Affine Varieties

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In algebraic geometry, our goal is to study geometric objects defined by polynomials. Specifically, given a field \(\mathbb{K}\) and a natural number \(n\), we are interested in sets of the form

\[Z(f)= \{(x_1, \ldots, x_n) \in \mathbb{A}^n \mid f(x_1, \ldots, x_n) = 0\},\qquad f\in \mathbb{K}[\x_1,\ldots, \x_n]\]

This is the set of solutions to the polynomial \(f\) in \(\mathbb{K}^n\). We usually take \(\mathbb{K}=\mathbb{C}\), and a particularly useful feature of this assumption is that \(\mathbb{C}\) is algebraically closed. However, in most cases this assumption is not very helpful, so we will work in a more general setting. Also, to avoid confusion, we will denote the variables of a polynomial \(f\) by bold letters such as \(\x\).

Definition of Affine Varieties

Definition 1 The affine \(n\)-space \(\mathbb{A}^n_\mathbb{K}\) defined over a field \(\mathbb{K}\) is the \(n\)-dimensional vector space \(\mathbb{K}^n\).

When the field \(\mathbb{K}\) is clear from context, we write simply \(\mathbb{A}^n\). We call the elements of the affine space

\[\mathbb{A}^n=\{(x_1,\ldots, x_n)\mid x_i\in \mathbb{K}\}\]

points, and each coordinate \(x_i\) the \(i\)-th coordinate. As mentioned above, the geometric objects we will examine are those represented by the zero set \(Z(f)\) of a polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\).

Definition 2 Given polynomials \(f_1, \ldots, f_k \in \mathbb{K}[\x_1, \ldots, \x_n]\), we define the affine algebraic set \(Z(f_1, \ldots, f_k)\) determined by them as

\[Z(f_1, \ldots, f_k) = \{x=(x_1, \ldots, x_n) \in \mathbb{A}^n \mid f_1(x) = \cdots = f_k(x) = 0\}\]

Among affine algebraic sets, those that cannot be expressed as the union of two strictly smaller affine algebraic sets are called affine varieties.

That is, an affine algebraic set is the set of common zeros of several polynomials \(f_1,\ldots, f_k\). More generally, for a subset \(S\) of \(\mathbb{K}[\x_1,\ldots, \x_n]\), we can define \(Z(S)\) similarly, and then by definition the ideal \((S)\) generated by \(S\) in \(\mathbb{K}[\x_1,\ldots, \x_n]\) satisfies the identity

\[Z(S)=Z((S))\]

Conversely, since the field \(\mathbb{K}\) is Noetherian, applying [Commutative Algebra] §Basic Notions, ⁋Theorem 12 (Hilbert basis theorem) repeatedly shows that \(\mathbb{K}[\x_1,\ldots, \x_n]\) is also Noetherian, and therefore any ideal \(\mathfrak{a}\) is generated by finitely many polynomials \(f_1,\ldots, f_k\), so that \(Z(\mathfrak{a})=Z(f_1,\ldots, f_k)\). Thus the affine algebraic sets in the sense of Definition 2 and the zero sets of ideals form exactly the same collection, so we need only consider affine algebraic sets defined by ideals \(\mathfrak{a}\).

In general, a space \(X\) is said to be irreducible if it cannot be written as the union of two proper closed subsets. ([Topology] §Dimension, ⁋Definition 6) Thus our definition means that we call an irreducible affine algebraic set an affine variety. This ensures that we deal with a single object that does not break into several pieces.

Example 3 Since most geometric objects we know are represented by polynomials, they all become examples of affine algebraic sets.

1. Consider the affine variety \(Z(\x^2+\y^2-1)\) defined in \(\mathbb{A}^2\). By definition, this set is the collection of points in \(\mathbb{A}^2\) satisfying the equation \(\x^2+\y^2-1=0\), so it represents the unit circle.
2. In general, for any affine space \(\mathbb{A}^n\) and any non-constant polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\), the affine algebraic set \(Z(f)\) is called a hypersurface.
3. Another important example is the twisted cubic defined in \(\mathbb{A}^3\). This is the curve defined by the two polynomials \(\y-\x^2\), \(\z-\x^3\) in \(\mathbb{A}^3\), and it is in one-to-one correspondence with \(\mathbb{A}^1\) via the parametrization \((t,t^2,t^3)\).
4. The affine space \(\mathbb{A}^n\) itself and the empty set are affine varieties. This is obvious from \(Z(0)=\mathbb{A}^n\) and \(Z(1)=\emptyset\). This will be used importantly when defining the Zariski topology in Proposition 4.

In the example above, we saw that familiar geometric objects can all be written as affine algebraic sets. However, to think of them as geometric objects, there must be a topological structure on them. Our only tool is polynomials, so we will use them to define a topology.

Proposition 4 The following hold.

1. \(Z(0) = \mathbb{A}^n\),
2. \(Z(1) = \emptyset\),
3. \(\mathfrak{a} \subseteq \mathfrak{b} \implies Z(\mathfrak{b}) \subseteq Z(\mathfrak{a})\),
4. \(\bigcap_{i\in I} Z(\mathfrak{a}_i) = Z\left(\sum_i \mathfrak{a}_i\right)\),
5. \(Z(\mathfrak{a}) \cup Z(\mathfrak{b}) = Z(\mathfrak{a} \cap \mathfrak{b}) = Z(\mathfrak{a}\mathfrak{b})\).

Proof

The first two statements were already examined in Example 3.

To prove the third result, suppose \(\mathfrak{a}\subseteq \mathfrak{b}\) and let \(x\in Z(\mathfrak{b})\). Then \(f(x) = 0\) for all \(f \in \mathfrak{b}\), and since \(\mathfrak{a} \subseteq \mathfrak{b}\), the desired identity holds for all elements of \(\mathfrak{a}\) as well.

For the fourth result, a point \(x\) belongs to all \(Z(\mathfrak{a}_i)\) if and only if \(f(x) = 0\) for all \(i\) and all \(f \in \mathfrak{a}_i\). This is equivalent to saying that every element of \(\sum_i \mathfrak{a}_i\) vanishes at \(x\).

Now let us show the last claim. First, let \(x\in Z(\mathfrak{a})\cup Z(\mathfrak{b})\). Then either \(x\in Z(\mathfrak{a})\) or \(x\in Z(\mathfrak{b})\), and in either case every element of \(\mathfrak{a}\cap \mathfrak{b}\) vanishes at \(x\), so \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\). Now suppose \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\). Evaluating at \(x\) an arbitrary element of \(\mathfrak{a}\mathfrak{b}\) of the form

\[f_1g_1+\cdots+ f_kg_k,\qquad f_i\in \mathfrak{a}, g_i\in \mathfrak{b}\]

since \(\mathfrak{a}\mathfrak{b}\subseteq \mathfrak{a}\cap \mathfrak{b}\), each \(f_ig_i\) vanishes at \(x\) and therefore their sum also vanishes. Finally, let \(x\in Z(\mathfrak{a}\mathfrak{b})\). If, contrary to the conclusion, \(x\not\in Z(\mathfrak{a})\cup Z(\mathfrak{b})\), then there exist suitable \(f\in \mathfrak{a}\) and \(g\in \mathfrak{b}\) such that \(f(x),g(x)\neq 0\). But if this were the case, then \(f(x)g(x)\neq 0\), contradicting the assumption that \(x\in Z(\mathfrak{a}\mathfrak{b})\).

First, the last result of the above proposition shows that for \(Z(\mathfrak{a}\mathfrak{b})\) to be an affine variety, one must have either \(Z(\mathfrak{a})\subseteq Z(\mathfrak{b})\) or \(Z(\mathfrak{b})\subseteq Z(\mathfrak{a})\). This provides good intuition for understanding algebraically what an affine variety is.

More importantly, by the above proposition, if we declare the affine algebraic sets defined on \(\mathbb{A}^n\) to be closed sets, then the conditions of [Topology] §Interior, Closure, and Boundary, ⁋Proposition 2 are all satisfied, and thus a topology on \(\mathbb{A}^n\) is uniquely determined. We call this the Zariski topology. By definition, any affine variety \(X\) is a closed subset of a suitable affine space \(\mathbb{A}^n\), and we can define the topology on \(X\) via the subspace topology induced from \(\mathbb{A}^n\).

As a special example, consider the Zariski topology on \(\mathbb{A}^1\): since any element of \(\mathbb{K}\) is the zero set of the linear polynomial \(\x-x\), any singleton is closed, and therefore any finite set is closed. However, any non-zero element of \(\mathbb{K}[\x]\) has at most finitely many roots, so in this topology (provided \(\mathbb{K}\) is not finite), the only closed set with infinitely many elements is \(\mathbb{K}\) itself. That is, the Zariski topology on \(\mathbb{A}^1\) is the cofinite topology, from which we observe that the Zariski topology need not be Hausdorff. More generally, an irreducible space cannot be Hausdorff unless it is a one-point space, and since affine varieties are all irreducible by our definition, no affine variety is a Hausdorff space. ([Topology] §Dimension, ⁋Proposition 7)

Now let us examine the open sets of the Zariski topology.

Definition 5 For a polynomial \(f \in \mathbb{K}[\x_1, \ldots, \x_n]\), we define the principal open set \(D(f)\) as

\[D(f) = \{x\in \mathbb{A}^n \mid f(x) \ne 0\} = \mathbb{A}^n \setminus Z(f)\]

The following proposition shows that principal open sets form a base for the affine variety. ([Topology] §Bases of a Topological Space, ⁋Definition 1)

Proposition 6 For any open set \(U\) of an affine variety \(X \subseteq \mathbb{A}^n\), there exists a family of principal open sets \(D(f_i)\) satisfying

\[U = \bigcup_i (D(f_i) \cap X)\]

Proof

By the definition of the Zariski topology, there exists a suitable ideal \(\mathfrak{a}\subseteq \mathbb{K}[\x_1,\ldots, \x_n]\) such that

\[X\setminus U=Z(\mathfrak{a})\cap X\]

Therefore

\[U = X \setminus (Z(\mathfrak{a}) \cap X) = X \cap (\mathbb{A}^n \setminus Z(\mathfrak{a}))\]

On the other hand, \(\mathbb{A}^n \setminus Z(\mathfrak{a})\) is the set of points where \(f(x) \ne 0\) for \(f \in \mathfrak{a}\), so

\[\mathbb{A}^n \setminus Z(\mathfrak{a}) = \bigcup_{f \in \mathfrak{a}} D(f)\]

Hence \(U = \bigcup_{f \in \mathfrak{a}} (D(f) \cap X)\).

In general, an open set of an affine variety need not be an affine variety, and indeed it is not. However, any non-empty principal open subset of an affine variety is necessarily an affine variety.

Proposition 7 For an affine variety \(X \subseteq \mathbb{A}^n\) and a polynomial \(f \in \mathbb{K}[\x_1, \ldots, \x_n]\) satisfying \(D(f)\cap X\neq\emptyset\), the set \(D(f) \cap X\) is an affine variety.

Proof

Choose a suitable ideal \(\mathfrak{a}\) such that \(X = Z(\mathfrak{a})\). We will represent \(D(f) \cap X\) as an affine variety in \(\mathbb{A}^{n+1}\). To this end, let the coordinates of \(\mathbb{A}^{n+1}\) be

\[\x_1,\ldots, \x_n,\y\]

and consider the ideal of \(\mathbb{K}[\x_1,\ldots, \x_n,\y]\)

\[\mathfrak{b}=\mathfrak{a}+(1-f\y)\]

Then

\[Z(\mathfrak{b})=\{(x_1,\ldots, x_n, y)\in \mathbb{A}^{n+1}\mid x\in X, 1-f(x)y=0\}\]

From the condition \(1-f(x)y=0\), we know that \(f(x)\neq 0\) and \(y=1/f(x)\). From this we obtain the following bijection

\[(x_1,\ldots, x_n,y)\mapsto (x_1,\ldots, x_n)\]

That this is a homeomorphism is obvious.

Finally, let us show that \(Z(\mathfrak{b})\) is irreducible. By assumption, \(D(f)\cap X\) is a non-empty open subset of \(X\). Then any two non-empty open sets of \(D(f)\cap X\) are also non-empty open sets of \(X\), so by the irreducibility of \(X\) and the second condition of [Topology] §Dimension, ⁋Proposition 7, they intersect. That is, \(D(f)\cap X\) is irreducible, and by the above homeomorphism, \(Z(\mathfrak{b})\) is also irreducible.

At this point, we should note that our definition of an affine variety strictly depends on the ambient space \(\mathbb{A}^n\). For example, the principal open set \(D(\x)\) of \(\mathbb{A}^1\) is, by the above proposition, an affine variety. However, we have already seen that the Zariski topology on \(\mathbb{A}^1\) is the cofinite topology, and therefore \(D(\x)\) cannot be defined as the zero set of polynomials in \(\mathbb{K}[\x]\). In fact, looking closely at the proof of Proposition 7, the fact that \(D(\x)\) is an affine variety is obtained through the isomorphism

\[D(\x)\cong Z(\x\y-1)\subseteq \mathbb{A}^2\]

This issue may confuse us somewhat when we understand regular functions, so we will revisit this problem in the relevant section.

Nullstellensatz

The \(Z\) examined in Proposition 4 sends algebraic objects, that is, polynomials in \(\mathbb{K}[\x_1,\ldots, \x_n]\), to geometric objects, that is, the zero sets defined by these polynomials. Conversely, we can also take a geometric object and assign algebraic objects to it.

Definition 8 For an arbitrary subset \(X \subseteq \mathbb{A}^n\), we define the subset \(I(X)\) of \(\mathbb{K}[\x_1,\ldots, \x_n]\) as

\[I(X) = \{f \in \mathbb{K}[\x_1, \ldots, \x_n] \mid f(a) = 0 \text{ for all } a \in X\}\]

Then it is obvious that for any subset \(X\), \(I(X)\) is an ideal of \(\mathbb{K}[\x_1,\ldots, \x_n]\). Moreover, the following holds.

Proposition 9 For subsets \(X,Y\) of \(\mathbb{A}^n\) and an arbitrary subset \(S\) of \(\mathbb{K}[\x_1,\ldots, \x_n]\), the following hold.

1. If \(X \subseteq Y\) then \(I(Y) \subseteq I(X)\).
2. \(I(\emptyset) = \mathbb{K}[\x_1, \ldots, \x_n]\).
3. If \(\mathbb{K}\) is infinite then \(I(\mathbb{A}^n) = (0)\).
4. \(X \subseteq Z(I(X))\) always holds.
5. \(S \subseteq I(Z(S))\) always holds.

Proof

1. If \(X \subseteq Y\) and \(f \in I(Y)\), then \(f(a) = 0\) for all \(a \in Y\). In particular \(f(a) = 0\) for all \(a \in X\), so \(f \in I(X)\).
2. Obvious.
3. If \(\mathbb{K}\) is infinite, the only polynomial vanishing at every point is the zero polynomial.
4. If \(a \in X\) and \(f \in I(X)\), then \(f(a) = 0\). That is, \(a \in Z(I(X))\).
5. If \(f \in S\) and \(a \in Z(S)\), then \(f(a) = 0\). That is, \(f \in I(Z(S))\).

That is, \(Z\) and \(I\) define an antitone Galois connection. ([Set Theory] §Filters, Ideals, and Galois Connections, ⁋Definition 6) Therefore, each of the two compositions \(ZI\) and \(IZ\) defines a closure operator. In the case of \(ZI\), this closure becomes the closure in the Zariski topology. This is because if \(X \subseteq Y = Z(J)\), then \(I(Z(J)) \subseteq I(X)\), and since \(J \subseteq I(Z(J))\) by condition 5 of Proposition 9, we get \(ZI(X) \subseteq Z(J) = Y\), so that \(ZI(X)\) is the smallest Zariski closed set containing \(X\). In the case of \(IZ\), this is not immediately visible; for this, we need the notion of the radical of an ideal. ([Commutative Algebra] §Properties of Localization, ⁋Corollary 8)

Theorem 10 (Nullstellensatz) Let \(\mathbb{K}\) be an algebraically closed field and let \(\mathfrak{a}\subseteq \mathbb{K}[\x_1,\ldots, \x_n]\) be an ideal. Then

\[I(Z(\mathfrak{a}))=\sqrt{\mathfrak{a}}\]

holds.

Proof

[Commutative Algebra] §Nullstellensatz, ⁋Proposition 6

Broadly speaking, this could be said to be a somewhat anticipated result, since from condition 5 of Proposition 4 we already have

\[Z(\mathfrak{a}^k)=Z(\mathfrak{a}\cap\cdots\cap \mathfrak{a})=Z(\mathfrak{a})\]

On the other hand, since \(\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}\) holds for any ideal \(\mathfrak{a}\), we know from the third condition of Proposition 4 that \(Z(\sqrt{\mathfrak{a}})\subseteq Z(\mathfrak{a})\). But by definition, for any \(f\in \sqrt{\mathfrak{a}}\), there exists a suitable \(r\) such that \(f^r\in \mathfrak{a}\). Therefore, if \(x\in Z(\mathfrak{a})\), then we must have \(x\in Z(\sqrt{\mathfrak{a}})\), and from this we know that \(Z(\mathfrak{a})=Z(\sqrt{\mathfrak{a}})\). That is, the radical of an ideal can be thought of as giving the standard method of obtaining the ideal when representing an affine algebraic set as the zero set of an ideal, and to distinguish the difference between them, one can define schemes.

Henceforth, to use Theorem 10 (Nullstellensatz) freely, we will assume unless otherwise mentioned that \(\mathbb{K}\) is an algebraically closed field.

Now, combining the fifth result of Proposition 4 with the above result, we can see that for \(Z(\mathfrak{a})\) to be an affine variety, \(\sqrt{\mathfrak{a}}\) must be a prime ideal. ([Commutative Algebra] §Basic Notions, ⁋Definition 10) That is, there is a Galois correspondence between the irreducible closed algebraic sets of \(\mathbb{A}^n\) and the prime ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\).

Coordinate Rings and Regularity

We can further extend the philosophy contained in the two-way correspondence \(Z\), \(I\). Specifically, the geometry of \(\mathbb{A}^n\) contains, by its definition, the elements of \(\mathbb{K}[\x_1,\ldots, \x_n]\). Conversely, taking an arbitrary point \(x=(x_1,\ldots, x_n)\) of \(\mathbb{A}^n\), we can think of the function that outputs the \(i\)-th coordinate as \(\x_i: x\mapsto x_i\), and from this perspective, all elements of \(\mathbb{K}[\x_1,\ldots, \x_n]\) can be viewed as (polynomial) functions defined on \(\mathbb{A}^n\).

More generally, we define the following.

Definition 11 The coordinate ring \(\mathbb{K}[X]\) of an affine variety \(X = Z(\mathfrak{a}) \subseteq \mathbb{A}^n\) is defined as

\[\mathbb{K}[X] = \mathbb{K}[\x_1, \ldots, \x_n] / I(X)=\mathbb{K}[\x_1, \ldots, \x_n] / \sqrt{\mathfrak{a}}\]

We call its elements regular functions defined on \(X\).

The reason for defining the coordinate ring is to implement the central philosophy of algebraic geometry, namely the correspondence between the geometric object \(X\) and the algebraic object \(\mathbb{K}[X]\). All geometric information of \(X\), such as points, relations among polynomial functions, inclusion relations of subsets, etc., is encoded in the ring structure of \(\mathbb{K}[X]\), and this becomes the foundation for later translating morphisms between affine varieties into coordinate ring homomorphisms.

If \(X\) is defined as the zero set of some polynomials, then \(\mathbb{K}[X]\) is the collection of the remaining polynomial functions after factoring out these polynomial relations. As in the case \(X=\mathbb{A}^n\) examined above, the elements of \(\mathbb{K}[X]\) can be thought of as functions defined on \(X\). Specifically, for an element \(\overline{f}\in \mathbb{K}[X]\), the following function

\[X\rightarrow \mathbb{K};\qquad x\mapsto f(x)\]

is well-defined. Here, by saying this function is well-defined, we mean that the value \(f(x)\) is unique regardless of the choice of representative of \(\overline{f}\), and this is possible precisely because \(X\) is defined as the zero set of \(I(X)\).

Now we are ready to extend the one-to-one correspondence between prime ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) and closed subvarieties of \(\mathbb{A}^n\), discussed earlier, to arbitrary affine varieties.

Proposition 12 Let \(X \subseteq \mathbb{A}^n\) be an affine variety over an algebraically closed field \(\mathbb{K}\). Then there is a one-to-one correspondence between the prime ideals of the coordinate ring \(\mathbb{K}[X]\) and the closed subvarieties of \(X\) as follows.

1. For a prime ideal \(\mathfrak{p} \subseteq \mathbb{K}[X]\), letting \(\tilde{\mathfrak{p}}\) be the preimage of \(\mathfrak{p}\) in \(\mathbb{K}[\x_1, \ldots, \x_n]\), the set \(Z(\tilde{\mathfrak{p}}) \subseteq X\) is a closed subvariety of \(X\).
2. For a closed subvariety \(Y \subseteq X\), the ideal \(I(Y)/I(X) \subseteq \mathbb{K}[X]\) is a prime ideal.

The correspondences \(\mathfrak{p} \mapsto Z(\tilde{\mathfrak{p}})\) and \(Y \mapsto I(Y)/I(X)\) are inverse to each other.

The proof of this is essentially obvious from the fourth isomorphism theorem.

Example 13 In the case of the affine varieties examined in Example 3, it can be shown that the ideals \((\x^2+\y^2-1)\) and \((\y-\x^2,\z-\x^3)\) are radical. Therefore, the coordinate ring of the unit circle \(X = Z(\x^2+\y^2-1)\) is \(\mathbb{K}[X] = \mathbb{K}[\x, \y]/(\x^2+\y^2-1)\), and the coordinate ring of the twisted cubic \(C\) is \(\mathbb{K}[C] = \mathbb{K}[\x, \y, \z]/(\y-\x^2, \z-\x^3) \cong \mathbb{K}[\x]\).

However, in general, the coordinate ring of a hypersurface \(Z(f)\) is \(\mathbb{K}[Z(f)] = \mathbb{K}[\x_1, \ldots, \x_n]/I(Z(f)) = \mathbb{K}[\x_1, \ldots, \x_n]/\sqrt{(f)}\), so when computing the coordinate ring, one must determine whether the given ideal is radical.

Meanwhile, we pointed out earlier that the definition of an affine variety actually depends on the (closed) embedding \(X\subseteq \mathbb{A}^n\), and a problem arising from this occurs here as well. Looking at the example of the principal open set \(X=D(\x)\) of \(\mathbb{A}^1\) used in that section, the correct coordinate ring of this affine variety must be computed not as a subset of \(\mathbb{A}^1\), but as a subset \(Z(\x\y-1)\) of \(\mathbb{A}^2\), and then

\[\mathbb{K}[X]=\mathbb{K}[\x,\y]/(\x\y-1)\cong \mathbb{K}[\x,1/\x]\]

Keeping this in mind, the following definition can also be understood.

Definition 14 For an arbitrary affine variety \(V\subseteq \mathbb{A}^k\) and a function \(f:V\rightarrow \mathbb{K}\) defined on it, we say that \(f\) is regular at a point \(p\in V\) if there exist a suitable open neighborhood \(D(h)\) of \(p\) and a polynomial \(g\) such that \(f=g/h\) holds on \(U\), where \(h\) is a polynomial that does not vanish on \(U=D(h)\).

Then under this definition, it would be natural to call a function that is regular at every point a regular function. When \(\mathbb{K}\) is algebraically closed, the proof that these two definitions Definition 11 and Definition 14 are equivalent can be somewhat tedious, but since the essential content is contained in the example examined above, we will omit the proof. The key to the proof is to obtain Definition 11 from Definition 14, which is achieved by gluing together the functions that appear in the form \(g/h\) on each \(D(h)\).

Morphisms Between Affine Varieties

Now we define morphisms between affine varieties. Since affine varieties are geometric objects defined by polynomials, it is reasonable that the functions between them should also be represented by polynomials.

Definition 15 A function \(\varphi:X \rightarrow Y\) between two affine varieties \(X \subseteq \mathbb{A}^n\) and \(Y \subseteq \mathbb{A}^m\) is called a morphism (or regular map) between them if there exist suitable polynomials \(f_1, \ldots, f_m \in \mathbb{K}[\x_1, \ldots, \x_n]\) such that

\[\varphi(a_1, \ldots, a_n) = (f_1(a), \ldots, f_m(a))\]

holds.

For example, we showed in Example 3 that the twisted cubic corresponds to \(\mathbb{A}^1\) via \(t\mapsto (t,t^2,t^3)\), and the above definition shows that this is a morphism between affine varieties.

Intuitively, since the \(\mathbb{K}[X]\) are functions defined on \(X\), if a morphism \(X\rightarrow Y\) is given, we can pull back the regular functions of \(Y\) to \(X\) via composition with this morphism. This is one direction from geometric morphism to algebraic morphism. More important is the reverse direction: that is, given a coordinate ring homomorphism \(\mathbb{K}[Y]\rightarrow \mathbb{K}[X]\), we can recover a geometric morphism \(X\rightarrow Y\) from it.

Proposition 16 A morphism \(\varphi: X \rightarrow Y\) induces a coordinate ring homomorphism \(\varphi^\ast: \mathbb{K}[Y] \rightarrow \mathbb{K}[X]\). Specifically, for \(\bar{g} \in \mathbb{K}[Y]\),

\[\varphi^\ast(\bar{g}) = \overline{g \circ \varphi}\]

holds.

Proof

First we must show that \(\varphi^\ast\) is well-defined. If \(g, h \in \mathbb{K}[\y_1, \ldots, \y_m]\) define the same function on \(Y\), then \(g - h \in I(Y)\). Since \(\varphi(X) \subseteq Y\), for all \(a \in X\) we have

\[(g \circ \varphi)(a) - (h \circ \varphi)(a) = (g - h)(\varphi(a)) = 0\]

That is, \(g \circ \varphi - h \circ \varphi \in I(X)\), and therefore \(\overline{g \circ \varphi} = \overline{h \circ \varphi}\).

That \(\varphi^\ast\) is a ring homomorphism is now obvious.

That is, a morphism \(\varphi: X \rightarrow Y\) induces a coordinate ring homomorphism \(\varphi^\ast: \mathbb{K}[Y] \rightarrow \mathbb{K}[X]\). This means that \(X\mapsto \mathbb{K}[X]\) is a contravariant functor from the category of affine varieties to \(\Ring\). ([Category Theory] §Functor, ⁋Definition 5)

Once the notion of a morphism is defined, the notion of an isomorphism naturally exists.

Definition 17 A morphism \(\varphi: X \rightarrow Y\) is called an isomorphism if there exists an inverse function \(\psi: Y \rightarrow X\) such that \(\psi\) is also a morphism.

For example, the morphism \(t\mapsto (t, t^2, t^3)\) from \(\mathbb{A}^1\) to the twisted cubic \(C\) is an isomorphism. This is because \((x,y,z)\mapsto x\) defines the inverse.

As we saw above, since \(X\mapsto \mathbb{K}[X]\) defines a contravariant functor from the category of affine varieties to \(\Ring\), it is obvious that isomorphic affine varieties have isomorphic coordinate rings. The following proposition shows that the converse also holds.

Proposition 18 A morphism \(\varphi: X \rightarrow Y\) is an isomorphism if and only if \(\varphi^\ast: \mathbb{K}[Y] \rightarrow \mathbb{K}[X]\) is a ring isomorphism.

In Proposition 16 we saw that a morphism \(\varphi: X \rightarrow Y\) induces a coordinate ring homomorphism \(\varphi^\ast: \mathbb{K}[Y] \rightarrow \mathbb{K}[X]\). Intuitively, \(\varphi^\ast\) corresponds a function \(g\) on \(Y\) to the function \(g \circ \varphi\) on \(X\), which is the operation of pulling back the geometric information of \(Y\) to \(X\). Therefore, if \(\varphi^\ast\) is an isomorphism, the functions of both coordinate rings correspond perfectly to each other, so it is natural to expect that \(X\) and \(Y\) are essentially the same geometrically. The following proof rigorously implements this intuition.

Proof of Proposition 18

It suffices to show the reverse direction. Suppose \(\varphi^\ast\) is an isomorphism. Then \(\psi^\ast = (\varphi^\ast)^{-1}: \mathbb{K}[X] \rightarrow \mathbb{K}[Y]\) exists.

Now let us define a morphism \(\theta: Y \rightarrow X\) from \(\psi^\ast\).

For each element \(\bar{\x}_i\) of

\[\mathbb{K}[X] = \mathbb{K}[\x_1, \ldots, \x_n]/I(X)\]

we can consider \(\psi^\ast(\bar{\x}_i) \in \mathbb{K}[Y]\). Write this as \(\bar{g}_i\), and consider some representatives \(g_i\) of these. Then we can define \(\theta: Y \rightarrow \mathbb{A}^n\) by \(\theta(y) = (g_1(y), \ldots, g_n(y))\), and by the definition of \(\mathbb{K}[Y]\) this does not depend on the choice of representatives \(g_i\). Now since \(\psi^\ast\) is well-defined, we have \(\theta(Y) \subseteq X\), and therefore \(\theta: Y \rightarrow X\) is a morphism. The remaining part is a simple computation.

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References

[Har] J. Harris, Algebraic Geometry: A First Course, Springer, 1992.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Zarieties in Projective Space, Springer, 2013.
[Ful] W. Fulton, Algebraic Curves, 2008. (Available online)