*Note:
These are notes live-tex’d from a graduate course in Algebraic Geometry taught by Philip Engel at the University of Georgia in Fall 2020. As such, any errors or inaccuracies are almost certainly my own.
*

dzackgarza@gmail.com

Last updated: 2020-11-25

- Gathmann’s Algebraic Geometry notes(Gathmann, 2019) https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2019/alggeom-2019.pdf

\[\begin{align*} V(I) && \text{The variety associated to an ideal } I {~\trianglelefteq~}k[x_1, \cdots, x_{n}] .\end{align*}\]

General idea: functions in a *coordinate ring* \(R[x_1, \cdots, x_n]/I\) will correspond to the geometry of the *variety* cut out by \(I\).

\(x^2 + y^2 - 1\) defines a circle, say, over \({\mathbb{R}}\)

\(y^2 = x^3-x\) gives an elliptic curve:

\(x^n+y^n-1\): does it even contain a \({\mathbb{Q}}{\hbox{-}}\)point? (Fermat’s Last Theorem)

\(x^2 + 1\), which has no \({\mathbb{R}}{\hbox{-}}\)points.

\(x^2 + y^2 + 1/{\mathbb{R}}\) vanishes nowhere, so its ring of functions is not \({\mathbb{R}}[x, y] / \left\langle{x^2 + y^2 + 1}\right\rangle\). The problem: \({\mathbb{R}}\) is not algebraically closed.

\(x^2 - y^2 = 0\) over \({\mathbb{C}}\) is not a manifold (no chart at the origin):

\(x+y+1/{\mathbb{F}}_3\), which has 3 points over \({\mathbb{F}}_3^2\), but \(f(x, y) = (x^3 - x)(y^3-y)\) vanishes at every point

Not possible when algebraically closed (is there nonzero polynomial that vanishes on every point in \({\mathbb{C}}\)?)

\(V(f) = {\mathbb{F}}_3^2\), so the coordinate ring is zero instead of \({\mathbb{F}}_3[x, y]/\left\langle{f}\right\rangle\) (addressed by scheme theory)

If \(f \in {\mathbb{R}}[x, y]\) is of degree \(d\), then[^actual_statement] \[\begin{align*} \pi_1 V(f) \subseteq {\mathbb{R}}^2 \leq 1 + {(d-1)(d-2) \over 2} \end{align*}\] [^actual_statement]: Actual statement: the number of connected components is bounded above by this quantity.

Take the curve \[\begin{align*} X = \left\{{(x, y, z) = (t^3, t^4, t^5) \in {\mathbb{C}}^3 {~\mathrel{\Big|}~}t\in {\mathbb{C}}}\right\} .\end{align*}\]

Then \(X\) is cut out by three equations:

- \(y^2 = xz\)
- \(x^2 = yz\)
- \(z^2 = x^2 y\)

Show that the vanishing locus of the first two equations above is \(X\cup L\) for \(L\) a line.

Compare to linear algebra: codimension \(d\) iff cut out by exactly \(d\) equations.

Given the Riemann surface \[\begin{align*} y^2 = (x-1)(x-2)\cdots(x-2n) ,\end{align*}\] how does one visualize its solution set?

On \({\mathbb{C}}\) with some slits, you can consistently choose a square root of the RHS.

Away from \(x=1, \cdots, 2n\), there are two solutions for \(y\) given \(x\).

After gluing along strips, obtain:

Let \(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) and \(R\) a ring containing ideals \(I, J\). Recall the definition of the *radical*:

The *radical* of an ideal \(I {~\trianglelefteq~}R\) is defined as \[\begin{align*} \sqrt{I} = \left\{{r\in R ~{\text{s.t.}}~r^k\in I \text{ for some } k\in {\mathbb{N}}}\right\} .\end{align*}\]

Let \[\begin{align*} I &= (x_1, x_2^2) \subset {\mathbb{C}}[x_1, x_2] \\ &= \left\{{ f_1 x_1 + f_2 x_2 ~{\text{s.t.}}~f_1, f_2 \in {\mathbb{C}}[x_1, x_2]}\right\} \end{align*}\]

Then \(\sqrt{I} = (x_1, x_2)\), since \(x_2^2 \in I \implies x_2 \in \sqrt{I}\).

Given \(f\in k[x_1, \cdots, x_n]\), take its value at \(a = (a_1, \cdots, a_n)\) and denote it \(f(a)\).

Set \(\deg(f)\) to be the largest value of \(i_1 + \cdots + i_n\) such that the coefficient of \(\prod x_j ^{i_j}\) is nonzero.

\(\deg(x_1 + x_2^2 + x_1 x_2^3) = 4\)

Affine \(n{\hbox{-}}\)space \({\mathbb{A}}^n = {\mathbb{A}}_k^n\) is defined as \(\left\{{(a_1, \cdots, a_n) {~\mathrel{\Big|}~}a_i \in k}\right\}\).

^{1}Let \(S\subset k[x_1, \cdots, x_n]\) to be a set of polynomials. Then define the

**affine variety**of \(S\) as \[\begin{align*} V(S) \mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{A}}^n ~{\text{s.t.}}~f(x) = 0}\right\} \subset {\mathbb{A}}^n \end{align*}\]

- \({\mathbb{A}}^n = V(0)\).
- For any point \((a_1, \cdots, a_n)\in {\mathbb{A}}^n\), then \(V(x_1 - a_1, \cdots, x_n - a_n) = \left\{{a_1, \cdots, a_n}\right\}\) uniquely determines the point.
- For any finite set \(r_1, \cdots, r_k \in {\mathbb{A}}^1\), there exists a polynomial \(f(x)\) whose roots are \(r_i\).

We may as well assume \(S\) is an ideal by taking the ideal it generates, \[\begin{align*} S\subseteq \left\langle{S}\right\rangle = \left\{{\sum g_i f_i {~\mathrel{\Big|}~}g_i \in k[x_1, \cdots, x_n],\, f_i\in S}\right\} .\end{align*}\] Then \(V(\left\langle{S}\right\rangle) \subset V(S)\).

Conversely, if \(f_1, f_2\) vanish at \(x\in {\mathbb{A}}^n\), then \(f_1 + f_2, gf_1\) also vanish at \(x\) for all \(g\in k[x_1, \cdots, x_n]\). Thus \(V(S) \subset V(\left\langle{S}\right\rangle)\).

\[\begin{align*} I+J &\mathrel{\vcenter{:}}=\left\{{f+g ~{\text{s.t.}}~f\in I,\, g\in J}\right\} \\ IJ &\mathrel{\vcenter{:}}=\left\{{\sum_{i=1}^N f_i g_i ~{\text{s.t.}}~f_i\in I,\, g_i\in J, N\in {\mathbb{N}}}\right\} \\ I+J = \left\langle{1}\right\rangle &\implies I\cap J = IJ && \text{(coprime or comaximal)} .\end{align*}\]

If \(I = \left\langle{a}\right\rangle\) and \(J = \left\langle{b}\right\rangle\), then \(I + J = \left\langle{a}\right\rangle + \left\langle{b}\right\rangle = \left\langle{a, b}\right\rangle\).

- If \(S_1 \subseteq S_2\) then \(V(S_1) \supseteq V(S_2)\).
- \(V(S_1) \cup V(S_2) = V(S_1 S_2) = V(S_1 \cap S_2)\).
- \(\bigcap V(S_i) = V\qty{\bigcup S_i}\).

We thus have a map \[\begin{align*} V: \left\{{\text{Ideals in } k[x_1, \cdots, x_n]}\right\} \to \left\{{\text{Affine varieties in } {\mathbb{A}}^n}\right\} .\end{align*}\]

Let \(X\subset {\mathbb{A}}^n\) be any set, then *the ideal of \(X\)* is defined as \[\begin{align*} I(X) \mathrel{\vcenter{:}}=\left\{{f\in k[x_1, \cdots, x_n] ~{\text{s.t.}}~f(x) = 0\, \forall x\in X}\right\} .\end{align*}\]

Let \(X\) be the union of the \(x_1\) and \(x_2\) axes in \({\mathbb{A}}^2\), then \[\begin{align*} I(X) = (x_1 x_2) = \left\{{x_1 x_2 g~{\text{s.t.}}~g\in k[x_1, x_2]}\right\} .\end{align*}\]

If \(X_1 \subset X_2\) then \(I(X_1) \subset I(X_2)\).

\(I(X)\) is a radical ideal, i.e. \(I(X) = \sqrt{I(X)}\).

This is because \(f(x)^k = 0 \forall x\in X\) implies \(f(x) = 0\) for all \(x\in X\), so \(f^k \in I(X)\) and thus \(f\in I(X)\).

Our correspondence is thus \[\begin{align*} \left\{{\substack{\text{Ideals in } k[x_1, \cdots, x_n]}}\right\} &\xrightarrow{V} \left\{{\substack{\text{Affine Varieties}}}\right\} \\ \left\{{\substack{\text{Radical Ideals}}}\right\} &\xleftarrow{I} \left\{{\substack{\text{?}}}\right\} .\end{align*}\]

For any affine variety \(X\), \[\begin{align*} V(I(X)) = X .\end{align*}\]

For any ideal \(J \subset k[x_1, \cdots, x_n]\), \[\begin{align*} I(V(J)) = \sqrt{J} .\end{align*}\]

Thus there is a bijection between radical ideals and affine varieties.

Recall the Hilbert Basis Theorem: any ideal in a finitely generated polynomial ring over a field is again finitely generated.

We need to show 4 inclusions, 3 of which are easy.

- \(X \subset V(I(X))\):

- If \(x\in X\) then \(f(x) = 0\) for all \(f\in I(X)\).
- So \(x\in V(I(X))\), since every \(f\in I(X)\) vanishes at \(x\).

- \(\sqrt{J} \subset I(V(J))\):

- If \(f\in \sqrt{J}\) then \(f^k \in J\) for some \(k\).
- Then \(f^k(x) = 0\) for all \(x\in V(J)\).
- So \(f(x) = 0\) for all \(x\in V(J)\).
- Thus \(f\in I(V(J))\).

- \(V(I(X)) \subset X\):

- Need to now use that \(X\) is an affine variety.
- Counterexample: \(X = {\mathbb{Z}}^2 \subset {\mathbb{C}}^2\), then \(I(X) = 0\). But \(V(I(X)) = {\mathbb{C}}^2 \not\subset {\mathbb{Z}}^2\).

- By (b), \(I(V(J)) \supset \sqrt{J} \supset J\).
- Since \(V({\,\cdot\,})\) is order-reversing, taking \(V\) of both sides reverses the containment.
- So \(V(I(V(J))) \subset V(J)\), i.e. \(V(I(X)) \subset X\).

Thus the hard direction that remains is

- \(I(V(J)) \subset \sqrt{J}\).

We’ll need the following important theorem:

Any finitely-generated field extension \(k_1 \hookrightarrow k_2\) is a finite extension of a purely transcendental extension, i.e. there exist \(t_1, \cdots, t_\ell\) such that \(k_2\) is finite over \(k_1(t_1, \cdots, t_\ell)\).

Noether normalization is perhaps more important than the Nullstellensatz!

Suppose \(k\) is algebraically closed and uncountable^{2} Then the maximal ideals in \(k[x_1, \cdots, x_n]\) are of the form \((x_1 - a_1, \cdots, x_n - a_n)\).

Let \({\mathfrak{m}}\) be a maximal ideal, then by the Hilbert Basis Theorem, \({\mathfrak{m}}= \left\langle{f_1, \cdots, f_r}\right\rangle\) is finitely generated. Let \(L = {\mathbb{Q}}[\left\{{c_i}\right\}]\) where the \(c_i\) are all of the coefficients of the \(f_i\) if \(\operatorname{ch}(K) = 0\), **or** \({\mathbb{F}}_p[\left\{{c_i}\right\}]\) if \(\operatorname{ch}(k) = p\). Then \(L\subset k\). Define \({\mathfrak{m}}_0 = {\mathfrak{m}}\cap L[x_1, \cdots, x_n]\). Note that by construction, \(f_i \in {\mathfrak{m}}_0\) for all \(i\), and we can write \({\mathfrak{m}}= {\mathfrak{m}}_0 \cdot k[x_1, \cdots, x_n]\).

\({\mathfrak{m}}_0\) is a maximal ideal.

If it were the case that \[\begin{align*} {\mathfrak{m}}_0 \subsetneq {\mathfrak{m}}_0' \subsetneq L[x_1, \cdots, x_n] ,\end{align*}\] then \[\begin{align*} {\mathfrak{m}}_0\cdot k[x_1, \cdots, x_n] \subsetneq {\mathfrak{m}}_0'\cdot k[x_1, \cdots, x_n] \subsetneq k[x_1, \cdots, x_n] .\end{align*}\]

So far, we’ve constructed a smaller polynomial ring and a maximal ideal in it. Thus \(L[x_1, \cdots, x_n]/{\mathfrak{m}}_0\) is a field that is finitely generated over either \({\mathbb{Q}}\) or \({\mathbb{F}}_p\). So \(L[x_1, \cdots, x_n]/{\mathfrak{m}}_0\) is finite over some \({\mathbb{Q}}(t_1, \cdots, t_n)\), and since \(k\) is uncountable, there exists an embedding \({\mathbb{Q}}(t_1, \cdots, t_n) \hookrightarrow k\).^{3}

This extends to an embedding of \(\phi: L[x_1, \cdots, x_n]/{\mathfrak{m}}_0 \hookrightarrow k\) since \(k\) is algebraically closed. Letting \(a_i\) be the image of \(x_i\) under \(\phi\), then \(f(a_1, \cdots, a_n) = 0\) by construction, \(f_i \in (x_i - a_i)\) implies that \({\mathfrak{m}}= (x_i - a_i)\) by maximality.

Recall Hilbert’s Nullstellensatz:

For any affine variety, \(V(I(X)) = X\).

For any ideal \(J{~\trianglelefteq~}k[x_1, \cdots, x_n]\), \(I(V(J)) = \sqrt{J}\).

So there’s an order-reversing bijection \[\begin{align*} \left\{{\substack{\text{Radical ideals } k[x_1, \cdots, x_n]}}\right\} \to{V({\,\cdot\,})}{I({\,\cdot\,})} \left\{{\substack{\text{Affine varieties in } {\mathbb{A}}^n}}\right\} .\end{align*}\]

In proving \(I(V(J)) \subseteq \sqrt{J}\), we had an important lemma (Noether Normalization): the maximal ideals of \(k[x_1, \cdots, x_n]\) are of the form \(\left\langle{x-a_1, \cdots, x-a_n}\right\rangle\).

If \(V(I)\) is empty, then \(I = \left\langle{1}\right\rangle\).

This is because no common vanishing locus \(\implies\) trivial ideal, so there’s a linear combination that equals 1.

The only ideals that vanish nowhere are trivial.

By contrapositive, suppose \(I\neq \left\langle{1}\right\rangle\). By Zorn’s Lemma, these exists a maximal ideals \({\mathfrak{m}}\) such that \(I \subset {\mathfrak{m}}\). By the order-reversing property of \(V({\,\cdot\,})\), \(V({\mathfrak{m}}) \subseteq V(I)\). By the classification of maximal ideals, \({\mathfrak{m}}= \left\langle{x-a_1, \cdots, x-a_n}\right\rangle\), so \(V({\mathfrak{m}}) = \left\{{a_1, \cdots, a_n}\right\}\) is nonempty.

Returning to the proof that \(I(V(J)) \subseteq \sqrt{J}\): let \(f\in V(I(J))\), we want to show \(f\in \sqrt{J}\). Consider the ideal \(\tilde J \mathrel{\vcenter{:}}= J + \left\langle{ft - 1}\right\rangle \subseteq k[x_1, \cdots, x_n, t]\).

\(f = 0\) on all of \(V(J)\) by the definition of \(I(V(J))\).

However, if \(f=0\), then \(ft-1 \neq 0\), so \(V(\tilde J) = V(G) \cap V(ft-1) = \emptyset\).

Applying the corollary \(\tilde J = (1)\), so \[\begin{align*} 1 = \left\langle{ft-1}\right\rangle g_0(x_1, \cdots, x_n, t) + \sum f_i g_i(x_1, \cdots, x_n, t) \end{align*}\] with \(f_i \in J\). Let \(t^N\) be the largest power of \(t\) in any \(g_i\). Thus for some polynomials \(G_i\), we have \[\begin{align*} f^N \mathrel{\vcenter{:}}=(ft-1) G_0(x_1, \cdots, x_n, ft) + \sum f_i G_i(x_1, \cdots, x_n, ft) \end{align*}\] noting that \(f\) does not depend on \(t\). Now take \(k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle\), so \(ft=1\) in this ring. This kills the first term above, yielding \[\begin{align*} f^N = \sum f_i G_i(x_1, \cdots, x_n, 1) \in k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle .\end{align*}\]

There is an inclusion \[\begin{align*} k[x_1, \cdots, x_n] \hookrightarrow k[x_1, \cdots, x_n, t]/\left\langle{ft-1}\right\rangle .\end{align*}\]

Why is this true?

Since this is injective, this identity also holds in \(k[x_1, \cdots, x_n]\). But \(f_i\in J\), so \(f\in \sqrt{I}\).

Consider \(k[x]\). If \(J\subset k[x]\) is an ideal, it is principal, so \(J = \left\langle{f}\right\rangle\). We can factor \(f(x) = \prod_{i=1}^k (x-a_i)^{n_i}\) and \(V(f) = \left\{{a_1, \cdots, a_k}\right\}\). Then \[\begin{align*} I(V(f)) = \left\langle{(x-a_1)(x-a_2)\cdots(x-a_k)}\right\rangle = \sqrt{J} \subsetneq J ,\end{align*}\] so this loses information.

Let \(J = \left\langle{x-a_1, \cdots, x-a_n}\right\rangle\), then \(I(V(J)) = \sqrt{J} = J\) with \(J\) maximal. Thus there is a correspondence \[\begin{align*} \left\{{\substack{\text{Points of } {\mathbb{A}}^n}}\right\} \iff \left\{{\substack{\text{Maximal ideals of }k[x_1, \cdots, x_n]}}\right\} .\end{align*}\]

\(I(X_1 \cup X_2) = I(X_1) \cap I(X_2)\).

\(I(X_1) \cap I(X_2) = \sqrt{I(X_1) + I(X_2)}\).

We proved (a) on the variety side.

For (b), by the Nullstellensatz, \(X_i = V(I(X_i))\), so \[\begin{align*} I(X_1\cap X_2) &= I\qty{ VI(X_1) \cap VI(X_2)} \\ &= IV\qty{I(X_1) + I(X_2)} \\ &= \sqrt{I(X_1) + I(X_2)} .\end{align*}\]

Example of property (b):

Take \(X_1 = V(y-x^2)\) and \(X_2 = V(y)\), a parabola and the \(x{\hbox{-}}\)axis.

Then \(X_1 \cap X_2 = \left\{{(0, 0)}\right\}\), and \(I(X_1) + I(X_2) = \left\langle{y-x^2, y}\right\rangle = \left\langle{x^2, y}\right\rangle\), but \[\begin{align*} I(X_1 \cap X_2) = \left\langle{x, y}\right\rangle = \sqrt{\left\langle{x^2, y}\right\rangle} \end{align*}\]

If \(f, g\in k[x_1, \cdots, x_n]\), and suppose \(f(x) = g(x)\) for all \(x\in {\mathbb{A}}^n\). Then \(f = g\).

Since \(f-g\) vanishes everywhere, \(f-g \in I({\mathbb{A}}^n) = I(V(0)) = \sqrt{0} = 0\).

More generally suppose \(f(x) = g(x)\) for all \(x\in X\), where \(X\) is some affine variety. Then by definition, \(f-g \in I(X)\), so a “natural” space of functions on \(X\) is \(k[x_1,\cdots, x_n]/I(X)\).

For an affine variety \(X\), the *coordinate ring of \(X\)* is \[\begin{align*} A(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_n]/ I(X) .\end{align*}\]

Elements \(f\in A(X)\) are called *polynomial* or *regular* functions on \(X\).

The constructions \(V({\,\cdot\,}), I({\,\cdot\,})\) work just as well for \(A(X)\) and \(X\).

Given any \(S\subset A(Y)\) for \(Y\) an affine variety, \[\begin{align*} V(S) = V_Y(S) \mathrel{\vcenter{:}}=\left\{{x\in Y ~{\text{s.t.}}~f(x) = 0\,\,\forall f\in S}\right\} .\end{align*}\]

Given \(X\subset Y\) a subset, \[\begin{align*} I(X) = I_Y(X) \mathrel{\vcenter{:}}=\left\{{f\in A(Y) ~{\text{s.t.}}~f(x) = 0\,\,\forall x\in X}\right\} \subseteq A(Y) .\end{align*}\]

For \(X\subset Y \subset {\mathbb{A}}^n\), we have \(I(X) \supset I(Y) \supset I({\mathbb{A}}^n)\), so we have maps

Let \(X\subset Y\) be an affine subvariety, then

\(A(X) = A(Y) / I_Y(X)\)

There is a correspondence \[\begin{align*} \left\{{\substack{\text{Affine subvarieties of }Y}}\right\} &\iff \left\{{\substack{\text{Radical ideals in }A(Y)}}\right\} \\ X &\mapsto I_Y(X) \\ V_Y(J) &\mapsfrom J .\end{align*}\]

Properties are inherited from the case of \({\mathbb{A}}^n\), see exercise in Gathmann.

Let \(Y = V(y-x^2) \subset {\mathbb{A}}^2/{\mathbb{C}}\) and \(X = \left\{{(1, 1)}\right\} = V(x-1, y-1)\subset {\mathbb{A}}^2/{\mathbb{C}}\).

Then there is an inclusion \(\left\langle{y-x^2}\right\rangle \subset \left\langle{x-1, y-1}\right\rangle\) (e.g. by Taylor expanding about the point \((1, 1)\)), and there is a map

Last time: \(V(I) = \left\{{x\in {\mathbb{A}}^n ~{\text{s.t.}}~f(x) = 0 \, \forall x\in I}\right\}\) and \(I(X) = \left\{{f\in k[x_1, \cdots, x_n] ~{\text{s.t.}}~f(x) = 0\, \forall x\in X}\right\}\). We proved the Hilbert Nullstellensatz \(I(V(J)) = \sqrt{J}\), defined the coordinate ring of an affine variety \(X\) as \(A(X) \mathrel{\vcenter{:}}= k[x_1, \cdots, x_n] / I(X)\), the ring of “regular” (polynomial) functions on \(X\).

Recall that a *topology* on \(X\) can be defined as a collection of “closed” subsets of \(X\) that are closed under arbitrary intersections and finite unions. A subset \(Y\subset X\) inherits a subspace topology with closed sets of the form \(Z\cap Y\) for \(Z\subset X\) closed.

Let \(X\) be an affine variety. The closed sets are affine subvarieties \(Y\subset X\).

We have \(\emptyset, X\) closed, since

- \(V_X(1) = \emptyset\),
- \(V_X(0) = X\)

Closure under finite unions: Let \(V_X(I), V_X(J)\) be closed in \(X\) with \(I, J \subset A(X)\) ideals. Then \(V_X(IJ) = V_X(I) \cup V_X(J)\).

Closure under intersections: We have \(\bigcap_{i\in \sigma} V_X(J) = V_X\qty{ \sum_{i\in \sigma} J_i}\).

There are few closed sets, so this is a “weak” topology.

Compare the classical topology on \({\mathbb{A}}^1/{\mathbb{C}}\) to the Zariski topology.

Consider the set \(A\mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{A}}^1/{\mathbb{C}}~{\text{s.t.}}~{\left\lVert {x} \right\rVert} \leq 1}\right\}\), which is closed in the classical topology.

But \(A\) is not closed in the Zariski topology, since the closed subsets are finite sets or the whole space.

Here the topology is in fact the cofinite topology.

Let \(f: {\mathbb{A}}^1/k\to {\mathbb{A}}^1/k\) be any injective map. Then \(f\) is necessarily continuous wrt the Zariski topology.

Thus the notion of continuity is too weak in this situation.

Consider \(X\times Y\) a product of affine varieties. Then there is a product topology where open sets are of the form \(\bigcup_{i=1}^n U_i \times V_i\) with \(U_i, V_i\) open in \(X, Y\) respectively.

This is the wrong topology! On \({\mathbb{A}}^1 \times{\mathbb{A}}^1 = {\mathbb{A}}^2\), the diagonal \(\Delta \mathrel{\vcenter{:}}= V(x-y)\) is closed in the Zariski topology on \({\mathbb{A}}^2\) but not in the product topology.

Consider \({\mathbb{A}}^2/{\mathbb{C}}\), so the closed sets are curves and points. Observation: \(V(x_1 x_2 ) \subset {\mathbb{A}}^2/{\mathbb{C}}\) decomposed into the union of the coordinate axes \(X_1 \mathrel{\vcenter{:}}= V(x_1)\) and \(X_2 \mathrel{\vcenter{:}}= V(x_2)\). The Zariski topology can detect these decompositions.

Let \(X\) be a topological space.

\(X\) is

*reducible*iff there exist nonempty proper closed subsets \(X_1 ,X_2 \subset X\) such that \(X = X_1 \cup X_2\). Otherwise, \(X\) is said to be*irreducible*.\(X\) is

*disconnected*if there exist \(X_1, X_2 \subset X\) such that \(X = X_1 {\coprod}X_2\). Otherwise, \(X\) is said to be*connected*.

\(V(x_1 x_2)\) is reducible but connected.

\({\mathbb{A}}^1/{\mathbb{C}}\) is *not* irreducible, since we can write \({\mathbb{A}}^1/{\mathbb{C}}= \left\{{{\left\lVert {x} \right\rVert} \leq 1}\right\} \cup\left\{{{\left\lVert {x} \right\rVert} \geq 1}\right\}\).

Let \(X\) be a disconnected affine variety with \(X = X_1 {\coprod}X_2\). Then \(A(X) \cong A(X_1) \times A(X_2)\).

We have \(X_1 \cup X_2 = X\), so \(I(X_1) \cap I(X_2) = I(X) = (0)\) in the coordinate ring \(A(X)\) (recalling that it is a quotient by \(I(X)\).)

Since \(X_1 \cap X_1 \emptyset\), we have \[\begin{align*} I(X_1 \cap X_2) = \sqrt{I(X_1) + I(X_2) } = I(\emptyset) = \left\langle{1}\right\rangle .\end{align*}\]

Thus \(I(X_1) + I(X_2) = \left\langle{1}\right\rangle\), and by the Chinese Remainder Theorem, the following map is an isomorphism: \[\begin{align*} A(X) \to A(X)/I(X_1) \times A(X) / I(X_2) .\end{align*}\]

But the codomain is precisely \(A(X_1) \times A(X_2)\).

An affine variety \(X\) is irreducible \(\iff\) \(A(X)\) is an integral domain.

\(\implies\): By contrapositive, suppose \(f_1, f_2 \in A(X)\) are nonzero with \(f_1 f_2 = 0\). Let \(X_i = V(f_i)\), then \(X= V(0) = V(f_1 f_2) = X_1 \cup X_2\) which are closed and proper since \(f_i \neq 0\).

\(\impliedby\): Suppose \(X\) is reducible with \(X = X_1 \cup X_2\) with \(X_i\) proper and closed. Define \(J_i \mathrel{\vcenter{:}}= I(X_i)\), and note \(J_i \neq 0\) because \(V(J_i) = V(I(X_i)) = X_i\) by part (a) of the Nullstellensatz.

So there exists a nonzero \(f_i \in J_i = I(X_i)\), so \(f_i\) vanishes on \(X_i\). But then \(V(f_1) \cup V(f_2) \supset X_1 \cup X_2 = X\), so \(X= V(f_1 f_2)\) and \(f_1 f_2 \in I(X) = \left\langle{0}\right\rangle\) and \(f_1 f_2 = 0\). So \(A(X)\) is not a domain.

Let \(X = \left\{{p_1, \cdots, p_d}\right\}\) be a finite set in \({\mathbb{A}}^n\). The Zariski topology on \(X\) is the discrete topology, and \(X = {\coprod}\left\{{p_i}\right\}\). So \[\begin{align*} A(X) = A({\coprod}\left\{{p_i}\right\}) = \prod_{i=1}^d A({\left\{{p_i}\right\}}) = \prod_{i=1}^d k[x_1, \cdots, x_n] / \left\langle{x_j - a_j(p_i)}\right\rangle_{j=1}^d .\end{align*}\]

Set \(V(x_1 x_2) = X\), then \(A(X) = k[x_1, x_2]/ \left\langle{x_1 x_2}\right\rangle\). This not being a domain (since \(x_1 x_2 = 0\)) corresponds to \(X = V(x_1) \cup V(x_2)\) not being irreducible.

\({\mathbb{A}}^2/k\) is irreducible since \(k[x_1, \cdots x_n]\) is a domain.

Let \(X_1\) be the \(xy\) plane and \(X_2\) be the line parallel to the \(y{\hbox{-}}\)axis through \({\left[ {0,0,1} \right]}\), and let \(X= X_1 {\coprod}X_2\). Then \(X_1 = V(z)\) and \(X_2 = V(x, z-1)\), and \(I(X) = \left\langle{z}\right\rangle \cdots \left\langle{x, z-1}\right\rangle= \left\langle{xz, z^2 - z}\right\rangle\).

Then the coordinate ring is given by \(A(X) = {\mathbb{C}}[x, y, z] / \left\langle{xz, z^2 - z}\right\rangle = {\mathbb{C}}[x, y, z] / \left\langle{z}\right\rangle \oplus {\mathbb{C}}[x, y,z] / \left\langle{x, z-1}\right\rangle\).

Recall that the Zariski topology is defined on an affine variety \(X = V(J)\) with \(J {~\trianglelefteq~}k[x_1, \cdots, x_n]\) by describing the closed sets.

\(X\) is irreducible if its coordinate ring \(A(X)\) is a domain.

There is a 1-to-1 correspondence \[\begin{align*} \left\{{\substack{\text{Irreducible subvarieties} \\ \text{of }X}}\right\} \iff \left\{{\substack{\text{Prime ideals} \\ \text{in }A(X)}}\right\} .\end{align*}\]

Suppose \(Y\subset X\) is an affine subvariety. Then \[\begin{align*} A(X) / I_X(Y) = A(Y) .\end{align*}\]

By NSS, there is a bijection between subvarieties of \(X\) and radical ideals of \(A(X)\) where \(Y\mapsto I_X(Y)\). A quotient is a domain iff quotienting by a prime ideal, so \(A(Y)\) is a domain iff \(I_X(Y)\) is prime.

Recall that \({\mathfrak{p}}{~\trianglelefteq~}R\) is prime when \(fg\in {\mathfrak{p}}\iff f\in {\mathfrak{p}}\) or \(g\in {\mathfrak{p}}\). Thus \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0\) in \(R/{\mathfrak{p}}\) implies \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = 0\) or \(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu = 0\) in \(R/{\mathfrak{p}}\), i.e. \(R/{\mathfrak{p}}\) is a domain.

Finally note that prime ideals are radical (easy proof).

Consider \({\mathbb{A}}^2/{\mathbb{C}}\) and some subvarieties \(C_i\):