Algebraic geometry (Notes)

Given \(A \subset k[x_1,\dots,x_n]\), the vanishing locus of \(A\), denoted by \(V(A)\) is the set \[ V(A) = \{a \in \mathbb A^n_k \mid f(a) = 0 \text{ for all } f \in A\}.\]

Thus the affine algebraic sets are precisely the sets of the form \(V(A)\) for some \(A\).

The following are affine algebraic sets

1. The empty set
2. Entire affine space
3. Single point

The following are not affine algebraic sets

1. The unit cube in \(\mathbb A^n_{{\mathbb R}}\)
2. Points with rational coordinates in \(\mathbb A^n_{{\mathbb C}}\)

Let \(A \subset k[x_1,\dots,x_n]\). Then we have \(V(A) = V(\langle A \rangle)\).

Let \(R\) be a ring. The following are equivalent

1. Every ideal of \(R\) is finitely generated.
2. Every infinite chain of ideals \[I_1 \subset I_2 \subset I_3 \subset \cdots\] stabilises.

A ring \(R\) satisfying the equivalent conditions of Proposition 1.4.1 is called Noetherian.

The following rings are Noetherian

1. \(R = {\mathbb Z}\)
2. \(R\) a field.

The ring of continuous functions on the interval is not Noetherian. #+begin_proof. Let \(I_n\) be the set of functions on \([0,1]\) that vanish on \([0,1/n]\). This forms an increasing chain of ideals that does not stabilise. #+end_proof

If \(R\) is Noetherian and \(I \subset R\) is any ideal, then \(R/I\) is Noetherian.

If \(R\) is Noetherian, then so is \(R[x]\)

\(k[x_1,\dots,x_n]\) is Noetherian.

Every affine algebraic subset of \(\mathbb A^n_k\) is the vanishing set of a finite set of polynomials.

The collection of affine algebraic subsets satisfies the three conditions above.

The Zariski topology on \(\mathbb A^1_k\) is the finite complement topology. The only closed sets are the finite sets (or the whole space). In other words, the only open sets are the complements of finite sets (or the empty set).

Every Zariski closed (open) subset of \(\mathbb A^n_{\mathbb C}\) is also closed (open) in the usual Euclidean topology. The converse is not true.

Let \(f\) be a polynomial function on \(\mathbb A^n_k\), viewed as a map \(f \colon \mathbb A^n_k \to \mathbb A^1_k\). Then \(f\) is continuous in the Zariski topology.

The Zariski topology has very few open sets, and as a result has terrible separation properties. It is not even Hausdorff (except in very small examples). Nevertheless, we will see that it is extremely useful. For one, it makes sense over every field!

Let \(S \subset \mathbb A^n_k\) be a set. The ideal vanishing on \(S\), denoted by \(I(S)\), is the set \[ I(S) = \{f \in k[x_1,\dots,x_n] \mid f(a) = 0 \text{ for all } a \in S\}\]

Recall that an ideal \(I \subset k[x_1,\dots,x_n]\) is radical if it has the property that whenever \(f^n \in I\) for some \(n > 1\), then \(f \in I\).

The set \(I(S)\) is a radical ideal of \(k[x_1,\dots,x_n]\).

1. \(I \subset R\) is radical if and only if \(R/I\) has no (non-zero) nilpotents.
2. All prime ideals are radical. In particular, all maximal ideals are radical.

Let \(I\) be an ideal, and set \(\sqrt I = \{f \mid f^n \in I \text{ for some } n > 0\}.\) Then \(\sqrt I\) is a radical ideal.

The ideal \(\sqrt I\) is called the radical of \(I\).

We have \(V(I) = V(\sqrt I)\).

We now state a string of important theorems, all called the “Nullstellensatz”, starting with the most comprehensive one.

Let \(k\) be an algebraically closed field. Then we have a bijection \[ \text{Radical ideals of \(k[x_1, \dots, x_n]\)} \leftrightarrow \text{Zariski closed subsets of \(\mathbb A^n_k\)}\] where the map from the left to the right is \(I \mapsto V(I)\) and the map from the right to the left is \(S \mapsto I(S)\). The correspondence is inclusion reversing.

Let \(k\) be an algebraically closed field and \(I \subset k[x_1,\dots,x_n]\) an ideal. If \(V(I) = \emptyset\), then \(I = (1)\).

Let \(k\) be an algebraically closed field. Then all the maximal ideals of \(k[x_1,\dots,x_n]\) are of the form \(\langle x_1-a_1, \dots, x_2-a_n\rangle\) for some \((a_1, \dots, a_n) \in \mathbb A^n_k\).

Theorem 1.6.8 says that we have a dichotomy: either a system of equations \(f_i = 0\) has a solution, or there exist polynomials \(g_i\) such that \[ \sum f_i g_i = 1.\]

Let \(k\) be an algebraically closed field and \(I \subset k[x_1,\dots,x_n]\) an ideal. If \(f\) is identically zero on \(V(I)\), then \(f^n \in I\) for some \(n\).

Let \(K\) be any field and let \(L\) be a finitely generated \(K\)-algebra. If \(L\) is a field, then it must be a finite extension of \(K\).

Let \(m \subset k[x_1,\dots,x_n]\) be a maximal ideal. Taking \(K = k\) and \(L = k[x_1,\dots,x_n]/m\) in Theorem 1.7.1, and using that \(k\) is algebraically closed, we get that the natural map \(k \to k[x_1,\dots,x_n]/m\) is an isomorphism. Let \(a_i \in k\) be the pre-image of \(x_i\) under this isomorphism. Then we have \(m = (x_1-a_1,\dots,x_n-a_n)\).

Suppose \(I\) is not the unit ideal. We show that \(V(I)\) is non-empty. To do so, we use that every proper ideal is contained in a maximal ideal.

We consider the system \(g = 0\) for \(g \in I\) and \(f \neq 0\). Notice that the last one is not an equation, but there is a trick that allows us to convert it into an equation. Let \(y\) be a new variable, and consider the polynomial ring \(k[x_1,\dots,x_n,y]\). In the bigger ring, consider the system of equations \(g = 0\) for \(g \in I\) and \(yf - 1 = 0\). By our assumption, this system of equations has no solutions.

We show that the maps \(I \to V(I)\) and \(S \to I(S)\) are mutual inverses. That is, we show that \(I(V(I)) = I\) if \(I\) is a radical ideal, and \(V(I(S)) = S\) if \(S\) is a Zariski closed subset of \(\mathbb A^n_k\).

Let us first show that for any ideal \(I\), we have \(I(V(I)) = \sqrt I\). Suppose \(f \in \sqrt I\), then \(f^n \in I\) for some \(n > 0\). But then \(f^n\) is identically zero on \(V(I)\), and hence so is \(f\); that is, \(f \in I(V(I))\). It remains to show that \(I(V(I)) \subset \sqrt I\). Let \(f \in I(V(I))\). Then \(f\) is identically zero on \(V(I)\). By 1.6.10, there is some \(n\) such that \(f^n \in I\), and hence \(f \in \sqrt I\).

Let us now show that \(V(I(S)) = S\). Since \(S\) is Zariski closed, we know that \(S = V(J)\) for some ideal \(J\). So \(I(S) = I(V(J)) = \sqrt{J}.\) But we know that \(V(J) = V(\sqrt{J})\), and hence \(V(I(S)) = S\). The proof of Theorem 1.6.7 is then complete.

We say that \(f\) is regular (or algebraic) at \(a\) if there exists a Zariski open set \(U \subset \mathbb A^n\) and polynomials \(p, q \in k[x_1,\dots,x_n]\) with \(q(a) \neq 0\) such that \[ f \equiv p/q \text{ on } S \cap U.\] We say that \(f\) is regular if it is regular at all points of \(S\).

In other words, \(f\) is regular at a point \(a\) if locally around \(a\) (in the Zariski topology), \(f\) can be expessed as a ratio of two polynomials. Although the definition of a regular function makes sense for \(S \subset \mathbb A^n\), we use it only in the context of quasi-affine varieties.

1. A constant function is regular.
2. Every polynomial function is regular.
3. Sums and products of regular functions are regular. So, the set of regular functions forms a ring. This ring contains a copy of \(k\), namely the constant functions.

We denote the ring of regular functions on \(S\) by \(k[S]\). This ring is a \(k\)-algebra.

Let \(f\) be a function on \(S\), and let \(\{U_i\}\) be an open cover of \(S\). If the restriction of \(f\) to each \(U_i\) is regular, then \(f\) is regular.

Let \(X \subset \mathbb A^n\) be a Zariski closed subset. Let \(f\) be a regular function on \(X\). Then there exists a polynomial \(P \in k[x_1,\dots,x_n]\) such that \(P(x) = f(x)\) for all \(x \in X\).

Let \(X \subset \mathbb A^n\) be any subset. We have a ring homomorphism \[ \pi \colon k[x_1,\dots,x_n] \to k[X],\] where a polynomial \(f\) is sent to the regular function it defines on \(X\).

Let \(X \subset \mathbb A^n\) be a closed subset. Then the ring homomorphism \(\pi \colon k[x_1,\dots,x_n] \to k[X]\) induces an isomorphism \[ k[x_1,\dots,x_n]/I(X) \xrightarrow{\sim} k[X].\]

We say that \(f\) is regular at a point \(a \in X\) if all its coordinate functions \(f_1, \dots, f_m\) are regular at \(a\). If \(f\) is regular at all points of \(X\), then we say that it is regular.

A regular map to \(\mathbb A^1\) is the same as a regular function.

Let \(U = \mathbb A^1 \setminus \{0\}\) and \(V = V(xy-1) \subset \mathbb A^2\). We have a regular function \(\phi \colon V \to U\) given by \(\phi(x,y) = x\). We have a regular function \(\psi \colon U \to V\) given by \(\psi(t) = (t,1/t)\). These functions are mutual inverses, and hence we have a (bi-regular) isomorphism \(U \cong V\).

1. The identity map is regular.
2. The composition of two regular maps is regular.
3. Regular maps are continuous (in the Zariski topology).

Let \(\phi \colon X \to Y\) be a regular map. If \(f\) is a regular function on \(Y\), then \(f \circ \phi\) is a regular function on \(X\).

As a result, we get a \(k\)-algebra homomorphism \(k[Y] \to k[X]\), often denoted by \(\phi^*\): \[ \phi^* (f) = f \circ \phi.\]

We thus get a (contravariant) functor from the category of (quasi-affine) varieties to \(k\)-algebras. On objects, it maps \(X\) to \(k[X]\). On morphisms, it maps \(\phi \colon X \to Y\) to \(\phi^* \colon Y \to X\). It is easy to check that this recipe respects composition. That is, if we have maps \(\phi \colon X \to Y\) and \(\psi \colon Y \to Z\), and if we let \(\psi\circ\phi \colon X \to Z\) be the composite, then \[ (\psi \circ \phi)^* = \phi^* \circ \psi^*.\]

If \(\phi \colon X \to Y\) is an isomorphism of varieties, then \(\phi^* \colon k[Y] \to k[X]\) is an isomorphism of \(k\)-algebras.

Let \(X \subset \mathbb A^n\) and \(Y \subset \mathbb A^m\) be Zariski closed, and let \(f \colon k[Y] \to k[X]\) be a homomorphism of \(k\)-algebras. Then there is a unique (regular) map \(\phi \colon X \to Y\) such that \(f = \phi^*\).

Let \(X = \mathbb A^1_k\) and \(Y = V(y^2-x^3) \subset \mathbb A^2_k\). We have a regular map \(f \colon X \to Y\) given by \(f(t) = (t^2, t^3)\). It is easy to check that \(f\) is a bijection, but not an isomorphism.

Let \(U_f \subset \mathbb A^n\) be the complement of \(V(f)\). Then \(U_f\) is isomorphic to an affine variety, namely the variety \(V(yf-1) \subset \mathbb A^{n+1}\), where \(y\) denotes the \((n+1)\)-th coordinate.

The previous proposition only applies to the complement of \(V(f)\) for a single \(f\)! The complement of \(V(I)\), in general, is not isomorphic to an affine variety. For example, the complement of the origin in \(\mathbb A^2\) is not isomorphic to an affine variety.

An algebraic variety is a topological space with a quasi-affine atlas.

A quasi-affine variety \(X\) is itself an algebraic variety. The atlas is the obvious one, consisting of the single chart \({\rm id} \colon X \to X\).

The projective \(n\)-space over a field \(k\), denoted by \(\mathbb P^n_k\), is the set of one-dimensional subspaces of \(k^{n+1}\).

Before describing how \(\mathbb P^n_k\) is an algebraic variety, let us build some intuition about projective space. For easy visualisations, it helps to take \(k = \mathbb R\) or \(k = \mathbb C\). A one dimensional subspace of \(k^{n+1}\) is also called a line. Note that, by this definition, a line must contain the origin.

Let us take \(n = 0\). Then there is a unique one-dimenional subspace of \(k^{n+1} = k\), so \(\mathbb P^0_k\) is just a single point.

Let us take \(n = 1\). Then \(\mathbb P^1_k\) is the set of lines (through the origin) in \(k^2\). Let us take \(k = \mathbb R\). Every line through the origin is uniquely determined by its slope, which can be any element of \(\mathbb R\), so it seems like \(\mathbb P^1_{\mathbb R}\) is just a copy of \(\mathbb R\). But the vertical line does not have a (finite) slope, so \(\mathbb P^1_{\mathbb R} = \mathbb R \cup \{\infty\}\). In other words, \(\mathbb P^1\) contains the usual real line, plus “a point at infinity”.

It can be more instructive to see this in a picture. Fix a horizontal line \(L\) at, say, \(y = -1\). Every line through the origin intersects \(L\) at a unique point, except the horizontal line. So if we discard the one point of \(\mathbb P^1_k\) corresponding to the horizontal line, the rest is just a copy of \(L\). If we had chosen a different reference line \(L\), for example, a vertical one, then we get a similar description of \(\mathbb P^1\) away from a single point. In fact, we can discard any one point of \(\mathbb P^1\), and the rest will be a copy of \(\mathbb R\).

Let us take \(n = 2\). Then \(\mathbb P^2_k\) is the set of lines (through the origin) in \(k^3\). We can use the same technique as before: fix a reference plane \(P\) at \(z = -1\). Then most lines are uniquely characterised by their intersection point with \(P\). The only exceptions are the lines parallel to \(z = -1\), that is, the lines lying in the plane \(z = 0\), which we miss. But these form a small projective space \(\mathbb P^1\). So we see that \(\mathbb P^2 = P \sqcup \mathbb P^1\).

A one-dimensional subspace of \(k^{n+1}\) is spanned by a non-zero vector \((a_0,\dots,a_n)\). Two vectors \((a_0,\dots,a_n)\) and \((b_0,\dots,b_n)\) span the same subspace if and only if there exists \(\lambda \in k^\times\) such that \[ (b_0,\dots,b_n) = (\lambda a_0, \dots, \lambda a_n).\] So, we can identify \(\mathbb P^n\) with the equivalence classes of non-zero vectors \((a_0, \dots, a_n)\) where two non-zero vectors are considered equivalent if one is a scalar multiple of the other. In other words, we have \[ \mathbb P^n_k = (\mathbb A^{n+1} \setminus 0)/{\rm scaling}.\] We denote the equivalence class of \((a_0,\dots,a_n)\) by \([a_0:\dots:a_n]\).

We give \(\mathbb P^n_k\) the quotient topology inherited from \(\mathbb A^{n+1} \setminus 0.\) That is, a set \(U \subset \mathbb P^n_k\) is open/closed if and only if its pre-image in \(\mathbb A^{n+1} \setminus 0\) is open/closed.

For example, consider the subset \(U_n\) of \(\mathbb P^n_k\) consisting of \([a_0:\dots:a_n]\) with \(a_n \neq 0\). Its preimage in the set of \((a_0,\dots,a_n) \in \mathbb A^{n+1} \setminus 0\) with \(a_n \neq 0\), which is a (Zariski) open set. Hence \(U_n\) is open in \(\mathbb P^n_k\). Likewise, \(U_0, U_1, \dots\) are also open. Note that we have \[ \mathbb P^n_k = U_0 \cup \dots \cup U_n;\] that is, the sets \(U_0, \dots, U_n\) form an open cover of \(\mathbb P^n\).

Consider a point \([a_0:\dots:a_n] \in U_0\), so that \(a_0 \neq 0\). By scaling by \(\lambda = a_0^{-1}\), we have a distinguished representative of this point of the form \([1:b_1:\dots:b_n]\), which we can think of as a point \((b_1,\dots,b_n) \in \mathbb A^n\). Thus, we have a bijection \(\phi_0 \colon U_0 \to \mathbb A^n\), and similarly \(\phi_1 U_i \to \mathbb A^n\).

1. The bijections \(\phi_i \colon U_i \to \mathbb A^n\) defined above are homeomorphisms.
2. The charts \(\phi_i \colon U_i \to \mathbb A^n\) are mutually compatible, and hence give an atlas on \(\mathbb P^n\).

Let \(X\) be an algebraic variety, and \(Y \subset X\) an open or closed subset. Then \(Y\) inherits the structure of an algebraic variety. To get, the atlas for \(Y\), let \(\phi_i \colon U_i \to V_i\) be an atlas for \(X\). For \(Y\), we just take \(\phi_i \colon U_i \cap Y \to \phi(U_i \cap Y).\)

Let \(F \in k[X_0,\dots,X_n]\) be a homogeneous polynomial. Let \(V(F) \subset \mathbb P^n\) be the set of points \(\{[a_0:\dots:a_n] \mid F(a_0,\dots,a_n) = 0\}\). Then \(V(F)\) is a closed subset.

Let \(I \subset k[X_0,\dots,X_n]\) be a homogeneous ideal.

Conversely, let \(X \subset \mathbb P^n\) be a closed subset. Then there exists a homogeneous ideal \(I \subset k[X_0,\dots,X_n]\) such that \(X = V(I)\).

Suppose \(I \subset k[X_0, \dots, X_n]\) is generated by (homogeneous) linear equations. Then \(V(I) \subset \mathbb A^{n+1}\) is a sub-vector space \(W \subset \mathbb A^{n+1}\), and \(V(I) \subset \mathbb P^n\) is naturally the projective space of \(W\). We call such \(V(I) \subset \mathbb P^n\) linear subspaces, or “lines”, “planes”, etc. See that any two distinct lines in \(\mathbb P^2\) intersect at a unique point, and through any two distinct points in \(\mathbb P^2\) passes a unique line.

Let \(X\) be an algebraic variety and \(f \colon X \to k\) a function. Let \(\phi_1 \colon U_1 \to V_1\) and \(\phi_2 \colon U_2 \to V_2\) be two compatible charts such that \(x\) lies in both \(U_1\) and \(U_2\). Denote the images of \(x\) in the two charts by \(v_1\) and \(v_2\). Consider the functions \(f \circ \phi_1^{-1} \colon V_1 \to k\) and \(f \circ \phi_2^{-1} \colon V_2 \to k\). Then the first is regular at \(v_1\) if and only if the second is regular at \(v_2\).

Let \(f \colon X \to k\) be a continuous function. We say that \(f\) is regular at \(x\) if for some (equivalently, for every) chart \(\phi \colon U \to V\) with \(x \in U\), the function \(f \circ \phi^{-1} \colon V \to k\) is regular at \(\phi(x)\). We say that \(f\) is regular on \(X\) if it is regular at all points \(x \in X\).

Let \(X\) and \(Y\) be algebraic varieties and \(f \colon X \to Y\) a continuous map. We say that \(f\) is regular at a point \(x \in X\) if for any (equivalently, for every) chart \(\phi \colon U \to V\) with \(x \in U\) and \(\psi \colon U' \to V'\) with \(f(x) \in U'\), the composite map \[ \psi \circ f \circ \phi^{-1} \colon V \dashrightarrow V'\] is regular at \(\phi(x)\).T The reason for the dashed arrow is that the domain of \(\psi \circ f \circ \phi^{-1}\) may not be all of \(V\), but only an open subset of \(V\). To be precise, the domain is \(\phi (U \cap f^{-1}(U'))\). But the domain contains \(\phi(x)\), so it makes sense to talk about the regularity at \(\phi(x)\).

See Figure 2 for a picture (the bottom arrow should be dashed).

Let \(X = \mathbb P^1\). Set \(f([X:Y]) = X/Y\). Then \(f\) is defined at all points except the point \([1:0]\), and is a regular function on \(\mathbb P^1 \setminus \{[1:0]\}\). More generally, let \(X = \mathbb P^n\) and let \(F, G \in k[X_0,\dots,X_n]\) be homogeneous polynomials of the same degree. The function \[ [X_0:\dots:X_n] \mapsto F(X_0, \dots, X_n)/G(X_0, \dots, X_n)\] is regular outside \(V(G)\).

Let \(X = \mathbb P^n\) and let \(F_0, \dots, F_m\) be homogeneous polynomials of the same degree. Let \(Z \subset \mathbb P^n\) be \(V(F_0, \dots, F_m)\). Then the formula \[ [X_0:\dots:X_n] \mapsto [F_0(X_0,\dots, X_n):\dots:F_m(X_0,\dots, X_n)]\] defines a regular map from \(X \setminus Z\) to \(\mathbb P^m\).

The natural map \(\mathbb A^{n+1} - 0 \to \mathbb P^n\) is regular.

Consider the \(n+1\)-dimensional \(k\)-vector space \(V\) spanned by \(X_0, \dots, X_n\). Pick any basis \(\ell_0, \dots, \ell_n\) of this vector space. Then we have a regular map

Explicitly, if we write \[ \ell_i = L_{i,0}X_0 + \dots + L_{i,n}X_n\] and write our homogenous vector as a column vector, then the map is \[ [X] \mapsto [LX].\] In other words, it is induced by the invertible linear map \(L \colon V \to V\). As a result, it has an inverse, induced by the inverse of the matrix \(M\): \[ [X] \mapsto [MX].\] In this way, we get an action of \(GL_n(k)\) on \(\mathbb P^n\). But notice that a matrix \(L\) and a scalar multiple \(\lambda L\) induce the same map on \(\mathbb P^n\). So the action descends to an action of the group \(PGL_n(k) = GL_n(k) / \rm{scalars}.\)

The previous example gave examples of regular functions on (strict) open subsets of the projective space. It turns out that there are no regular functions on \(\mathbb P^n\) other than the constant functions!

The identity map is regular. The composition of two regular maps is regular.

The image of \(v_n\) is a closed subset of \(C\) of \(\mathbb P^n\). If we denote the homogeneous coordinates on \(\mathbb P^n\) by \([U_0:\dots: U_n]\), then \(C\) is cut out by the equations \[\{U_iU_j - U_k U_\ell \mid 0 \leq i,j,k,l \leq n \text{ and } i+j = k+\ell.\]

The map \(v_n \colon \mathbb P^1 \to C\) is in fact an isomorphism.

The proposition above generalises to all dimensions. Consider the \(k\)-vector space of degree \(n\) homogeneous polynomials in \(X_0, \dots, X_m\). It has dimension \(N = {n+m \choose m}\). Choosing a basis gives a map \(\mathbb P^m \to \mathbb P^N\). The image of this map is a closed subvariety \(Z\) and the map \(\mathbb P^m \to Z\) is an isomorphism. The equations of \(Z\) and the description of the inverse map are analogous to the \(m = 1\) case, but (understandably) somewhat more cumbersome.

Let \(k\) be an algebraically closed field of characteristic \(\neq 2\) and let \(q\) be a quadratic form on a \(k\)-vector space \(V\). Then there exists a basis \(X_0, \dots, X_n\) for \(V\) such that \[ q(X_0, \dots, X_n) = X_0^2 + \dots + X_\ell^2.\]

The form is called non-degenerate if \(\ell = n\).

Let \(Q\) be a non-degenate conic in \(\mathbb P^2\). Then \(Q\) is isomorphic to \(\mathbb P^1\).

What do the degenerate conics in \(\mathbb P^2\) look like?

The two projection maps \(X \times Y \to X\) and \(X \times Y \to Y\) are regular. A map \(\phi \colon Z \to X \times Y\) is regular if and only if the two component maps \(\phi_1 \colon Z \to X\) and \(\phi_2 \colon Z \to Y\) are regular.

If you have seen some category theory (in particular, Yoneda’s lemma), you will see that the above proposition characterises the product “uniquely up to a unique isomorphism.”

The rank 1 locus \(Z \subset \mathbb P^N\) is closed, and the Segre map \(\mathbb P^n \times \mathbb P^m \to Z\) is a bi-regular isomorphism.

A projective variety is a variety isomorphic to a closed subset of projective space. A quasi-projective variety is a variety isomorphic to an open subset of a projective variety.

Every quasi-affine variety is quasi-projective.

If \(X\) and \(Y\) are (quasi)-projective, then so is \(X \times Y\).

The Segre embedding of \(\mathbb P^1 \times \mathbb P^1\) lives in \(\mathbb P^3\).

Consider the diagonal map \(\Delta \colon \mathbb P^n \to \mathbb P^n \times \mathbb P^n.\) The image of \(\Delta\) is a closed subset. If we use homogeneous coordinates \([X_0:\dots:X_n]\) and \([Y_0 : \dots :Y_n]\) on the two copies of \(\mathbb P^n\), then the image is the vanishing set of the bi-homogeneous polynomials \[ X_iY_j - X_j Y_i \text{ for } 0 \leq i,j \leq n. \]

Algebraic varieties \(X\) for which the image of the diagonal map \(\Delta \colon X \to X \times X\) is closed are called separated. This condition is analogous to the Hausdorff condition in topology. Not all varieties are separated, but all quasi-projective varieties are.

All quasi-projective varieties are separated.

The collection of charts \(\{\phi_I\}\) gives an atlas on the Grassmannian.

The image of the Plucker embedding \(p\) is a closed subset of \(\mathbb P^N\) and the map \(p\) is an isomorphism onto the image.

The space \(X = V(xy) \subset \mathbb A^2\) is reducible (in the Zariski topology). We can write \(X\) as the union \(V(x) \cup V(y)\). On the other hand, we will soon see that \(X = V(xy-1)\) is irreducible (the real picture is misleading!).

In the usual Euclidean topology, we rarely encounter irreducible spaces. In fact, it is not hard to show that \(X \subset \mathbb R^n\) is irreducible (in the Euclidean topology) if and only if \(X\) is a single point. But irreducibility turns out to be an important notion in algebraic geometry.

The following are equivalent

1. \(X\) is irreducible.
2. Every non-empty open subset of \(X\) is dense.
3. Any two non-empty open subsets of \(X\) have a non-empty intersection.

1. Suppose \(U \subset X\) is dense. Then \(U\) is irreducible if and only if \(X\) is irreducible.
2. If \(f \colon X \to Y\) is a surjective continuous map and \(X\) is irreducible, then \(Y\) is irreducible.

For affine varieties, irreducibility is (unsurprisingly) related to a well-known property of the ring of regular functions.

Let \(X \subset \mathbb A^n\) be a Zariski closed subset. Then the following are equivalent.

1. \(X\) is irreducible.
2. \(I(X)\) is a prime ideal.
3. \(k[X]\) is an integral domain.

The Grassmannians (and in particular, the projective spaces) are irreducible.

Let \(X\) and \(Y\) be irreducible varieties. Then \(X \times Y\) is irreducible.

Let \(X \subset \mathbb P^n\) be an irreducible subset. Then the cone \(C \subset \mathbb A^{n+1}\) over \(X\) is closed.

A topological space \(X\) is Noetherian if every nested sequence of closed subsets \[ X \supset X_1 \supset X_2 \supset X_3 \supset \cdots\] stabilises.

A consequence of the Hilbert basis theorem is that every affine variety is Noetherian. It is easy to check that if \(X\) has a finite open cover by Noetherian topological spaces, then \(X\) is Noetherian. As a result, every algebraic variety of finite type is Noetherian. (A variety is of finite type if it has an atlas consisting of finitely many charts.)

Let \(X\) be a Noetherian topological space. We can write \[ X = X_1 \cup \dots \cup X_n,\] where \(X_i \subset X\) are irreducible closed subsets with \(X_i \not \subset X_j\) for \(i \neq j\). Furthermore, this decomposition is unique (up to permutation of the factors).

The factors \(X_i\) are called irreducible components of \(X\).

Let \(X = V(f) \subset \mathbb A^n\). Then the unique decomposition of \(X\) into irreducible components corresponds precisely to the unique factorisation of \(f\) into prime factors.

In the following, all varieties are assumed to be irreducible and separated.

1. Any variety is birational to any of its open subsets.
2. The affine space \(\mathbb A^n\), the projective space \(\mathbb P^n\), any product \(\mathbb P^a \times \mathbb P^b\) with \(a+b = n\) (and any triple product etc.) are in the same birational isomorphism class.
3. The group of biregular automorphisms of \(\mathbb P^n\) turns out to be quite easy to understand—it is just \({\rm PGL}_{n+1}\)—but the group of birational automorphisms is huge and very poorly understood (except when \(n = 1\), where it agrees with the biregular automorphisms group by one of the homework questions). Here is an example of a birational automorphism of \(\mathbb P^2\), called a ’Cremona transformation’: \[ \phi \colon [X:Y:Z] \mapsto [1/X:1/Y:1/Z]. \]

Let \(X\) be an irreducible variety. The set of rational maps \(X \dashrightarrow \mathbb A^1 = k\) is naturally a ring. But in fact, it is actually a field, called the fraction field of \(X\), and is denoted by \(k(X)\). < If \(X\) is affine, then we really do have \[ k(X) = \operatorname{frac} k[X].\]

It is easy to check that a birational isomorphism \(f \colon X \dashrightarrow Y\) induces an isomorphism of fields over \(k\): \[ f^* \colon k(Y) \to k(X).\] (and conversely).

Let \(X\) be irreducible and \(Y \subset X\) a proper closed subset. For any \(y \in Y\), we have \({\rm krdim}_yY < {\rm krdim}_yX\).

Let \(X\) be any variety, \(f\) a regular function on \(X\), and \(Y = V(f)\) the zero locus of \(f\). For any \(y \in Y\), we have \({\rm sldim}_y Y \geq {\rm sldim}_y X - 1\).

Slogan: Slicing by 1 function cuts down the dimension by at most 1.

There are instances where the inequality is strict.

Let \(f \colon X \to Y\) be a dominant map of irreducible varieties. Then \({\rm trdim} Y \leq {\rm trdim} X\).

Let \(X\) be an algebraic variety and \(x \in X\) a point. Then we have \[ {\rm krdim}_x X = {\rm sldim}_x X.\] Furthermore, if \(X\) is irreducible, then both are equal to \({\rm trdim} X\).

We denote the dimension by \({\rm dim}_x X\). The theorem says that if \(X\) is irreducible then this number does not depend on \(x\). If \(X\) is reducible, then it is easy to see (using the Krull dimension) that \({\rm dim}_x X\) is the maximum of the dimensions of the irreducible components of \(X\) that contain \(x\). A variety is equidimensional if \({\rm dim}_x X\) is the same for all \(x \in X\). This is the same as saying that all irreducible components of \(X\) have the same dimension.

We will not prove this theorem. Its proper place is a course in commutative algebra. The famous book “Commutative Algebra” by Atiyah and MacDonald has an excellent exposition (in the last chapter), where they also give a fourth equivalent definition.

For irreducible \(X\) and \(Y\), we have \[{\rm dim} (X \times Y) = {\rm dim} X + {\rm dim} Y. \]

The dimension of \(\mathbb A^1\) is 1 (you should be able to check this using any of the definitions). As a result, the dimension of \(\mathbb A^n\) is \(n\). Consequently, the dimension of \(\mathbb P^n\) is \(n\) and the dimension of \({\rm Gr}(m,n)\) is \(m(n-m)\).

Let \(f \in k[x_1, \dots, x_n]\) be non-zero. Then \(V(f) \subset \mathbb A^n\) is equidimensional of dimension \((n-1)\). Conversely, any closed \(X \subset \mathbb A^n\) which is equidimensional of dimension \((n-1)\) has the form \(V(f)\) for some \(f \in k[x_1, \dots, x_n]\).

Let \(F \in k[X_0, X_1, \dots, X_n]\) be homogeneous and non-zero. Then \(V(F) \subset \mathbb P^n\) is equidimensional of dimension \((n-1)\). Conversely, any closed \(X \subset \mathbb P^n\) which is equidimensional of dimension \((n-1)\) has the form \(V(F)\) for some homogeneous \(F \in k[X_0, \dots, X_n]\).

Let \(X \subset \mathbb P^n\) be closed of dimension \(r \geq 1\) and let \(F \in k[X_0, \dots, X_n]\) be homogeneous of positive degree. Then \(X \cap V(F)\) is non-empty and of dimension at least \(r-1\).

In \(\mathbb P^n\), a collection of at most \(n\) homogeneous forms (of positive degree) have a non-empty intersection.

Suppose \(n > m\). Then there are no non-constant regular maps from \(\mathbb P^n\) to \(\mathbb P^m\).

The proof relies on the following fact about maps from one projective space to another.

Let \(U \subset \mathbb P^n\) be an open subset and \(\phi \colon U \to \mathbb P^m\) a regular function. Then there exist homogeneous functions \(F_0, \dots, F_m \in k[X_0, \dots, X_n]\) of the same degree such that they have no common zero on \(U\) and for every \(u \in U\), we have \[ \phi(u) = [F_0(u): \dots :F_m(u)]\]

Suppose we have a regular map \(\phi \colon \mathbb P^n \to \mathbb P^m\). By Proposition 8.5.8, there exist \(F_0, \dots, F_m\) such that they have no common zero and \(\phi =[F_0:\dots:F_m]\). By Corollary 8.5.6 this is impossible if \(m < n\).

Let \(f \colon X \to Y\) be a dominant map between irreducible varieties. Then for every \(x \in X\) with \(y = f(x)\), we have \[ {\rm dim}_x f^{-1}(y) \geq {\rm dim} X - {\rm dim} Y.\] Furthermore, there exists a non-empty open \(U \subset Y\) such that for every \(y \in U\), the fiber \(f^{-1}(y)\) is non-empty and equidimensional of dimension \({\rm dim} X - {\rm dim} Y\).

That is, for almost all \(y \in Y\), the dimension of the fiber is the difference in the dimensions, as expected. But there may be some points in \(Y\) whose fiber has a different dimension. But in this case, the dimension can only be bigger, not smaller.

The proof of the theorem uses transcendental dimension. The proof is straightforward, but a bit technical, so I am skipping it. See Chapter 1, Section 6.3 of Shafarevich for the proof.

Let us construct an example where the dimension does actually jump. Consider \[ f \colon \mathbb A^2 \to \mathbb A^2\] defined by \[ f (x,y) = (xy, y). \] For all \((a,b)\) such that \(b \neq 0\), the fiber is a single point (dimension 0). But over the point \((0,0)\), the fiber is a copy of \(\mathbb A^1\) (dimension 1).

Theorem 8.6.1 is used very often in finding dimensions. Here is a typical example.

Let \(\mathbb A^{n \times n}\) be the affine space of \(n \times n\) matrices, and given \(r \in \{0,1,\dots,n\}\), let \(X_r \subset \mathbb A^n\) be the set of matrices of rank at most \(r\). The subset \(X_r\) is Zariski closed (it is the vanishing locus of all \((r+1) \times (r+1)\))-minors, and it is not hard to check that it is irreducible. What is its dimension?

Consider \(P \subset \mathbb A^{n \times n} \times {\rm Gr}(n-r,n)\) consisting of \((M, V)\) (where \(M\) is an \(n \times n\) matrix and \(V \subset k^n\) is an \(n-r\) dimensional subspace) such that \(M v = 0\) for all \(v \in V\). That is, the restriction of the linear map \(M \colon k^n \to k^n\) to \(V\) is zero.

Claim 1: \(P\) is a Zariski closed subset.

We can prove that \(P\) is also irreducible, but let us skip this for now.

Claim 2. The dimension of \(P\) is \(r(2n-r)\).

Claim 3. The dimension of \(X_r\) is \(r(2n-r)\).

Let \(X\) be irreducible. Then we have a natural inclusion \(O_{X,x} \subset k(X)\) and \(O_{X,x}\) is the set of rational functions on \(X\) which are defined at \(x\).

In particular, if \(X\) is affine and irreducible, it is easy to calculate the ring of germs.

Let \(X\) be irreducible and affine. Then the ring \(O_{X,x} \subset {\rm frac\ } k[X]\) is given by \[ O_{X,x} = \left\{\frac{f}{g} \mid f \in k[X], g \in k[X], g(x) \neq 0. \right\}\] That is, in the denominator, we are only allowed to have functions which are not zero at \(x\).

Here is another explicit description of the local ring for an affine.

Let \(X \subset \mathbb A^n\) be the closed subset with \(I(X) = \langle f_1, \dots, f_r \rangle.\) Let \(x = (a_1,\dots, a_n) \in X\). Then \(O_{X,x}\) is the quotient of \(O_{\mathbb A^n, x}\) by the ideal generated by \(f_1, \dots, f_r\).

The construction of the local ring is functorial. That is, if we have a regular map \(f \colon X \to Y\) such that \(y = f(x)\), then pull-back of functions induces a \(k\)-algebra homomorphism \[ f^* \colon O_{Y,y} \to O_{X,x}.\] If \(f\) is a local isomorphism—that is, if there exist opens \(U \subset X\) and \(V \subset Y\) containing \(x\) and \(y\), respectively, such that \(f\) induces an isomorphism \(f \colon U \to V\)—then \(f^*\) is an isomorphism.

Let \(m \subset O_{X,x}\) be the set of germs \(f\) such that \(f(x) = 0\). Equivalently, let \(m\) be the kernel of the map \[ O_{X,x} \to k \] that sends \(f\) to \(f(x)\). Then \(m\) is a maximal ideal. It is not hard to see that this is the only maximal ideal of \(O_{X,x}\).

The ring \(O_{X,x}\) has a unique maximal ideal \(m\), which consists of functions that vanish at \(x\).

A local ring is a ring with a unique maximal ideal. We just proved that \(O_{X,x}\) is a local ring. Local rings are intensely studied in commutative algebra, mostly because they arise as rings of germs in geometry.

A tangent vector to \(X\) at \(x\) is a \(k\)-algebra homomorphism \[ v \colon O_{X,x} \to k[\epsilon]/\epsilon^2.\]

Let us understand this concretely when \(X\) is affine, say \(X \subset \mathbb A^n\) closed. Let \(I(X) = \langle f_1, \dots, f_r \rangle\). Then \(X\) is the set of \(k\)-valued solutions of the system of equations

Let \(x = (a_1,\dots,a_n) \in X\). We have a bijection between \(k\)-algebra homomorphisms \(O_{X,x} \to k[\epsilon]/\epsilon^2/\) and \(k[\epsilon]/\epsilon^2\)-valued solutions of the system (\eqref{org7252932}) based at \((a_1,\dots,a_n)\), that is, solutions of the form \((a_1+b_1\epsilon, \dots, a_n + b_n \epsilon)\).

In the proof of 9.2.2, we saw that the “constant term” of \(v(f)\) must be \(f(x)\), that is \(v\) must have the form \[ v(f) = f(x) + \epsilon \cdot \delta(f)\] where \(\delta \colon O_{X,x} \to k\) is some function. Since \(v\) is a ring homomorphism, it satisfies \[v(f+g) = v(f) + v(g) \text{ and } v(fg) = v(f)v(g). \] In terms of \(\delta\), these become

Furthermore, for a constant function \(c\), we have \(v(c) = c\), and hence

Equation (\eqref{org2187d85}) should remind you of the sum and product rule for derivatives. Maps \(\delta \colon O_{X,x} \to k\) satisfying these equation are called derivations. If they also satisfy equation (\eqref{org3731762}), then they are called \(k\)-derivations or derivations over \(k\). This indicates that the elements of \(k\) in \(O_{X,x}\) are to be treated as “constants”. Denote by \({\rm Der}_k(O_{X,x})\) the set of \(k\)-derivations of \(O_{X,x}\). Note that derivations can be added and multiplied by scalars (elements of \(k\)), which makes \({\rm Der}_k(O_{X,x})\) a \(k\)-vector space.

We saw that a \(k\)-algebra homomorphism \(v \colon O_{X,x} \to k\) gives a \(k\)-derivation \(\delta \colon O_{X,x} \to k\). Conversely, it is easy to check that a \(k\)-derivation \(\delta \colon O_{X,x} \to k\) gives a \(k\)-algebra homomorphism \(v(f) = f(x) + \epsilon \cdot \delta(f)\). Thus, a tangent vector to \(X\) at \(x\) is equivalent to a \(k\)-derivation of \(O_{X,x}\).

Geometrically, the correspondance between curves and derivations is as follows. A curve in a space gives a recipe to differentiate a function; this is the directional derivative of the function in the direction of the curve. But to define the directional derivative, we don’t need an actual curve, an “infinitesimal curve” will do. There is no way (that I know of) to make this precise in (differential) geometry, but it can be made perfectly precise in algebraic geometry using the ring \(k[\epsilon]/\epsilon^2\).

Let \(m \subset O_{X,x}\) be the maximal ideal. A derivation \(\delta \colon O_{X,x} \to k\) restricted to \(m\) gives a \(k\)-linear map \[\delta \colon m \to k\] that takes \(m^2\) to \(0\), and hence gives a map \[\overline{\delta} \colon m/m^2 \to k. \] Conversely, any \(k\)-linear map \(w \colon m/m^2 \to k\) gives a derivation \(\delta \colon O_{X,x} \to k\) defined by \[ \delta(f) = w(f - f(x)),\] where \(f(x)\) denotes the constant function on \(X\) with value \(f(x)\). Thus, we get an isomorphism of vector spaces \[ {\rm Der}_k (O_{X,x}) \cong {\rm Hom}(m/m^2, k).\] The space \({\rm Hom}(m/m^2,k)\) is called the Zariski tangent space and \(m/m^2\) is called the Zariski cotangent space to \(X\) at \(x\).

Let \(X \subset \mathbb A^n\) be affine with \(I(X) = \langle f_1, \dots, f_r \rangle\) and let \(x = (a_1,\dots,a_n)\) be a point of \(X\). We know that \(O_{X,x}\) is the quotient of \(O_{\mathbb A^n, x}\) by \(\langle f_1,\dots,f_r \rangle\). Let us denote the maximal ideal of \(O_{\mathbb A^n,x}\) by \(\mathfrak m\). Then \(\mathfrak m\) is generated by \(\langle x_1-a_1, \dots, x_n-a_n\rangle\) and its square \({\mathfrak m}^2\) is generated by the pairwise products. As a result, \(\mathfrak m/\mathfrak m^2\) has the \(k\)-basis \((x_1-a_1, \dots, x_n-a_n)\). To get \(m/m^2\), we need to further quotient by the polynomials \(f_1,\dots,f_r\). Let \(\overline f_1, \dots, \overline f_r\) denote the images of \(f_1,\dots,f_r\) in \(\mathfrak m/{\mathfrak m}^2\). Then \[ m/m^2 = \langle x_1-a_1, \dots, x_n-a_n\rangle / \langle \overline f_1, \dots, \overline f_r\rangle.\] But what are these mysterious \(\overline f_1, \dots, \overline f_r\). They are not mysterious at all! We have \[ \overline f_i = \frac{\partial f_i}{\partial x_1}(a_1,\dots,a_n) \cdot (x_1-a_1) + \dots + \frac{\partial f_i}{\partial x_n}(a_1,\dots,a_n) (x_n - a_n).\]

Let \(T_xX\) denote the tangent space of \(X\) at \(x\).

We have \({\rm dim} T_x X \geq {\rm dim}_x X\).

We say that \(X\) is smooth or non-singular at \(x\) if \[{\rm dim}_x X = {\rm dim}\ T_x X.\]

Affine spaces, projective spaces, and Grassmannians are smooth at all points. So are their open subsets.

\(X = V(f) \subset \mathbb A^n\) is smooth at \(x\) if and only if at least one of the partial derivatives of \(f\) is non-zero at \(x\).

The Fermat cubic \(V(X^3+Y^3+Z^3) \subset \mathbb P^2\) is smooth at every point on it.

We say that a variety \(X\) is complete if for any \(Y\), the projection map \[ \pi \colon X \times Y \to Y\] is closed.

Why is this a big deal? Let us consider an example, one we have seen in the homework. Let \(V\) be the vector space of homogeneous polynomials of degree \(d\) in \(X_0, X_1, X_2\) and let \(\Delta \subset V\) be the set of polynomials \(F\) that have a singularity at some point \(p \in \mathbb P^2\). (This means that all three partials of \(F\) vanish at \(p\)). That is, \[ \Delta = \{F \mid \exists p \text{ such that } \frac{\partial F}{\partial X_i} (p) = 0 \text{ for } i = 0,1,2\}.\] We want to prove that \(\Delta \subset V\) is closed. Let us eliminate the existential quantifier by considering the set \[ Z = \{(F,p) \mid \frac{\partial F}{\partial X_i}(p) = 0 \text{ for } i = 0,1,2. \} \subset V \times \mathbb P^2.\] It is easy to see that \(Z\) is closed: it is defined by polynomial equations in the coefficients of \(F\) and the coordinates of \(p\). By definition, \(\Delta\) is the image of \(Z\) under the projection map \(V \times \mathbb P^2 \to V\). Since \(\mathbb P^2\) is projective, hence complete, the image is closed.

The upshot is that Theorem 10.3 allows us to eliminate existential quantifiers as long as they are quantified over a complete variety. Note that the resulting statements about closedness can be extremely non-trivial. The fact that \(\Delta \subset V\) is closed means that there is a system of polynomials in the coefficient of \(F\) that detects whether \(F\) has a singularity. (In the homework, you proved that \(\Delta\) has codimension 1, which shows that the system consists of just one equation.)

Here are some more examples of sets that we can show are closed by the same reasoning.

1. The subset of \({\rm Gr}(2, 4) \times {\rm Gr}(2, 4)\) consisting of \((V,W)\) such that \(V \cap W\) is non-zero.
2. Let \(\mathbb P V\) be the projective space of surfaces of degree \(d\) in \(\mathbb P^3\). The subset of \(\mathbb P V\) consisting of surfaces that contain a line.

Intuitively, what does it mean that \(\pi \colon X \times Y \to Y\) is closed? Suppose you have a family of points \((x_t, y_t) \in X \times Y\) such that \(\lim_{t \to 0} y_t\) exists in \(Y\). Then \(\lim_{t \to 0} x_t\) must exist in \(X\). That is, “points cannot escape to infinity in the \(X\)-direction.”

We have the following very useful criterion for irreducibility in the context of closed maps.

Let \(X\) be a connected projective variety. Then the only regular functions on \(X\) are the constant functions.