$V(y - 2x + 1) \subseteq \mathbb{A}^2$ is the line $y = 2x - 1$. This is an algebraic set defined by a single linear polynomial.

$V(y - x^2) \subseteq \mathbb{A}^2$ is the standard parabola. As an affine variety, it is isomorphic to $\mathbb{A}^1$ via $t \mapsto (t, t^2)$.

$V(x^2 + y^2 - 1) \subseteq \mathbb{A}^2_\mathbb{C}$. Over $\mathbb{C}$ this is irreducible. It is isomorphic to $\mathbb{A}^1 \setminus \{0\}$ via the rational parametrization

$t \mapsto \left(\frac{1-t^2}{1+t^2},\; \frac{2t}{1+t^2}\right).$

Over $\mathbb{R}$, this is the familiar circle; over $\mathbb{C}$, it is a punctured affine line (the point at infinity is "missing").

Every conic in $\mathbb{A}^2$ is $V(ax^2 + bxy + cy^2 + dx + ey + f)$. Over an algebraically closed field, the classification by the discriminant $b^2 - 4ac$ gives:

• Nondegenerate ($b^2 - 4ac \neq 0$): a smooth curve, isomorphic to $\mathbb{A}^1 \setminus \{0\}$.
• Degenerate: a pair of lines ($V(xy)$), a double line ($V(x^2)$), or a single point.

All nondegenerate conics over $k = \bar{k}$ are isomorphic — there is no distinction between "ellipse," "parabola," and "hyperbola" in algebraic geometry over $\mathbb{C}$.

$V(y^2 - x^2(x+1)) \subseteq \mathbb{A}^2$. This curve has a node (self-intersection) at the origin: both partial derivatives vanish there, and the tangent cone is $V(y^2 - x^2) = V(y-x) \cup V(y+x)$, i.e., two distinct tangent lines.

The parametrization $t \mapsto (t^2 - 1, t(t^2 - 1))$ shows this curve is rational, but it is not isomorphic to $\mathbb{A}^1$ (the map is not injective: $t = 1$ and $t = -1$ both map to the origin).

$V(y^2 - x^3) \subseteq \mathbb{A}^2$. This curve has a cusp at the origin: the tangent cone is $V(y^2) = V(y)$ (a double line — only one tangent direction).

The parametrization $t \mapsto (t^2, t^3)$ gives a bijection $\mathbb{A}^1 \to V(y^2 - x^3)$. The map is a bijection but not an isomorphism: the inverse map would need $t = y/x$, which is not a regular function at the origin. The coordinate ring $k[t^2, t^3] \cong k[x,y]/(y^2-x^3)$ is not isomorphic to $k[t]$.

The twisted cubic is the image of the map $\mathbb{A}^1 \to \mathbb{A}^3$ given by

$t \mapsto (t, t^2, t^3).$

It is the algebraic set $V(y - x^2,\; z - x^3) = V(y - x^2,\; z - xy)$. In fact, this is the intersection of three quadric surfaces:

$V(y - x^2) \cap V(z - xy) \cap V(xz - y^2),$

but it cannot be defined as a set-theoretic intersection of only two hypersurfaces (it is not a complete intersection). This is a fundamental example in Hartshorne (Ex. I.2.12).

$V(xy) \subseteq \mathbb{A}^2$ is the union of the $x$-axis $V(y)$ and the $y$-axis $V(x)$. This is reducible: $V(xy) = V(x) \cup V(y)$. The ideal $(xy)$ is not prime (since $x \cdot y \in (xy)$ but $x \notin (xy)$ and $y \notin (xy)$).

$V(x^2 + y^2 + 1) \subseteq \mathbb{A}^2_\mathbb{R}$ is empty — no real solution exists. But $V(x^2 + y^2 + 1) \subseteq \mathbb{A}^2_\mathbb{C}$ is a nonempty curve. This illustrates why we work over algebraically closed fields: the Nullstellensatz fails over $\mathbb{R}$.

On $\mathbb{A}^1 = k$, the nonzero polynomials in $k[x]$ have finitely many roots. So:

• Closed sets: $\varnothing$, $\mathbb{A}^1$, and finite subsets of $\mathbb{A}^1$.
• Open sets: $\varnothing$, $\mathbb{A}^1$, and cofinite subsets.

This is the cofinite topology. Any bijection $k \to k$ that maps finite sets to finite sets is automatically a homeomorphism — the Zariski topology on $\mathbb{A}^1$ carries much less information than the variety structure.

For $P = (a_1,\ldots,a_n) \in \mathbb{A}^n$:

$I(\{P\}) = (x_1 - a_1, \ldots, x_n - a_n).$

This is a maximal ideal. The weak Nullstellensatz says that every maximal ideal of $k[x_1,\ldots,x_n]$ (for $k = \bar{k}$) is of this form. The points of $\mathbb{A}^n$ biject with the maximal ideals of $k[x_1,\ldots,x_n]$.

For $Y = V(y^2 - x^3)$:

$I(Y) = (y^2 - x^3).$

The polynomial $y^2 - x^3$ is irreducible, so $(y^2 - x^3)$ is already a prime ideal. No extra generators sneak in — $I(V(f)) = (f)$ whenever $f$ is irreducible.

$V(x^2) = V(x) = \{0\} \times \mathbb{A}^1$ in $\mathbb{A}^2$, but $(x^2) \neq (x)$. The ideal $(x^2)$ is not radical: $x \notin (x^2)$ but $x^2 \in (x^2)$. The Nullstellensatz says $I(V(\mathfrak{a})) = \sqrt{\mathfrak{a}}$, so $I(V(x^2)) = (x)$. Varieties cannot "see" nilpotent thickening — that requires the language of schemes.

| Algebraic set | Ideal | Prime? | Irreducible? | |---|---|---|---| | $V(y - x^2)$ | $(y - x^2)$ | Yes | Yes — a variety | | $V(xy)$ | $(xy)$ | No | No — $V(x) \cup V(y)$ | | $V(y^2 - x^3)$ | $(y^2-x^3)$ | Yes | Yes — singular but irreducible | | $V(x^2 + y^2 - 1)$ over $\mathbb{C}$ | $(x^2+y^2-1)$ | Yes | Yes | | $V(y^2 - x^2)$ | $(y-x)(y+x)$ | No | No — two lines | | $\mathbb{A}^n$ | $(0)$ | Yes | Yes | | $\{(0,0)\}$ in $\mathbb{A}^2$ | $(x,y)$ | Yes | Yes (a point is irreducible) |

In $\mathbb{A}^3$:

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

• $V(x)$ is the $yz$-plane (a variety of dimension 2).
• $V(y,z)$ is the $x$-axis (a variety of dimension 1).

The irreducible components have different dimensions — this is allowed.

$Y = V(y - x^2)$, so $I(Y) = (y - x^2)$ and

$k[Y] = k[x, y]/(y - x^2) \cong k[x]$

via $y \mapsto x^2$. The variety $Y$ is isomorphic to $\mathbb{A}^1$, and its coordinate ring is a polynomial ring in one variable — confirming this.

$Y = V(y^2 - x^3)$, so

$k[Y] = k[x,y]/(y^2 - x^3) \cong k[t^2, t^3] \subsetneq k[t].$

This is the subring of $k[t]$ generated by $t^2$ and $t^3$. It is an integral domain but not integrally closed: $t = t^3/t^2$ is in the fraction field but not in $k[t^2,t^3]$. The failure of normality reflects the cusp singularity.

$Y = V(x^2 + y^2 - 1)$ over $\mathbb{C}$. Then

$k[Y] = \mathbb{C}[x,y]/(x^2+y^2-1).$

Using $u = x + iy$ and $v = x - iy$, we get $uv = x^2+y^2 = 1$, so $k[Y] \cong \mathbb{C}[u, u^{-1}]$, the ring of Laurent polynomials — confirming that $Y \cong \mathbb{A}^1 \setminus \{0\}$.

$Y = V(y^2 - x^2(x+1))$. Then

$k[Y] = k[x,y]/(y^2 - x^2(x+1)).$

Setting $t = y/x$ (which is well-defined on $Y \setminus \{0\}$), we get $t^2 = x + 1$, so $x = t^2 - 1$, $y = t(t^2-1)$. Thus $k[Y] \cong k[t^2-1, t(t^2-1)]$, which is not normal (the element $t$ is in the fraction field but not in $k[Y]$). The normalization of $Y$ is $\mathbb{A}^1$, and the normalization map $t \mapsto (t^2-1, t^3-t)$ identifies $t=1$ and $t=-1$ to produce the node.

If $X \subseteq \mathbb{A}^m$ and $Y \subseteq \mathbb{A}^n$ are affine varieties, then $X \times Y \subseteq \mathbb{A}^{m+n}$ is also an affine variety, and

$k[X \times Y] \cong k[X] \otimes_k k[Y].$

For example, $\mathbb{A}^1 \times \mathbb{A}^1 = \mathbb{A}^2$, and $k[\mathbb{A}^2] = k[x] \otimes_k k[y] = k[x,y]$.

Warning: The Zariski topology on $X \times Y$ is not the product topology. In $\mathbb{A}^1 \times \mathbb{A}^1 = \mathbb{A}^2$, the diagonal $\{(a,a)\} = V(x-y)$ is Zariski-closed, but it is not closed in the product of the cofinite topologies (because the cofinite topology on a product requires closed sets to be finite unions of products of finite sets).