Affine Varieties

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 the sets given by the equation

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

This is the collection of solutions in \(\mathbb{K}^n\) to the polynomial \(f\). We usually set \(\mathbb{K}=\mathbb{C}\), and what is particularly nice about this assumption is the fact that \(\mathbb{C}\) is algebraically closed. However, since in most cases this assumption is not of much help, we will use a more general setting. Also, to avoid confusion, the variables of the polynomial \(f\) will be denoted by upright letters such as \(\x\).

Definition of Affine Varieties

Definition 1 Defined over a field \(\mathbb{K}\), the affine \(n\)-space \(\mathbb{A}^n_\mathbb{K}\) means the \(n\)-dimensional vector space \(\mathbb{K}^n\).

When \(\mathbb{K}\) can be omitted from context, we write it as \(\mathbb{A}^n\). 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 that, for a polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\), appear as the zero set \(Z(f)\).

Definition 2 For 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 collection of common zeros of several polynomials \(f_1,\ldots, f_k\). More generally, for a subset of \(\mathbb{K}[\x_1,\ldots, \x_n]\), say \(S\), we can define \(Z(S)\) similarly to the definition above, and then by definition, the ideal generated by \(S\) in \(\mathbb{K}[\x_1,\ldots, \x_n]\), \((S)\), satisfies the equation

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

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 \(Z(\mathfrak{a})=Z(f_1,\ldots, f_k)\) holds. That is, the affine algebraic sets in the sense of Definition 2 and the zero sets of ideals form exactly the same collection, and therefore we only need to care about the affine algebraic sets defined by ideals \(\mathfrak{a}\).

In general, for a space \(X\) to be irreducible means that \(X\) cannot be expressed 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 is in order to deal only with a single object that does not geometrically 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. In \(\mathbb{A}^2\), consider the affine variety \(Z(\x^2+\y^2-1)\). By definition, this set is the collection of points satisfying the equation \(\x^2+\y^2-1=0\) in \(\mathbb{A}^2\), 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 in \(\mathbb{A}^3\) by the two polynomials \(\y-\x^2\), \(\z-\x^3\), which, via the parametrization \((t,t^2,t^3)\), is in one-to-one correspondence with \(\mathbb{A}^1\).
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 is crucially used when defining the Zariski topology in Proposition 4.

In the example above, we saw that, as sets, 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. Since our only tool is polynomials, we will use them to define a topological structure.

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 for all \(f \in \mathfrak{b}\), we have \(f(x) = 0\), and since \(\mathfrak{a} \subseteq \mathfrak{b}\), in particular the desired identity holds for all elements of \(\mathfrak{a}\) as well.

For the fourth result, that a point \(x\) belongs to all \(Z(\mathfrak{a}_i)\) means that for all \(i\) and all \(f \in \mathfrak{a}_i\), we have \(f(x) = 0\). This is equivalent to saying that every element of \(\sum_i \mathfrak{a}_i\) is 0 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 \(\mathfrak{a}\cap \mathfrak{b}\) evaluated at \(x\) is \(0\), so \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\) holds.
Now suppose \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\). Evaluating an arbitrary element of \(\mathfrak{a}\mathfrak{b}\)

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

at \(x\), since \(\mathfrak{a}\mathfrak{b}\subseteq \mathfrak{a}\cap \mathfrak{b}\), each \(f_ig_i\) at \(x\) is \(0\), and therefore their sum is also \(0\).
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 some \(f\in \mathfrak{a}\) and some \(g\in \mathfrak{b}\) such that \(f(x),g(x)\neq 0\). But if this holds, then \(f(x)g(x)\neq 0\), which contradicts 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 serves as good intuition for examining 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 all the conditions of [Topology] §Interior, Closure, and Boundary, ⁋Proposition 2 are 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 via the subspace topology of the topology defined on \(\mathbb{A}^n\), we can define the topology on \(X\).

As a special example, looking at the Zariski topology on \(\mathbb{A}^1\), any element of \(\mathbb{K}\) is the zero set of the linear polynomial \(\x-x\), so any singleton is a closed set, and therefore any finite set is a closed set. However, any element of \(\mathbb{K}[\x]\) other than \(0\) has at most finitely many solutions, so in this topology (as long as \(\mathbb{K}\) is not a finite set), 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 can 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 with at least two points 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 \(D(f)\) by

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

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

Proposition 6 For an affine variety \(X \subseteq \mathbb{A}^n\) and any open set \(U\) in it,

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

holds for some family of principal open sets \(D(f_i)\).

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,

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

Meanwhile, \(\mathbb{A}^n \setminus Z(\mathfrak{a})\) is the set of points such that for \(f \in \mathfrak{a}\), \(f(x) \ne 0\), so

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

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

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

Proposition 7 For an affine variety \(X \subseteq \mathbb{A}^n\), if \(D(f)\cap X\neq\emptyset\) for a polynomial \(f \in \mathbb{K}[\x_1, \ldots, \x_n]\), then \(D(f) \cap X\) 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 representing \(\mathbb{A}^{n+1}\) 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

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

Now, 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 fact that \(X\) is irreducible and the second condition of [Topology] §Dimension, ⁋Proposition 7, the two 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, strictly speaking, our definition of an affine variety depends on the ambient space \(\mathbb{A}^n\). For example, in \(\mathbb{A}^1\), the principal open set \(D(\x)\) 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 a result obtained through the isomorphism

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

This issue may cause some confusion when understanding 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 of \(\mathbb{K}[\x_1,\ldots, \x_n]\), \(I(X)\), 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 of \(\mathbb{A}^n\), \(X,Y\), and an arbitrary subset of \(\mathbb{K}[\x_1,\ldots, \x_n]\), \(S\), 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 for all \(a \in Y\), we have \(f(a) = 0\). In particular, for all \(a \in X\), we have \(f(a) = 0\), 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 is actually 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 obvious; 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] §The Nullstellensatz, ⁋Proposition 6

Broadly speaking, since from condition 5 of Proposition 4 we already have

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

this could be said to be a somewhat anticipated result.

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, given 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 a standard method of obtaining this ideal when representing an affine algebraic set as the zero set of an ideal, and to distinguish 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, by definition, the elements of \(\mathbb{K}[\x_1,\ldots, \x_n]\) contain the geometry of \(\mathbb{A}^n\). 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 these 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 a well-defined function. Here, that this function is well-defined means 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 \(\mathfrak{p} \subseteq \mathbb{K}[X]\), if \(\tilde{\mathfrak{p}}\) is the preimage of \(\mathfrak{p}\) in \(\mathbb{K}[\x_1, \ldots, \x_n]\), then \(Z(\tilde{\mathfrak{p}}) \subseteq X\) is a closed subvariety of \(X\).
2. For a closed subvariety \(Y \subseteq X\), \(I(Y)/I(X) \subseteq \mathbb{K}[X]\) 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 it becomes

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

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 \(U=V\cap D(h)\) of \(p\) and a polynomial \(g\) such that on \(U\), \(f=g/h\) holds. Here, \(h\) is a polynomial that on \(U\) does not become \(0\).

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 suitably gluing together, on each \(D(h)\), the functions that appear in the form \(g/h\).

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 For two affine varieties \(X \subseteq \mathbb{A}^n\) and \(Y \subseteq \mathbb{A}^m\), a function \(\varphi:X \rightarrow Y\) between them is called a morphism (or regular map) 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 via \(t\mapsto (t,t^2,t^3)\), the twisted cubic corresponds to \(\mathbb{A}^1\), and the above definition shows that this is a morphism between affine varieties.

Intuitively, since elements of \(\mathbb{K}[X]\) 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 morphisms to algebraic morphisms. More important is the reverse direction: that is, the fact that 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 \(\overline{g \circ \varphi} = \overline{h \circ \varphi}\).

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

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)

Meanwhile, once the notion of a morphism has been defined, naturally the notion of an isomorphism exists.

Definition 17 A morphism \(\varphi: X \rightarrow Y\) is 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, \(\varphi^\ast\) maps 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 geometrically, too, the structures of \(X\) and \(Y\) are essentially the same. 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, \(\theta(Y) \subseteq X\), 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: Varieties in Projective Space, Springer, 2013.
[Ful] W. Fulton, Algebraic Curves, 2008. (Available online)