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

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

These are the sets of roots of a polynomial \(f\) in \(\mathbb{K}^n\). We usually take \(\mathbb{K}=\mathbb{C}\), and the especially convenient feature of this assumption is that \(\mathbb{C}\) is algebraically closed. However, in most cases this assumption is not particularly helpful, so we shall work in a more general setting. Also, to avoid confusion, we denote the variables of a polynomial \(f\) by upright letters such as \(\x\).

Definition of Affine Varieties

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

When \(\mathbb{K}\) is understood from the 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 shall study are those represented by the zero set \(Z(f)\) of a polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\).

Definition 2 For polynomials \(f_1, \ldots, f_k \in \mathbb{K}[\x_1, \ldots, \x_n]\), the affine algebraic set \(Z(f_1, \ldots, f_k)\) defined by them is

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

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

In other words, 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 \(Z(S)\) in the same manner, and then by definition the ideal \((S)\) generated by \(S\) in \(\mathbb{K}[\x_1,\ldots, \x_n]\) satisfies

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

Therefore, it suffices to consider only affine algebraic sets defined by ideals \(\mathfrak{a}\).

In general, a space \(X\) is called irreducible if it cannot be written as a union of two proper open subsets. ([Topology] §Dimension, ⁋Definition 6) Thus our definition says that an irreducible affine algebraic set is called an affine variety. This is so that we may deal, geometrically, with a single connected object.

Example 3 Most geometric objects we know can be expressed by polynomials, so they all furnish examples of affine varieties.

1. Consider the affine variety \(Z(\x^2+\y^2-1)\) in \(\mathbb{A}^2\). By definition, this set consists of the points in \(\mathbb{A}^2\) satisfying \(\x^2+\y^2-1=0\), so it represents the unit circle.
2. In general, for any affine space \(\mathbb{A}^n\) and any polynomial \(f\in \mathbb{K}[\x_1,\ldots, \x_n]\), the set \(Z(f)\) defines a hypersurface.
3. Another important example is the twisted cubic in \(\mathbb{A}^3\). This is the curve defined by the two polynomials \(\y-\x^2\) and \(\z-\x^3\), 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 fact will be used crucially in Proposition 4 when defining the Zariski topology.

In the examples above, we saw that familiar geometric objects can all be written, as sets, as affine algebraic sets. However, to regard them as geometric objects, we need a topological structure on them. Our only tool is polynomials, so we 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. \(\displaystyle\bigcap_{i\in I} Z(\mathfrak{a}_i) = Z\left(\sum_i \mathfrak{a}_i\right)\),
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 discussed 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 equation holds in particular for every element of \(\mathfrak{a}\).

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

Now let us prove the last claim. First, suppose \(x\in Z(\mathfrak{a})\cup Z(\mathfrak{b})\). 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 \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\).
Next, suppose \(x\in Z(\mathfrak{a}\cap \mathfrak{b})\). That evaluating any element

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

of \(\mathfrak{a}\mathfrak{b}\) at \(x\) yields \(0\) is obvious.
Finally, suppose \(x\in Z(\mathfrak{a}\mathfrak{b})\). If, contrary to the conclusion, \(x\not\in Z(\mathfrak{a})\cup Z(\mathfrak{b})\), then there exist \(f\in \mathfrak{a}\) and \(g\in \mathfrak{b}\) with \(f(x),g(x)\neq 0\). But 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 algebraic variety, we must have \(\mathfrak{a}=(1)\) or \(\mathfrak{b}=(1)\). This provides one useful piece of intuition for understanding algebraically what an algebraic variety is.

More importantly, the above proposition implies that if we declare the affine algebraic sets in \(\mathbb{A}^n\) to be the closed sets, then the conditions of §Interior, Closure, and Boundary, ⁋Proposition 2 are all satisfied, and hence a topology on \(\mathbb{A}^n\) is uniquely determined. This is called the Zariski topology. By definition, any affine variety \(X\) is a closed subset of some affine space \(\mathbb{A}^n\), and we define the topology on \(X\) as the subspace topology inherited from \(\mathbb{A}^n\).

As a concrete example, consider the Zariski topology on \(\mathbb{A}^1\). Any element of \(\mathbb{K}\) is the zero set of the linear polynomial \(\x-x\), so every singleton is closed, and therefore every finite set is closed. However, any nonzero element of \(\mathbb{K}[\x]\) has at most finitely many roots, so in this topology (provided \(\mathbb{K}\) is infinite) the only infinite closed set 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, and since affine varieties are 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 \(f \in \mathbb{K}[\x_1, \ldots, \x_n]\), the principal open set \(D(f)\) is defined by

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

The next proposition shows that principal open sets form a basis for an affine variety. (§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 \(D(f_i)\) such that

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

Proof

By the definition of the Zariski topology, there exists an ideal \(\mathfrak{a}\subset \mathbb{K}[\x_1,\ldots, \x_n]\) 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 subset of an affine variety need not be an affine variety, and indeed it typically is not. However, a principal open subset of an affine variety is always an affine variety.

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

Proof

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

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

and consider the ideal

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

in \(\mathbb{K}[\x_1,\ldots, \x_n,\y]\). 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 see that \(f(x)\neq 0\) and \(y=1/f(x)\). This yields the bijection

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

That this is a homeomorphism is obvious.

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, so \(D(\x)\) cannot be defined as the zero set of polynomials in \(\mathbb{K}[\x]\). In fact, examining the proof of Proposition 7, we see that the fact that \(D(\x)\) is an affine variety follows from the isomorphism

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

This issue can confuse us somewhat when we study regular functions, so we shall revisit it in the relevant section.

Nullstellensatz

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

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

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

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

Proposition 9 For subsets \(X,Y\) of \(\mathbb{A}^n\) and any subset \(I\) 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. \(I \subseteq I(Z(I))\) 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 \(f(a) = 0\). Hence \(a \in Z(I(X))\).
5. If \(f \in I\) and \(a \in Z(I)\) then \(f(a) = 0\). Hence \(f \in I(Z(I))\).

Thus, \(Z\) and \(I\) define an antitone Galois connection. (§Filters, Ideals, and Galois Connections, ⁋Definition 6) Therefore, the compositions \(ZI\) and \(IZ\) of the two operators are each closure operators. In the case of \(ZI\), this closure is precisely the closure in the Zariski topology. Indeed, if \(X \subseteq Y = Z(J)\), then \(I(Z(J)) \subseteq I(X)\), and by condition 5 of Proposition 9 we have \(J \subseteq I(Z(J))\), so \(ZI(X) \subseteq Z(J) = Y\), making \(ZI(X)\) the smallest Zariski closed set containing \(X\). In the case of \(IZ\), this is not immediately obvious; to see it, we need the notion of the radical of an ideal. (§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

§Nullstellensatz, ⁋Proposition 6

Viewed broadly, this result is to some extent anticipated by condition 5 of Proposition 4, since

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

Meanwhile, since \(\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}\) holds for any ideal \(\mathfrak{a}\), the third condition of Proposition 4 yields \(Z(\sqrt{\mathfrak{a}})\subseteq Z(\mathfrak{a})\). But by definition, for any \(f\in \sqrt{\mathfrak{a}}\) there exists \(r\) such that \(f^r\in \mathfrak{a}\). Therefore, if \(x\in Z(\mathfrak{a})\) then \(x\in Z(\sqrt{\mathfrak{a}})\) as well, and so \(Z(\mathfrak{a})=Z(\sqrt{\mathfrak{a}})\). That is, the radical of an ideal gives the standard way to recover the ideal when an affine algebraic set is represented as the zero set of an ideal, and to distinguish the differences between them one introduces the notion of a scheme.

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

Coordinate Rings and Regularity

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

More generally, we make the following definition.

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

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

Its elements are called regular functions defined on \(X\).

The reason for defining the coordinate ring is to realize the core philosophy of algebraic geometry: the correspondence between the geometric object \(X\) and the algebraic object \(\mathbb{K}[X]\). All geometric information of \(X\)—its points, the relations among polynomial functions, inclusion relations of subsets, and so on—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 polynomial functions with these polynomial relations factored out. Just as in the case \(X=\mathbb{A}^n\) examined above, 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 function

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

is well defined. Here, that this function is well defined means that the value \(f(x)\) is independent of the choice of representative of \(\overline{f}\), and this is possible precisely because \(X\) is defined as the zero set of \(I(X)\).

We are now ready to extend the one-to-one correspondence between prime ideals of \(\mathbb{K}[\x_1,\ldots, \x_n]\) and closed subvarieties of \(\mathbb{A}^n\) to an arbitrary affine variety.

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

1. For a prime ideal \(\mathfrak{p} \subset \mathbb{K}[X]\), 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 set \(I(Y)/I(X) \subset \mathbb{K}[X]\) 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, one can show 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 in fact depends on the (closed) embedding \(X\subseteq \mathbb{A}^n\), and the problem caused by this arises here as well. Consider 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 the subset \(Z(\x\y-1)\) of \(\mathbb{A}^2\), yielding

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

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

Definition 14 For any 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 an open neighborhood \(D(h)\) of \(p\) and a polynomial \(g\) such that \(f=g/h\) on \(U\). Here \(h\) is a polynomial that does not vanish on \(U=D(h)\).

Then under this definition, it is natural to call a function regular at every point a regular function. The proof that these two definitions, Definition 11 and Definition 14, are equivalent may be somewhat tedious, but since the essential content is contained in the example examined above, we shall omit it. The heart of the proof is to obtain Definition 11 from Definition 14, which is done by gluing together functions of 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 functions between them should also be given 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 polynomials \(f_1, \ldots, f_m \in \mathbb{K}[\x_1, \ldots, \x_n]\) such that

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

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

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

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

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

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

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

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

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

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

Definition 17 A morphism \(\varphi: X \to Y\) is called an isomorphism if there exists an inverse map \(\psi: Y \to X\) 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 its inverse.

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

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

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

Proof of Proposition 18

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

Now let us define a morphism \(\theta: Y \to X\) 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 choose any representatives \(g_i\). Then we can define \(\theta: Y \to \mathbb{A}^n\) 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 \(g_i\). Since \(\psi^\ast\) is well defined, \(\theta(Y) \subseteq X\), and therefore \(\theta: Y \to X\) is a morphism. The remainder is a straightforward calculation.

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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)