Hypersurface $V(f)$, $f$ irred. | $n-1$ | $1$ | | A point $\{P\}$ | $0$ | $n$ | | A line | $1$ | $n-1$ | | $V(x_1,\ldots,x_r)$ ($r$ coordinate hyperplanes) | $n - r$ | $r$ |

Codimension-1 subvarieties are hypersurfaces. Codimension-$n$ subvarieties (in $\mathbb{A}^n$) are points.

Let $Y = V(y)$ and $Z = V(y - x)$ in $\mathbb{A}^2$. Both have dimension 1. Their intersection is $V(y, y-x) = V(x,y) = \{(0,0)\}$, which has dimension 0. We check:

$\dim(Y \cap Z) = 0 \geq 1 + 1 - 2 = 0.$

Equality holds --- this is a "transverse" intersection.

$Y = V(y)$ and $Z = V(y-1)$ in $\mathbb{A}^2$. Then $Y \cap Z = V(y, y-1) = \varnothing$. The theorem gives $\dim W \geq 0$ for any component $W$, but the intersection is empty, so there is no contradiction.

In $\mathbb{P}^2$, these lines do meet at the point at infinity $[1:0:0]$ (homogenize: $Y = 0$ and $Y - Z = 0$ give $Z = 0$, $Y = 0$).

Let $Y = V(z) \subseteq \mathbb{A}^3$ (a plane, dimension 2) and $Z = V(x,y) \subseteq \mathbb{A}^3$ (the $z$-axis, dimension 1). Then $Y \cap Z = V(x,y,z) = \{(0,0,0)\}$, dimension 0. The bound gives:

$\dim(Y \cap Z) = 0 \geq 2 + 1 - 3 = 0.$

Again, equality holds.

Let $Y = V(x,y) \subseteq \mathbb{A}^4$ (a plane, dimension 2) and $Z = V(x,z) \subseteq \mathbb{A}^4$ (another plane, dimension 2). Then

$Y \cap Z = V(x,y,z) \cong \mathbb{A}^1, \quad \dim(Y \cap Z) = 1.$

The lower bound is $2 + 2 - 4 = 0$, but the actual dimension is 1 --- the intersection is "excess" by 1. This happens because $Y$ and $Z$ share the common direction along the $w$-axis and are not "in general position."

Let $f = x^2 + y^2 + z^2 \in k[x,y,z]$. The ideal $(f)$ is prime (since $f$ is irreducible over $\mathbb{C}$, using the fact that $x^2 + y^2 + z^2 = (x + iy + \sqrt{-1-1}\, z)(\cdots)$ --- actually, $x^2+y^2+z^2$ is irreducible over $\mathbb{C}$ because it cannot factor as a product of two linear forms). By the Hauptidealsatz, the minimal prime over $(f)$ is $(f)$ itself with height 1. So $\dim V(f) = 3 - 1 = 2$.

In $\mathbb{A}^3$, consider $V(x, y)$: this is the $z$-axis. The ideal $(x,y)$ has height 2 in $k[x,y,z]$, and $\dim V(x,y) = 1 = 3 - 2$. The generalized principal ideal theorem guarantees $\operatorname{ht}(x,y) \leq 2$, and indeed equality holds here.

Now consider $V(x^2, xy)$. We have $V(x^2, xy) = V(x) \cup V(x, y) = V(x)$, which is the entire $yz$-plane of dimension 2. The minimal prime over $(x^2, xy)$ is $(x)$, which has height 1 --- well within the bound of 2.

Consider $V(xy, xz) \subseteq \mathbb{A}^3$. The ideal $(xy, xz)$ is generated by 2 elements, so the generalized Hauptidealsatz guarantees each component has codimension $\leq 2$. The decomposition is

$V(xy, xz) = V(x) \cup V(y,z).$

Here $V(x)$ has codimension 1 and $V(y,z)$ has codimension 2. Both are within the bound. The variety $V(x)$ has codimension strictly less than 2 --- the two equations $xy = 0$ and $xz = 0$ do not impose independent conditions along $V(x)$.

The twisted cubic $C \subseteq \mathbb{A}^3$ is the image of $t \mapsto (t, t^2, t^3)$. Its coordinate ring is $k[C] = k[x,y,z]/(y-x^2, z-x^3) \cong k[x]$, so

$\dim C = \dim k[x] = 1.$

Alternatively, $K(C) = k(x)$, so $\operatorname{tr.deg}_k K(C) = 1$. The twisted cubic is a curve --- a 1-dimensional subvariety of $\mathbb{A}^3$ with codimension 2.

Note that $C$ is defined by the ideal $(y-x^2, z-x^3)$, which has height 2 in $k[x,y,z]$, consistent with $\dim C = 3 - 2 = 1$.

The rational normal curve of degree $d$ is the image of $\nu_d : \mathbb{P}^1 \to \mathbb{P}^d$ given by

$[s:t] \mapsto [s^d : s^{d-1}t : \cdots : t^d].$

Since $\nu_d$ is an isomorphism onto its image, $\dim \nu_d(\mathbb{P}^1) = \dim \mathbb{P}^1 = 1$. The rational normal curve is a curve of codimension $d - 1$ in $\mathbb{P}^d$.

The Grassmannian $\mathbb{G}(r, n)$ parametrizes $r$-dimensional subspaces of $k^n$. An $r$-dimensional subspace is specified by an $r \times n$ matrix of rank $r$, modulo the action of $\operatorname{GL}_r$. The space of $r \times n$ matrices has dimension $rn$, and $\operatorname{GL}_r$ has dimension $r^2$, so

$\dim \mathbb{G}(r,n) = rn - r^2 = r(n - r).$

Concrete cases:

| Grassmannian | Parametrizes | Dimension | |---|---|---| | $\mathbb{G}(1, n) = \mathbb{P}^{n-1}$ | Lines through origin in $k^n$ | $n - 1$ | | $\mathbb{G}(2, 4)$ | Lines in $\mathbb{P}^3$ | $2 \cdot 2 = 4$ | | $\mathbb{G}(2, 5)$ | Lines in $\mathbb{P}^4$ | $2 \cdot 3 = 6$ | | $\mathbb{G}(3, 6)$ | Planes in $\mathbb{P}^5$ | $3 \cdot 3 = 9$ |

The Grassmannian $\mathbb{G}(2,4)$ sits inside $\mathbb{P}^5$ via the Plucker embedding, where it is a quadric hypersurface. Check: $\dim \mathbb{G}(2,4) = 4 = \dim \mathbb{P}^5 - 1 = 5 - 1$. This is consistent with $\mathbb{G}(2,4)$ being a hypersurface in $\mathbb{P}^5$.

A complete flag in $k^n$ is a sequence

$0 = V_0 \subsetneq V_1 \subsetneq V_2 \subsetneq \cdots \subsetneq V_n = k^n$

where $\dim V_i = i$. The flag variety $\operatorname{Fl}(n) = \operatorname{Fl}(1,2,\ldots,n-1;\, n)$ parametrizes all such flags. It is a projective variety of dimension

$\dim \operatorname{Fl}(n) = \frac{n(n-1)}{2}.$

This can be computed by a "tower of fibrations": choosing $V_1 \subseteq V_2 \subseteq \cdots$ successively, one gets a tower

$\operatorname{Fl}(n) \to \mathbb{G}(n-1, n) \to \mathbb{G}(n-2, n) \to \cdots$

At each step, the fiber is a projective space. Alternatively, $\operatorname{Fl}(n) = \operatorname{GL}_n / B$ where $B$ is the Borel subgroup of upper triangular matrices, and $\dim \operatorname{Fl}(n) = \dim \operatorname{GL}_n - \dim B = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$.

For $n = 3$: $\dim \operatorname{Fl}(3) = 3$, parametrizing flags $0 \subsetneq V_1 \subsetneq V_2 \subsetneq k^3$ (a point in a line in a plane in 3-space).

More generally, a partial flag variety $\operatorname{Fl}(d_1, \ldots, d_s;\, n)$ parametrizing flags $V_{d_1} \subsetneq \cdots \subsetneq V_{d_s} \subsetneq k^n$ has dimension

$\dim \operatorname{Fl}(d_1, \ldots, d_s;\, n) = \sum_{i=1}^{s} d_i(d_{i+1} - d_i)$

where $d_{s+1} = n$.

Let $M_{m \times n}(k) \cong \mathbb{A}^{mn}$ be the space of $m \times n$ matrices, and define

$\mathcal{M}_r = \{A \in M_{m \times n}(k) \mid \operatorname{rank}(A) \leq r\}.$

This is the determinantal variety defined by the vanishing of all $(r+1) \times (r+1)$ minors of a generic $m \times n$ matrix. Its dimension is

$\dim \mathcal{M}_r = r(m + n - r).$

To see this: a matrix of rank exactly $r$ is determined by its image (an $r$-dimensional subspace of $k^m$, parametrized by $\mathbb{G}(r,m)$ of dimension $r(m-r)$) together with a surjective linear map $k^n \to k^r$ (which, modulo the choice of basis, contributes $rn - r^2 = r(n-r)$ parameters). Total: $r(m-r) + r(n-r) = r(m+n-2r)$, but we must also account for the $r^2$ parameters in $\operatorname{GL}_r$. A cleaner computation:

$\dim \mathcal{M}_r = \dim\{(A,V) \mid \operatorname{Im}(A) \subseteq V,\; \dim V = r\} = \dim \mathbb{G}(r,m) + rn = r(m-r) + rn = r(m+n-r).$

Special cases:

• $r = 0$: $\mathcal{M}_0 = \{0\}$, dimension 0. Check: $0 \cdot (m+n-0) = 0$.
• $r = 1$: $\mathcal{M}_1$ has dimension $m + n - 1$. For $2 \times 2$ matrices ($m = n = 2$): $\dim \mathcal{M}_1 = 1 \cdot 3 = 3$. This is the locus $V(\det) \subseteq \mathbb{A}^4$, a hypersurface of dimension 3 in $\mathbb{A}^4$. Consistent!
• $r = \min(m,n) - 1$: the variety $V(\det)$ (for square matrices $m = n$) has dimension $n^2 - 1$, which is codimension 1 in $\mathbb{A}^{n^2}$.

For the space $M_n(k) \cong \mathbb{A}^{n^2}$ of $n \times n$ matrices, the locus of singular matrices is

$V(\det) = \mathcal{M}_{n-1} = \{A \mid \det(A) = 0\}.$

Since $\det$ is an irreducible polynomial of degree $n$ in the $n^2$ matrix entries, $V(\det)$ is an irreducible hypersurface of dimension $n^2 - 1$.

Let us verify using the determinantal formula: $\dim \mathcal{M}_{n-1} = (n-1)(n + n - (n-1)) = (n-1)(n+1) = n^2 - 1$.

For $n = 2$: $\det = ad - bc$ on $\mathbb{A}^4$, giving a 3-dimensional hypersurface. The singular locus of $V(\det)$ is $\{a = b = c = d = 0\}$ (the zero matrix), a single point.

For $n = 3$: $V(\det) \subseteq \mathbb{A}^9$ has dimension 8. It is singular along $\mathcal{M}_1$ (matrices of rank $\leq 1$), which has dimension $3 + 3 - 1 = 5$.

Let $X = V(xz - y) \subseteq \mathbb{A}^3$, which is isomorphic to $\mathbb{A}^2$ via $(x,z) \mapsto (x, xz, z)$, so $\dim X = 2$. Consider the projection $f : X \to \mathbb{A}^1$ given by $f(x,y,z) = x$.

For $a \neq 0$: $f^{-1}(a) = \{(a, y, z) \mid az - y = 0\} = \{(a, az, z) \mid z \in k\} \cong \mathbb{A}^1$, dimension 1.

For $a = 0$: $f^{-1}(0) = \{(0, y, z) \mid 0 \cdot z - y = 0\} = \{(0, 0, z) \mid z \in k\} \cong \mathbb{A}^1$, dimension 1.

In this case, every fiber has dimension $1 = 2 - 1$. The fiber dimension is constant.

The blow-up of $\mathbb{A}^2$ at the origin is

$\operatorname{Bl}_0(\mathbb{A}^2) = \{((x,y), [s:t]) \in \mathbb{A}^2 \times \mathbb{P}^1 \mid xt - ys = 0\}.$

This is an irreducible variety of dimension 2. The natural projection $\pi : \operatorname{Bl}_0(\mathbb{A}^2) \to \mathbb{A}^2$ is a morphism with:

• For $(x,y) \neq (0,0)$: $\pi^{-1}(x,y) = \{((x,y), [x:y])\}$, a single point (dimension 0).
• For $(x,y) = (0,0)$: $\pi^{-1}(0,0) = \{((0,0), [s:t])\} \cong \mathbb{P}^1$, dimension 1.

The generic fiber dimension is $2 - 2 = 0$, and the special fiber over the origin jumps to dimension 1. The set where fiber dimension $\geq 1$ is $\{(0,0)\}$, which is indeed closed.

Consider the map $f : M_{m \times n}(k) \to \mathbb{G}(r, m)$ sending a matrix $A$ of rank $r$ to its image $\operatorname{Im}(A) \in \mathbb{G}(r,m)$. (This is defined on the locally closed subset of matrices of rank exactly $r$.)

The fiber over a fixed $r$-plane $V \subseteq k^m$ is the set of all $m \times n$ matrices with image $V$, which is the set of $n$-tuples of vectors in $V$, i.e., $\operatorname{Hom}(k^n, V) \cong k^{rn}$ (modulo the constraint that the map is surjective onto $V$, which is an open condition). So the fiber has dimension $rn$ (as the open subset of matrices of rank $r$ in $k^{rn}$).

The total dimension: $\dim \mathbb{G}(r,m) + rn = r(m-r) + rn = r(m+n-r)$, matching our earlier computation.

Consider the incidence variety

$\Sigma = \{(p, \ell) \in \mathbb{P}^2 \times (\mathbb{P}^2)^* \mid p \in \ell\}$

where $(\mathbb{P}^2)^*$ is the dual projective plane parametrizing lines. Then $\Sigma \subseteq \mathbb{P}^2 \times \mathbb{P}^2$ is defined by $a_0 x_0 + a_1 x_1 + a_2 x_2 = 0$ where $[x_0:x_1:x_2]$ are coordinates on $\mathbb{P}^2$ and $[a_0:a_1:a_2]$ are coordinates on $(\mathbb{P}^2)^*$.

Projection to $\mathbb{P}^2$: The fiber over a point $p$ is the set of lines through $p$, which is $\mathbb{P}^1$. So every fiber has dimension 1, and $\dim \Sigma = \dim \mathbb{P}^2 + 1 = 3$.

Projection to $(\mathbb{P}^2)^*$: The fiber over a line $\ell$ is the set of points on $\ell$, which is $\mathbb{P}^1$. Again every fiber has dimension 1, and $\dim \Sigma = \dim (\mathbb{P}^2)^* + 1 = 3$.

Both projections give $\dim \Sigma = 3$, which is consistent.