The Algebra–Geometry Dictionary

Does this problem of having different ideals represent the empty variety go away if the field k is algebraically closed? It does in the one-variable case when the ring is k[x]. To see this, recall from §5 of Chapter  1 that any ideal I in k[x] can be generated by a single polynomial because k[x] is a principal ideal domain. So we can write \(I =\langle \, f\rangle\) for some polynomial f ∈ k[x]. Then V(I) is the set of roots of f; i.e., the set of a ∈ k such that f(a) = 0. But since k is algebraically closed, every nonconstant polynomial in k[x] has a root. Hence, the only way that we could have \(\mathbf{V}(I) = \varnothing\) would be to have f be a nonzero constant. In this case, 1∕f ∈ k. Thus, \(1 = (1/f) \cdot f \in I\), which means that \(g = g \cdot 1 \in I\) for all g ∈ k[x]. This shows that I = k[x] is the only ideal of k[x] that represents the empty variety when k is algebraically closed.

A wonderful thing now happens: the same property holds when there is more than one variable. In any polynomial ring, algebraic closure is enough to guarantee that the only ideal which represents the empty variety is the entire polynomial ring itself. This is the Weak Nullstellensatz, which is the basis of (and is equivalent to) one of the most celebrated mathematical results of the late nineteenth century, Hilbert’s Nullstellensatz. Such is its impact that, even today, one customarily uses the original German name Nullstellensatz: a word formed, in typical German fashion, from three simpler words: Null ( = Zero), Stellen ( = Places), Satz ( = Theorem).

Once we prove the claim, an easy induction gives elements \(a_{1},\ldots,a_{n} \in k\) such that \(I_{x_{n}=a_{n},\ldots,x_{1}=a_{1}} \subsetneq k\). But the only ideals of k are {0} and k (Exercise 3), so that \(I_{x_{n}=a_{n},\ldots,x_{1}=a_{1}} =\{ 0\}\). This implies \((a_{1},\ldots,a_{n}) \in \mathbf{V}(I)\). We conclude that \(\mathbf{V}(I)\neq \varnothing\), and the theorem will follow.

To prove the claim, there are two cases, depending on the size of \(I \cap k[x_{n}]\).

Case 1. \(I \cap k[x_{n}]\neq \{0\}\). Let \(f \in I \cap k[x_{n}]\) be nonzero, and note that f is nonconstant, since otherwise \(1 \in I \cap k[x_{n}] \subseteq I\), contradicting \(I\neq k[x_{1},\ldots,x_{n}]\).

We claim that the \(\bar{g}_{i}\) form a Gröbner basis of \(I_{x_{n}=a}\). Assuming the claim, it follows that \(1\notin I_{x_{n}=a}\) since no \(\mbox{ LT}(\bar{g}_{i})\) can divide 1. Thus \(I_{x_{n}=a}\neq k[x_{1},\ldots,x_{n-1}]\), which is what we want to show.

In the special case when \(k = \mathbb{C}\), the Weak Nullstellensatz may be thought of as the “Fundamental Theorem of Algebra for multivariable polynomials”—every system of polynomials that generates an ideal strictly smaller than \(\mathbb{C}[x_{1},\ldots,x_{n}]\) has a common zero in \(\mathbb{C}^{n}\).

To see this, let \(\{g_{1},\ldots,g_{t}\}\) be a Gröbner basis of \(I =\langle 1\rangle\). Thus, \(1 \in \langle \mbox{ LT}(I)\rangle =\langle \mbox{ LT}(g_{1}),\ldots,\mbox{ LT}(g_{t})\rangle\), and then Lemma  2 of Chapter  2, §4 implies that 1 is divisible by some LT(g _i ), say LT(g ₁). This forces LT(g ₁) to be constant. Then every other LT(g _i ) is a multiple of that constant, so that \(g_{2},\ldots,g_{t}\) can be removed from the Gröbner basis by Lemma  3 of Chapter  2, §7. Finally, since LT(g ₁) is constant, g ₁ itself is constant since every nonconstant monomial is > 1 (see Corollary  6 of Chapter  2, §4). We can multiply by an appropriate constant to make g ₁ = 1. Our reduced Gröbner basis is thus {1}.

To summarize, we have the following consistency algorithm: if we have polynomials \(f_{1},\ldots,f_{s} \in \mathbb{C}[x_{1},\ldots,x_{n}]\), we compute a reduced Gröbner basis of the ideal they generate with respect to any ordering. If this basis is {1}, the polynomials have no common zero in \(\mathbb{C}^{n}\); if the basis is not {1}, they must have a common zero. Note that this algorithm works over any algebraically closed field.

If we are working over a field k which is not algebraically closed, then the consistency algorithm still works in one direction: if {1} is a reduced Gröbner basis of \(\langle \,f_{1},\ldots,f_{s}\rangle\), then the equations \(f_{1} = \cdots = f_{s} = 0\) have no common solution. The converse is not true, as shown by the examples preceding the statement of the Weak Nullstellensatz.

Inspired by the Weak Nullstellensatz, one might hope that the correspondence between ideals and varieties is one-to-one provided only that one restricts to algebraically closed fields. Unfortunately, our earlier example V(x) = V(x ²) = { 0} works over any field. Similarly, the ideals \(\langle x^{2},y\rangle\) and \(\langle x,y\rangle\) (and, for that matter, \(\langle x^{n},y^{m}\rangle\) where n and m are integers greater than one) are different but define the same variety: namely, the single point \(\{(0,0)\} \subseteq k^{2}\). These examples illustrate a basic reason why different ideals can define the same variety (equivalently, that the map V can fail to be one-to-one): namely, a power of a polynomial vanishes on the same set as the original polynomial. The Hilbert Nullstellensatz states that over an algebraically closed field, this is the only reason that different ideals can give the same variety: if a polynomial f vanishes at all points of some variety V(I), then some power of f must belong to I.

To further explore the relation between ideals and varieties, it is natural to recast Hilbert’s Nullstellensatz in terms of ideals. Can we characterize the kinds of ideals that appear as the ideal of a variety? In other words, can we identify those ideals that consist of all polynomials which vanish on some variety V? The key observation is contained in the following simple lemma.

Thus, an ideal consisting of all polynomials which vanish on a variety V has the property that if some power of a polynomial belongs to the ideal, then the polynomial itself must belong to the ideal. This leads to the following definition.

Rephrasing Lemma 1 in terms of radical ideals gives the following statement.

On the other hand, Hilbert’s Nullstellensatz tells us that the only way that an arbitrary ideal I can fail to be the ideal of all polynomials vanishing on V(I) is for I to contain powers f ^m of polynomials f which are not in I—in other words, for I to fail to be a radical ideal. This suggests that there is a one-to-one correspondence between affine varieties and radical ideals. To clarify this and get a sharp statement, it is useful to introduce the operation of taking the radical of an ideal.

We are now ready to state the ideal-theoretic form of the Nullstellensatz.

It has become a custom, to which we shall adhere, to refer to Theorem 6 as the Nullstellensatz with no further qualification. The most important consequence of the Nullstellensatz is that it allows us to set up a “dictionary” between geometry and algebra. The basis of the dictionary is contained in the following theorem.

As a consequence of this theorem, any question about varieties can be rephrased as an algebraic question about radical ideals (and conversely), provided that we are working over an algebraically closed field. This ability to pass between algebra and geometry will give us considerable power.

The existence of these algorithms follows from the work of HERMANN (1926) [see also MINES, RICHMAN, and RUITENBERG (1988) and SEIDENBERG (1974, 1984) for more modern expositions]. More practical algorithms for finding radicals follow from the work of GIANNI, TRAGER and ZACHARIAS (1988), KRICK and LOGAR (1991), and EISENBUD, HUNEKE and VASCONCELOS (1992). These algorithms have been implemented in CoCoA, Singular, and Macaulay2, among others. See, for example, Section 4.5 of GREUEL and PFISTER (2008).

For now, we will settle for solving the more modest radical membership problem. To test whether \(f \in \sqrt{I}\), we could use the ideal membership algorithm to check whether f ^m  ∈ I for all integers m > 0. This is not satisfactory because we might have to go to very large powers of m, and it will never tell us if \(f\notin \sqrt{I}\) (at least, not until we work out a priori bounds on m). Fortunately, we can adapt the proof of Hilbert’s Nullstellensatz to give an algorithm for determining whether \(f \in \sqrt{\langle \,f_{1 },\ldots, f_{s}\rangle }\).

Proposition 8, together with our earlier remarks on determining whether 1 belongs to an ideal (see the discussion of the consistency problem in §1), immediately leads to the following radical membership algorithm: to determine if \(f \in \sqrt{\langle \,f_{1 },\ldots, f_{s}\rangle } \subseteq k[x_{1},\ldots,x_{n}]\), we compute a reduced Gröbner basis of the ideal \(\langle \,f_{1},\ldots,f_{s},1 - yf\rangle \subseteq k[x_{1},\ldots,x_{n},y]\) with respect to some ordering. If the result is {1}, then \(f \in \sqrt{I}\). Otherwise, \(f\notin \sqrt{I}\).

If we have f expressed as a product of irreducible polynomials, then it is easy to write down the radical of the principal ideal generated by f.

In view of Proposition 9, we make the following definition:

Thus, f _red is the polynomial f with repeated factors “stripped away.” So, for example, if f = (x + y ²)³(x − y), then f _red = (x + y ²)(x − y). Note that f _red is only unique up to a constant factor in k.

The usefulness of Proposition 9 is mitigated by the requirement that f be factored into irreducible factors. We might ask if there is an algorithm to compute f _red from f without factoring f first. It turns out that such an algorithm exists.

To state the algorithm, we will need the notion of a greatest common divisor of two polynomials.

It is easy to show that \(\mathrm{gcd}(\,f,g)\) exists and is unique up to multiplication by a nonzero constant in k (see Exercise 9). Unfortunately, the one-variable algorithm for finding the gcd (i.e., the Euclidean Algorithm) does not work in the case of several variables. To see this, consider the polynomials xy and xz in k[x, y, z]. Clearly, \(\mathrm{gcd}(xy,xz) = x\). However, no matter what term ordering we use, dividing xy by xz gives 0 plus remainder xy and dividing xz by xy gives 0 plus remainder xz. As a result, neither polynomial “reduces” with respect to the other and there is no next step to which to apply the analogue of the Euclidean Algorithm.

Nevertheless, there is an algorithm for calculating the gcd of two polynomials in several variables. We defer a discussion of it until the next section after we have studied intersections of ideals. For the purposes of our discussion here, let us assume that we have such an algorithm. We also remark that given polynomials f ₁, …, f _s  ∈ k[x ₁, …, x _n ], one can define \(\mathrm{gcd}(\,f_{1},f_{2},\ldots,f_{s})\) exactly as in the one-variable case. There is also an algorithm for computing \(\mathrm{gcd}(\,f_{1},f_{2},\ldots,f_{s})\).

Using this notion of gcd, we can now give a formula for computing the radical of a principal ideal.

It is worth remarking that for fields which do not contain \(\mathbb{Q}\), the above formula for f _red may fail (see Exercise 13).

Ideals are algebraic objects and, as a result, there are natural algebraic operations we can define on them. In this section, we consider three such operations: sum, intersection, and product. These are binary operations: to each pair of ideals, they associate a new ideal. We shall be particularly interested in two general questions which arise in connection with each of these operations. The first asks how, given generators of a pair of ideals, one can compute generators of the new ideals which result on applying these operations. The second asks for the geometric significance of these algebraic operations. Thus, the first question fits the general computational theme of this book; the second, the general thrust of this chapter. We consider each of the operations in turn.

The following corollary is an immediate consequence of Proposition 2.

To see what happens geometrically, let \(I =\langle x^{2} + y\rangle\) and \(J =\langle z\rangle\) be ideals in \(\mathbb{R}[x,y,z]\). We have sketched V(I) and V(J) on the next page. Then \(I + J =\langle x^{2} + y,z\rangle\) contains both x ² + y and z. Thus, the variety V(I + J) must consist of those points where both x ² + y and z vanish, i.e., it must be the intersection of V(I) and V(J).

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The same line of reasoning generalizes to show that addition of ideals corresponds geometrically to taking intersections of varieties.

An analogue of Theorem 4 stated in terms of generators was given in Lemma  2 of Chapter  1, §2.

The following proposition guarantees that the product of ideals does indeed correspond geometrically to the operation of taking the union of varieties.

In what follows, we will often write the product of ideals as IJ rather than I ⋅ J.

As in the case of sums, the set of ideals is closed under intersections.

Note that we always have \(IJ \subseteq I \cap J\) since elements of IJ are sums of polynomials of the form fg with f ∈ I and g ∈ J. But the latter belongs to both I (since f ∈ I) and J (since g ∈ J). However, IJ can be strictly contained in \(I \cap J\). For example, if \(I = J =\langle x,y\rangle\), then \(IJ =\langle x^{2},xy,y^{2}\rangle\) is strictly contained in \(I \cap J = I =\langle x,y\rangle\) (\(x \in I \cap J\), but \(x\notin IJ)\).

This computation is easy precisely because we were given factorizations of f and g into irreducible polynomials. In general, such factorizations may not be available. So any algorithm which allows one to compute intersections will have to be powerful enough to circumvent this difficulty.

Nevertheless, there is a nice trick that reduces the computation of intersections to computing the intersection of an ideal with a subring (i.e., eliminating variables), a problem which we have already solved. To state the theorem, we need a little notation: if I is an ideal in k[x ₁, …, x _n ] and f(t) ∈ k[t] a polynomial in the single variable t, then f(t)I denotes the ideal in k[x ₁, …, x _n , t] generated by the set of polynomials \(\{f(t) \cdot h\mid h \in I\}\). This is a little different from our usual notion of product in that the ideal I and the ideal generated by f(t) in k[t] lie in different rings: in fact, the ideal \(I \subseteq k[x_{1},\,\ldots,\,x_{n}]\) is not an ideal in k[x ₁, …, x _n , t] because it is not closed under multiplication by t. When we want to stress that a polynomial h ∈ k[x ₁, …, x _n ] involves only the variables x ₁, …, x _n , we write h = h(x). Along the same lines, if we are considering a polynomial g in k[x ₁, …, x _n , t] and we want to emphasize that it can involve the variables x ₁, …, x _n as well as t, we will write g = g(x, t). In terms of this notation, \(f(t)I =\langle \, f(t)h(x)\mid h(x) \in I\rangle\). So, for example, if f(t) = t ² − t and \(I =\langle x,y\rangle\), then the ideal f(t)I in k[x, y, t] contains (t ² − t)x and (t ² − t)y. In fact, it is not difficult to see that f(t)I is generated as an ideal by (t ² − t)x and (t ² − t)y. This is a special case of the following assertion.

We can generalize both of the examples above by introducing the following definition.

This result, together with our algorithm for computing the intersection of two ideals immediately gives an algorithm for computing the least common multiple of two polynomials: to compute the least common multiple of two polynomials f, g ∈ k[x ₁, …, x _n ], we compute the intersection \(\langle \,f\rangle \cap \langle g\rangle\) using our algorithm for computing the intersection of ideals. Proposition 13 assures us that this intersection is a principal ideal (in the exercises, we ask you to prove that the intersection of principal ideals is principal) and that any generator of it is a least common multiple of f and g.

This algorithm for computing least common multiples allows us to clear up a point which we left unfinished in §2: namely, the computation of the greatest common divisor of two polynomials f and g. The crucial observation is the following.

We should point out that the gcd algorithm just described is rather cumbersome. In practice, more efficient algorithms are used [see DAVENPORT, SIRET and TOURNIER (1993)].

Having dealt with the computation of intersections, we now ask what operation on varieties corresponds to the operation of intersection on ideals. The following result answers this question.

Thus, the intersection of two ideals corresponds to the same variety as the product. In view of this and the fact that the intersection is much more difficult to compute than the product, one might legitimately question the wisdom of bothering with the intersection at all. The reason is that intersection behaves much better with respect to the operation of taking radicals: the product of radical ideals need not be a radical ideal (consider IJ where I = J), but the intersection of radical ideals is always a radical ideal. The latter fact is a consequence of the next proposition.

We have already encountered a number of examples of sets which are not varieties. Such sets arose very naturally in Chapter  3, where we saw that the projection of a variety need not be a variety, and in the exercises in Chapter  1, where we saw that the (set-theoretic) difference of varieties can fail to be a variety.

This proposition leads to the following definition.

We note the following properties of Zariski closure.

A natural example of Zariski closure is given by elimination ideals. We can now prove the first assertion of the Closure Theorem (Theorem  3 of Chapter  3, §2).

The conclusion of Theorem 4 can be stated as \(\mathbf{V}(I_{l}) = \overline{\pi _{l}(V )}\). In general, if V is a variety, then we say that a subset \(S \subseteq V\) is Zariski dense in V if \(V = \overline{S}\), i.e., V is the Zariski closure of S. is Thus Theorem 4 tells us that π _l (V) is Zariski dense in V(I _l ) when the field is algebraically closed.

One context in which we encountered sets that were not varieties was in taking the difference of varieties. For example, let V = V(I) where \(I \subseteq k[x,y,z]\) is the ideal \(\langle xz,yz\rangle\) and W = V(J) where \(J =\langle z\rangle\). Then we have already seen that V is the union of the (x, y)-plane and the z-axis. Since W is the (x, y)-plane, \(V \setminus W\) is the z-axis with the origin removed [because the origin also belongs to the (x, y)-plane]. We have seen in Chapter  1 that this is not a variety. The z-axis [i.e., V(x, y)] is the Zariski closure of \(V \setminus W\).

We could ask if there is a general way to compute the ideal corresponding to the Zariski closure \(\overline{V \setminus W}\) of the difference of two varieties V and W. The answer is affirmative, but it involves two new algebraic constructions on ideals called ideal quotients and saturations.

We begin with the first construction.

The algebraic properties of ideal quotients and methods for computing them will be discussed later in the section. For now, we want to explore the relation between ideal quotients and the Zariski closure of a difference of varieties.

In Proposition 7, note that V(I + J) from part (i) matches up with \(V \cap W\) in part (ii) since \(\mathbf{V}(I + J) = \mathbf{V}(I) \cap \mathbf{V}(J)\). So it is natural to ask if \(\mathbf{V}(I:J)\) in part (i) matches up with \(\overline{V \setminus W}\) in part (ii). This is equivalent to asking if the inclusion \(\overline{\mathbf{V}(I)\setminus \mathbf{V}(J)} \subseteq \mathbf{V}(I:J)\) in part (iii) is an equality.

However, if we replace J with J ², then a computation similar to the above gives \(I:J^{2} =\langle y - 1\rangle\), so that \(\mathbf{V}(I:J^{2}) = \overline{\mathbf{V}(I)\setminus \mathbf{V}(J)}\). In general, higher powers may be required, which leads to our second algebraic construction on ideals.

Later in the section we will discuss further algebraic properties of saturations and how to compute them. For now, we focus on their relation to geometry.

In some situations, saturations can be replaced with ideal quotients. For example, the proof of Theorem 10 yields the following corollary when the ideal I is radical.

The following proposition takes care of some simple properties of ideal quotients and saturations.

When the field is algebraically closed, the reader is urged to translate parts (i) and (iii) of the proposition into terms of varieties (upon which they become clear).

The following proposition will help us compute ideal quotients and saturations.

We now turn to the question of how to compute generators of the ideal quotient \(I:J\) and saturation \(I:J^{\infty }\), given generators of I and J. Inspired by (5), we begin with the case when J is generated by a single polynomial.

This theorem, together with our procedure for computing intersections of ideals and equation (5), immediately leads to an algorithm for computing a basis of an ideal quotient: given \(I =\langle \, f_{1},\ldots,f_{r}\rangle\) and \(J =\langle g_{1},\ldots,g_{s}\rangle\), to compute a basis of \(I:J\), we first compute a basis for \(I:g_{i}\) for each i. In view of Theorem 14, this means computing a basis \(\{h_{1},\ldots,h_{p}\}\) of \(\langle \,f_{1},\ldots,f_{r}\rangle \cap \langle g_{i}\rangle\). Recall that we do this via the algorithm for computing intersections of ideals from §3. Using the division algorithm, we divide each of basis element h _j by g _i to get a basis for \(I:g_{i}\) by part (i) of Theorem 14. Finally, we compute a basis for \(I:J\) by applying the intersection algorithm s − 1 times, computing first a basis for \(I:\langle g_{1},g_{2}\rangle = (I:g_{1}) \cap (I:g_{2})\), then a basis for \(I:\langle g_{1},g_{2},g_{3}\rangle = (I:\langle g_{1},g_{2}\rangle ) \cap (I:g_{3})\), and so on.

Similarly, we have an algorithm for computing a basis of a saturation: given \(I =\langle \, f_{1},\ldots,f_{r}\rangle\) and \(J =\langle g_{1},\ldots,g_{s}\rangle\), to compute a basis of \(I:J^{\infty }\), we first compute a basis for \(I:g_{i}^{\infty }\) for each i using the method described in part (ii) of Theorem 14. Then by (5), we need to intersect the ideals \(I:g_{i}^{\infty }\), which we do as above by applying the intersection algorithm s − 1 times.

We have already seen that the union of two varieties is a variety. For example, in Chapter  1 and in the last section, we considered V(xz, yz), which is the union of a line and a plane. Intuitively, it is natural to think of the line and the plane as “more fundamental” than V(xz, yz). Intuition also tells us that a line or a plane are “irreducible” or “indecomposable” in some sense: they do not obviously seem to be a union of finitely many simpler varieties. We formalize this notion as follows.

Thus, V(xz, yz) is not an irreducible variety. On the other hand, it is not completely clear when a variety is irreducible. If this definition is to correspond to our geometric intuition, it is clear that a point, a line, and a plane ought to be irreducible. For that matter, the twisted cubic V(y − x ², z − x ³) in \(\mathbb{R}^{3}\) appears to be irreducible. But how do we prove this? The key is to capture this notion algebraically: if we can characterize ideals which correspond to irreducible varieties, then perhaps we stand a chance of establishing whether a variety is irreducible.

The following notion turns out to be the right one.

If we have set things up right, an irreducible variety will correspond to a prime ideal and conversely. The following theorem assures us that this is indeed the case.

It is an easy exercise to show that every prime ideal is radical. Then, using the ideal-variety correspondence between radical ideals and varieties, we get the following corollary of Proposition 3.

In fact, the above argument holds much more generally.

With a little care, the above argument extends still further to show that any variety defined by a rational parametrization is irreducible.

The simplest variety in k ⁿ given by a parametrization consists of a single point, {(a ₁, …, a _n )}. In the notation of Proposition 5, it is given by the parametrization in which each f _i is the constant polynomial \(f_{i}(t_{1},\ldots,t_{m}) = a_{i},1 \leq i \leq n\). It is clearly irreducible and it is easy to check that \(\mathbf{I}(\{(a_{1},\ldots,a_{n})\}) =\langle x_{1} - a_{1},\ldots,x_{n} - a_{n}\rangle\) (see Exercise 7), which implies that the latter is prime. The ideal \(\langle x_{1} - a_{1},\ldots,x_{n} - a_{n}\rangle\) has another distinctive property: it is maximal in the sense that the only ideal which strictly contains it is the whole ring k[x ₁, …, x _n ]. Such ideals are important enough to merit special attention.

In order to streamline statements, we make the following definition.

Thus, an ideal is maximal if it is proper and no other proper ideal strictly contains it. We now show that any ideal of the form \(\langle x_{1} - a_{1},\ldots,x_{n} - a_{n}\rangle\) is maximal.

Note that Propositions 9 and 10 together imply that \(\langle x_{1} - a_{1},\ldots,x_{n} - a_{n}\rangle\) is prime in k[x ₁, …, x _n ] even if k is not infinite. Over an algebraically closed field, it turns out that every maximal ideal corresponds to some point of k ⁿ .

Note the proof of Theorem 11 uses the Weak Nullstellensatz. It is not difficult to see that it is, in fact, equivalent to the Weak Nullstellensatz.

We have the following easy corollary of Theorem 11.

Thus, we have extended our algebra–geometry dictionary. Over an algebraically closed field, every nonempty irreducible variety corresponds to a proper prime ideal, and conversely. Every point corresponds to a maximal ideal, and conversely.

We can use Zariski closure to characterize when a variety is irreducible.

Let us make a final comment about terminology. Some references, such as HARTSHORNE (1977), use the term “variety” for what we call an irreducible variety and say “algebraic set” instead of variety. When reading other books on algebraic geometry, be sure to check the definitions!

In the last section, we saw that irreducible varieties arise naturally in many contexts. It is natural to ask whether an arbitrary variety can be built up out of irreducibles. In this section, we explore this and related questions.

We begin by translating the ascending chain condition (ACC) for ideals (see §5 of Chapter  2) into the language of varieties.

We can use Proposition 1 to prove the following basic result about the structure of affine varieties.

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Both in the above example and the case of V(xz, yz), it appears that the decomposition of a variety into irreducible pieces is unique. It is natural to ask whether this is true in general. It is clear that, to avoid trivialities, we must rule out decompositions in which the same irreducible piece appears more than once, or in which one irreducible piece contains another. This is the aim of the following definition.

With this definition, we can now prove the following uniqueness result.

The uniqueness part of Theorem 4 guarantees that the irreducible components of V are well-defined. We remark that the uniqueness is false if one does not insist that the union be finite. (A plane P is the union of all the points on it. It is also the union of some line in P and all the points not on the line—there are infinitely many lines in P with which one could start.) This should alert the reader to the fact that although the proof of Theorem 4 is easy, it is far from vacuous: one makes subtle use of finiteness (which follows, in turn, from the Hilbert Basis Theorem).

Here is a result that relates irreducible components to Zariski closure.

Theorems 2 and 4 can also be expressed purely algebraically using the one-to-one correspondence between radical ideals and varieties.

We can also use ideal quotients from §4 to describe the prime ideals that appear in the minimal representation of a radical ideal.

We should mention that Theorems 6 and 7 hold for any field k, although the proofs in the general case are different (see Corollary 10 of §8).

For an example of these theorems, consider the ideal \(I =\langle xz - y^{2},x^{3} - yz\rangle\). Recall that the variety V = V(I) was discussed earlier in this section. For the time being, let us assume that I is radical (eventually we will see that this is true). Can we write I as an intersection of prime ideals?

The answer to all three questions is yes, and descriptions of the algorithms can be found in the works of HERMANN (1926), MINES, RICHMAN, and RUITENBERG (1988), and SEIDENBERG (1974, 1984). As in §2, the algorithms in these articles are very inefficient. However, the work of GIANNI, TRAGER and ZACHARIAS (1988) and EISENBUD, HUNEKE and VASCONCELOS (1992) has led to more efficient algorithms. See also Chapter  8 of BECKER and WEISPFENNING (1993) and §4.4 of ADAMS and LOUSTAUNAU (1994).

This section will complete the proof of the Closure Theorem from Chapter  3, §2. We will use many of the tools introduced in this chapter, including the Nullstellensatz, Zariski closures, saturations, and irreducible components.

We begin by recalling the basic situation. Let k be an algebraically closed field, and let π _l : k ⁿ  → k ^n−l is projection onto the last n − l components. If V = V(I) is an affine variety in k ⁿ , then we get the l-th elimination ideal \(I_{l} = I \cap k[x_{l+1},\ldots,x_{n}]\). The first part of the Closure Theorem, which asserts that V(I _l ) is the Zariski closure of π _l (V) in k ^n−l , was proved earlier in Theorem 4 of §4.

The remaining part of the Closure Theorem tells us that π _l (V) fills up “most” of V(I _l ). Here is the precise statement.

This is slightly different from the Closure Theorem stated in §2 of Chapter  3. There, we assumed \(V \neq \varnothing\) and asserted that \(\mathbf{V}(I_{l})\setminus W \subseteq \pi _{l}(V )\) for some \(W \subsetneq \mathbf{V}(I_{l})\). In Exercise 1 you will prove that Theorem 1 implies the version in Chapter  3.

The proof of Theorem 1 will use the following notation. Rename \(x_{l+1},\ldots,x_{n}\) as \(y_{l+1},\ldots,y_{n}\) and write \(k[x_{1},\ldots,x_{l},y_{l+1},\ldots,y_{n}]\) as \(k[\mathbf{x},\mathbf{y}]\) for \(\mathbf{x} = (x_{1},\ldots,x_{l})\) and \(\mathbf{y} = (y_{l+1},\ldots,y_{n})\). Also fix a monomial order > on k[x, y] with the property that x ^α  > x ^β implies x ^α  > x ^β y ^γ for all γ. The product order described in Exercise  9 of Chapter  2, §4 is an example of such a monomial order. Another example is given by lex order with x ₁ > ⋯ > x _l  > y _l+1 > ⋯ > y _n .

An important tool in proving Theorem 1 is the following result.

Part (ii) of Theorem 2 is related to the Extension Theorem from Chapter  3. Compared to the Extension theorem, part (ii) is simultaneously stronger (the Extension Theorem assumes l = 1, i.e., just one variable is eliminated) and weaker [part (ii) requires the nonvanishing of all relevant leading coefficients, while the Extension Theorem requires just one].

For our purposes, Theorem 2 has the following important corollary.

Hence, to complete the proof of the Closure Theorem, we need to explore what happens when the difference \(\mathbf{V}(I_{l})\setminus \mathbf{V}\big(\prod _{g_{i}\in G\setminus k[\mathbf{y}]}\,c_{i}\big)\) is not Zariski dense in V(I _l ). The following proposition shows that in this case, the original variety V = V(I) decomposes into varieties coming from strictly bigger ideals.