Jan 2, 2011 - functions. For example, the curve in the last example is rational. 1.2 Algebraic Sets and Ideals. 1.2.1 Definition. If S is any set of p...

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Algebraic Sets 1.1

Affine Space

In elementary geometry, one considered figures with coordinates in some Cartesian power of the real numbers. As our starting point in algebraic geometry, we will consider figures with coordinates in the Cartesian power of some fixed field k. 1.1.1 Definition. Let k be a field, and let An (k) = {(a1 , . . . , an ) | a1 , . . . , an ∈ k}. When the field is clear, we will shorten An (k) to An . We will refer to An as affine n-space. In particular, A1 is called the affine line, and A2 is called the affine plane. From the algebraic point of view, the most natural functions to consider on An are those defined by evaluating a polynomial in k[x1 , . . . , xn ] at a point. Analogously, the simplest geometric figures in An are the zero sets of a single polynomial. 1.1.2 Definition. If f ∈ k[x1 , . . . , xn ], a point p = (a1 , . . . , an ) ∈ An such that f (p) = f (a1 , . . . , an ) = 0 is called a zero of f and V(f ) = {p ∈ An | f (p) = 0} is called the zero set or zero locus of f . If f is non-constant, V(f ) is called the hypersurface defined by f . A hypersurface in An is also called an affine surface, in order to distinguish it from hypersurfaces in other ambient spaces. 1.1.3 Examples. (i) In R1 , V(x2 + 1) = ∅, but in C1 , V(x2 + 1) = {±i}. Generally, if n = 1 then V(F ) is the set of roots of F in k. If k is algebraically closed and F is non-constant then V(F ) is non-empty. (ii) In Z1p , by Fermat’s Little Theorem, V(xp − x) = Z1p . (iii) By Fermat’s Last Theorem, if n > 2 then V(xn + y n − 1) is finite in Q2 . 1

(iv) In R2 , V(x2 +y 2 −1) = the unit circle in R2 , and in Q2 it gives the rational points on the unit circle. Notice the circle admits a parameterization by rational functions as follows: 2t 1 − t2 , , t ∈ R. (x, y) = 1 + t2 1 + t2 When t ∈ Z then we get a point in Q2 . Remark. A rational curve is a curve that admits a parameterization by rational functions. For example, the curve in the last example is rational.

1.2

Algebraic Sets and Ideals

1.2.1 Definition. If S is any set of polynomials in k[x1 , . . . , xn ], we define \ V(S) = {p ∈ An | f (p) = 0 for all f ∈ S} = V(f ) f ∈S

If S = {f1 , . . . , fn } then we may write V(f1 , . . . , fn ) for V(S). A subset X ⊆ An is an (affine) algebraic set if X = V(S) for some S ⊆ k[x1 , . . . , xn ] 1.2.2 Examples. (i) For any a, b ∈ k, {(a, b)} is an algebraic set in k2 since {(a, b)} = V(x − a, y − b). (ii) In R2 , V(y−x2 , x−y 2 ) is only 2 points, but in C2 it is 4 points. Generally, B´ezout’s Theorem tells us that the number of intersection points of a curve of degree m with a curve of degree n is mn in projective space over an algebraically closed field. (iii) The twisted cubic is the rational curve {(t, t2 , t3 ) | t ∈ R} ⊆ R3 . It is an algebraic curve; indeed, it is easy to verify that it is V(y − x2 , z − x3 ). (iv) Not all curves in R2 are algebraic. For example, let X = {(x, y) | y − sin x = 0} and suppose that X is algebraic, so that X = V(S) for some S ⊆ R[x, y]. Then there is F ∈ S such that F 6= 0 and so \ X = V(S) = V(f ) ⊆ V(F ). f ∈S

Notice that the intersection of X with any horizontal line y − c = 0 is infinite for −1 ≤ c ≤ 1. Choose c such that F (x, c) is not the zero polynomial and notice that the number of solutions to F (x, c) = 0 is finite, so X cannot be algebraic.

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Remark. The method used in the last example works in more generality. Suppose that C is an algebraic affine plane curve and L is a line not contained C. Then L ∩ C is either ∅ or a finite set of points. 1.2.3 Proposition. The algebraic sets in A1 are ∅, finite subsets of A1 , and A1 itself. Proof: Clearly these sets are all algebraic. Conversely, the zero set of any non-zero polynomial is finite, so if S contains a non-zero polynomial F then V(S) ⊆ V(F ) is finite. If S = ∅ or S = {0} then V(S) = A1 . Before we continue, we recall some notation. If R is a ring and S ⊆ R, then hSi denotes the ideal generated by S 1 . If S = {s1 , . . . , sn }, then we denote hSi by hs1 , . . . , sn i. 1.2.4 Proposition. (i) If S ⊆ T ⊆ k[x1 , . . . , xn ] then V(T ) ⊆ V(S). (ii) If S ⊆ k[x1 , . . . , xn ] then V(S) = V(hSi), so every algebraic set is equal to V(I) for some ideal I. Proof: (i) Since S ⊆ T , V(T ) =

\

V(f ) ⊆

f ∈T

\

V(f ) = V(S).

f ∈S

(ii) From (i), V(hSi) ⊆ V(S). If x ∈ V(S) and f ∈ I then we can write f as f = gq f1 + · · · + gm fm , where fi ∈ S and gi ∈ k[x1 , . . . , xn ]. Then f (x) = g1 (x)f1 (x) + · · · + gm (x)fm (x) = 0 since x ∈ V(S).

Since every algebraic set is the zero set of an ideal of polynomials, we now turn our attention to ideals in polynomial rings. If a ring R is such that all of its ideals are finitely generated it is said to be Noetherian 2 . For example, all fields are Noetherian. The Hilbert Basis Theorem states that all polynomial rings with coefficients in a Noetherian ring are Noetherian. 1 The

ideal generated by S is the intersection of all ideals containing S. More concretely, ( n ) X hSi = ak sk : a1 , . . . , an ∈ R and s1 , . . . , sn ∈ S . k=1

2 Some

readers may be more familiar with a different definition of Noetherian in terms of ascending chains of ideals. This definition is equivalent to ours by Proposition A.0.10.

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1.2.5 Theorem (Hilbert Basis Theorem). If R is Noetherian, then R[x1 , . . . , xn ] is Noetherian.

Proof: See Appendix A.

An important geometric consequence of the Hilbert Basis Theorem is that every algebraic set is the zero set of a finite set of polynomials. 1.2.6 Corollary. Every algebraic set X in An is the zero set of a finite set of polynomials.

Proof: k[x1 , . . . , xn ] is Noetherian, so if X = V (S), then X = V (hSi) = V (S 0 ), where S 0 is a finite subset of k[x1 , . . . , xn ] that generates hSi. Tn Remark. Since V(f1 , . . . , fn ) = k=1 V(fk ), the preceding corollary shows that every algebraic set is the intersection of finitely many hypersurfaces.

1.2.7 Proposition. S T (i) If {Iα } is a collection of ideals then V( α Iα ) = α V(Iα ), so the intersection of any collection of algebraic sets is an algebraic set. (ii) If I and J are ideals then V(IJ) = V (I) ∪ V(J), so the finite union of algebraic sets is an algebraic set.3 (iii) V(0) = An , V(1) = ∅, and V(x1 − a1 , . . . , xn − an ) = {(a1 , . . . , an )}, so any finite set of points is algebraic.

Proof: (i) We have ! V

[

Iα

α

=

\

V(f ) =

\ \

V(f ) =

α f ∈Iα

f ∈∪α Iα

\

V(Iα ).

α

3 Recall

that the product of I and J is the ideal generated by products of an element from I and an element from J. More concretely, ( IJ =

n X

) ak bk : a1 , . . . , an ∈ I and b1 , . . . , bn ∈ J

k=1

4

.

(ii) Since (gh)(x) = 0 if and only if g(x) = 0 or h(x) = 0, \ V(IJ) = V(f ) f ∈IJ

=

\

V(gh)

g∈I,h∈J

=

\

V(g) ∪ V(h)

g∈I,h∈J

=

\ g∈I

V(g) ∪

\

V(h)

h∈J

= V(I) ∪ V(J). (iii) This is clear.

Remark. Note that finiteness of the union in property (ii) is required; for example, consider Z in R. It is not an algebraic set, because a polynomial over a field can only have finitely many roots, but it is the union of (infinitely many) algebraic sets, namely V(x − n) for n ∈ Z. The properties in Proposition 1.2.7 allow us to define a topology4 on An whose closed sets are precisely the algebraic sets. 1.2.8 Definition. The topology on An whose closed sets are precisely the algebraic sets is called the Zariski topology. Remark. When k is one of Q, R, or C, the Zariski topology is weaker than the usual metric topology, as polynomial functions are continuous, so their zero sets are closed. However, in each of these cases, the Zariski topology is strictly weaker than the metric topology. For example, Z is closed in the usual topology of each of Q, R, or C, but is not algebraic and thus is not closed in the Zariski topology. 1.2.9 Example. The non-empty open sets in the Zariski topology on the affine line A1 are precisely the complements of finite sets of points. However, this is not true for An when k is infinite and n > 1. For example, V(x2 + y 2 − 1), the unit circle in R2 , is closed but is not finite. Moreover, note that the Zariski topology on An is Hausdorff5 if and only if k is finite, in which case it is identical to the discrete topology. 4 A topology on a set X is a collection τ of subsets of X that satisfies the following properties: (i) ∅, X ∈ τ , S (ii) if Gi ∈ τ for every i ∈ I then i∈I Gi ∈ τ , (iii) if G1 , G2 ∈ τ then G1 ∩ G2 ∈ τ . The sets in τ are said to be open, and their complements are said to be closed. 5 Recall that a topology is said to be Hausdorff if distinct points always have disjoint open neighbourhoods.

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We have associated an algebraic subset of An to any ideal in k[x1 , . . . , xn ] by taking the common zeros of its members. We would now like to do the converse and associate an ideal in k[x1 , . . . , xn ] to any subset of An . 1.2.10 Definition. Given any subset X ⊆ An we define I(X) to be the ideal of X, I(X) = {f ∈ k[x1 , . . . , xn ] | f (p) = 0 for all p ∈ X}. 1.2.11 Examples. (i) The following ideals of k[x] correspond to the algebraic sets of A1 : I(∅) = h1i, I({a1 , . . . , an }) = h(x − a1 ) · · · (x − an )i, and ( 0 if k is infinite, 1 I(A ) = n hxp − xi if k has pn elements. Note that if X ⊆ A1 is infinite then k is infinite and I(X) = 0. (ii) In A2 , I({(a, b)}) = hx − a, y − bi. Clearly hx − a, y − bi ⊆ I({(a, b)}), so we need only prove the reverse inequality. Assume that f ∈ I({(a, b)}). By the division algorithm, there is g(x, y) ∈ k[x, y] and r(y) ∈ k[y] such that f (x, y) = (x − a)g(x, y) + r(y). But 0 = f (a, b) = r(b), so y − b divides r(y) and we can write we can write r(y) = (y − b)h(y), and hence f = (x − a)g + (y − b)h ∈ hx − a, y − bi.

1.2.12 Proposition. (i) If X ⊆ Y ⊆ An then I(Y ) ⊆ I(X). (ii) I(∅) = k[x1 , . . . , xn ]. I({(a1 , . . . , an )}) = hx1 −a1 , . . . , xn −an i for any point (a1 , . . . , an ) ∈ An . I(An ) = 0 if k is infinite. (iii) S ⊆ I(V(S)) for any set of polynomials S ⊆ k[x1 , . . . , xn ]. X ⊆ V(I(X)) for any set of points X ⊆ An . (iv) V(I(V(S))) = V(S) for any set of polynomials S ⊆ k[x1 , . . . , xn ]. I(V(I(X))) = I(X) for any set of points X ⊆ An . Proof: (i) If f is zero on every point of Y then it is certainly zero on every point of X. (ii) See example 1.2.11. (iii) These follow from the definitions of I and V. 6

(iv) From (iii), V(S) ⊆ V(I(V(S))), and by Proposition 1.2.4 (i), V(I(V(S))) ⊆ V(S) since S ⊆ V(I(S)). Therefore V(S) = V(I(V(S))). The proof of the second part is similar.

Remarks. (i) As is shown in the proof of part (ii) of the last proposition, the ideal hx1 − a1 , . . . , xn − an i of any point (a1 , ..., an ) ∈ An is maximal. (ii) Equality does not always hold in part (iii) of the last proposition, as shown by the following examples: (a) Consider I = hx2 + 1i ⊆ R[x]. Then 1 ∈ / I, so I 6= R[x]. But V(I) = ∅, so I(V(I)) = R[x] % I. (b) Consider X = [0, 1] ⊆ R. Then I(X) = 0 and V(I(X)) = R % X. These examples also show that not every ideal of k[x1 , . . . , xn ] is the ideal of a set of points and that not every subset of An is algebraic. We have a correspondence between subsets of An and ideals of k[x1 , . . . , xn ] given by X 7→ I(X) and I 7→ V (I). By part (iv) of the last proposition, this correspondence is one-to-one when restricted to algebraic sets and ideals of sets of points. Given that not every subset of An is algebraic and not every ideal of k[x1 , . . . , xn ] is the ideal of a set of points, we would like to examine the smallest algebraic set containing an arbitary subset of An and the smallest ideal of a set of points containing an arbitrary ideal of k[x1 , . . . , xn ]. 1.2.13 Definition. Let X ⊆ An and I ⊆ k[x1 , . . . , xn ] be an ideal. The closure of X (in the Zariski topology) is the smallest algebraic set containing X (i.e. the smallest closed set containing X), and is denoted X. The closure of I is the smallest ideal of a set of points that contains I, and is denoted I. If I = I, we say that I is closed . Remark. Note that I is the ideal of a set of points if and only if I = I. 1.2.14 Proposition. (i) If X ⊆ An , then X = V(I(X)). (ii) If I ⊆ k[x1 , . . . , xn ] is an ideal, then I = I(V(I)). Proof: We will only prove (i), as the proof of (ii) is very similar. By part (iii) of Proposition 1.2.12, we have X ⊆ V(I(X)) Since V(I(X)) is an algebraic set, X ⊆ V(I(X)). Conversely, since X ⊆ X, V(I(X)) ⊆ V(I(X)). By part (ii) of Proposition 1.2.7, we have V(I(X)) = X, because X is an algebraic set. Therefore, V(I(X)) ⊆ X, showing that X = V(I(X)). 7

1.2.15 Examples. (i) If X = (0, 1) ⊆ R, then the closure of X in the metric topology is [0, 1], whereas the closure of X in the Zariski topology is R. (ii) If k is infinite and X ⊆ A1 is any infinite set of points then X = A1 . In particular, the Zariski closure of any non-empty open set is the whole line, or every non-empty open set is Zariski dense in the affine line. (iii) Let I = hx2 i. Then I = I(V(I)) = hxi, so that I 6= I and I is not an ideal of a set of points.

1.3

Radical Ideals and the Nullstellensatz

In the previous section, we examined algebraic sets and ideals of sets of points. We saw that every algebraic set is the zero set of a finite set of polynomials. In this section, we will look for an intrinsic description of ideals of sets of points. We have already seen that not every ideal is the ideal of a set of points. Intuitively, an ideal I of k[x1 , . . . , xn ] is the ideal of a set of points whenever its generators intersect with the smallest possible multiplicity. However, since the multiplicity of any intersection is lost when we take the zero set of an ideal, as sets do not have any way of keeping track of multiplicity, we should not expect to get it back when we again take the ideal of that zero set. 1.3.1 Examples. (i) Let I = hx2 +y 2 −1, xi ⊆ R[x, y]. The set V(x2 +y 2 −1) is the unit circle, and V(x) is the vertical line through the origin. The line intersects the circle twice, each time with “multiplicity one”. Therefore, our intuition would lead us to think that I is a closed ideal. This is correct, as I = I(V(I)) = I({(0, −1), (0, 1)}) = hx, y 2 − 1i = I. (ii) Let I = hx2 + y 2 − 1, x − 1i ⊆ R[x, y]. The set V(x2 + y 2 − 1) is the unit circle, and V(x − 1) is the vertical line that is tangent to the circle at (1, 0). Because it only intersects the circle at one point, the intersection is with “multiplicity two”. Therefore, our intuition would lead us to think that I is not a closed ideal. This is indeed the case, as I = I(V(I)) = I({(1, 0)}) = hx − 1, yi = 6 I. The zero sets of the generators of I are a vertical line through (1, 0) and a horizontal line through the origin, which intersect once at the point (1, 0) with “multiplicity one”, again confirming our intuition. Algebraically, if I = I(X) for some X ⊆ An then √ I is radical. Recall that an ideal I is radical if I is equal to its radical ideal I, √ I = {a ∈ R | an ∈ I for some n > 0}. 8

Equivalently, I is radical if the following condition holds: an ∈ I implies that a ∈ I for all a ∈ R and n > 0. (See Proposition A.0.15.) 1.3.2 Examples. (i) If X ⊆ An then I(X) is radical, because f (x) = 0 whenever f n (x) = 0. (ii) Every prime ideal is radical. For a proof, see Proposition A.0.16. However, not every proper radical ideal is prime. For example, the ideal hx(x − 1)i = I({0, 1}) of k[x] is radical, but it is not prime. (iii) Let I = hx2 + y 2 − 1, x − 1i ⊆ R[x, y]. Then y 2 ∈ I, because y 2 = (x2 + y 2 − 1) − (x + 1)(x − 1), but y ∈ / I, simply because of the degrees of the y terms in the generators. Hence I is not radical. We already examined this example geometrically above. (iv) Let I = hy − x2 , y − x3 i. If u = x(x − 1), then u2 = [(y − x2 ) − (y − x3 )](x − 1) ∈ I, but u ∈ / I, because of the degrees of the x terms in the generators. Hence I is not radical. Geometrically, V(y − x2 ) is an upwards parabola through the origin, and V(y − x3 ) intersects it twice, at the origin and at the point (1, 1). There are only two points of intersection, yet the degrees of the polynomials involved imply that there should be three, including multiplicity. Thus one of the points of intersection (in fact, the origin) has “multiplicity two”. We saw in the first of the above examples that if I is the ideal of a set of points then I is radical. Is the converse true? That is, if I is radical is it true that I = I? √ 1.3.3 Proposition. If I is an ideal of k[x1 , . . . , xn ], then I ⊆ I ⊆ I. In particular, a closed ideal is radical. √ √ Proof: Clearly, I ⊆ I. Suppose f ∈ I. Then f n ∈ I for some n ≥ 1. Since f n (x) = 0 if and only if f (x) = 0, √ we have f ∈ I(V(I)). By Proposition 1.2.14, I = I(V(I)), so f ∈ I. Therefore, I ⊆ I. √ It follows from the previous proposition that if I = I then I = I if and √ only if √ I = I(V(I)). However, if k is not algebraically closed, it often happens that I 6= I(V(I)): 9

1.3.4 Example. The polynomial x2 + 1 ∈ R[x] is irreducible, so the ideal hx2 + 1i is maximal. Hence it is radical, and it is obviously proper. However, I(V(x2 + 1)) = I(∅) = k[x] so hx2 + 1i is not an ideal of a set of points. Clearly, x2 + 1 can be replaced by any irreducible polynomial of degree at least 2 in any non-algebraically closed field. However, the lack of algebraic closure in the base field is actually √ necessary for a counterexample. If the base field is algebraically closed, I = I. This result is due to Hilbert and is known as the Nullstellensatz, which is German for “zero points theorem”. 1.3.5 Theorem (Nullstellensatz). Suppose√k is algebraically closed, and let √ I ⊆ k[x1 , . . . , xn ] be an ideal. Then √ I(V(I)) = I, so I = I and I is the ideal of a set of points if and only if I = I. Proof: To come soon.

A related question is the characterization of maximal ideals of k[x1 , . . . , xn ]. We have seen that the ideal of a single point (a1 , . . . , an ) ∈ An is the maximal ideal hx1 − a1 , . . . , xn − an i. Are all maximal ideals of k[x1 , . . . , xn ] of this form? Again, the example of hx2 + 1i in R[x] shows this to be false in general. However, this is true when k is algebraically closed. Indeed, if I is a maximal ideal of k[x1 , . . . , xn ] then I is radical, so by the Nullstellensatz I is the ideal of a set of points. Since I is a maximal ideal and taking zero sets reverses inclusions, V(I) is a non-empty minimal algebraic set, which must consist of a single point (a1 , . . . , an ) ∈ An .

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Appendix A

Some Ring Theory A.0.6 Definition. A principal ring is a ring for which every ideal is generated by a single element. A principal integral domain is called a principal ideal domain, or PID for short. A.0.7 Proposition. k[x] is a PID. Proof: Since k[x] is clearly an integral domain, we only need to show that it is principal. Let I be an ideal of k[x], and let f be a monic polynomial of minimum degree in I. First, we show that f is unique, i.e. if g is another monic polynomial in I such that deg(g) = deg(f ), then f = g. Let h = f − g. Then h ∈ I, and since deg(h) < deg(f ) we must have h = 0, so g = f . We now show that I = hf i. Since f ∈ I, we have hf i ⊆ I. To establish the reverse inclusion, fix g ∈ I. By the division algorithm, there exist q, r ∈ k[x] such that r is monic, g = qf + r, and either r = 0 or deg(r) < deg(f ). Since I is an ideal, r = g − qf ∈ I. By the minimality of the degree of f , we can not have deg(r) < deg(f ), so r = 0. Therefore, g = qf and g ∈ hf i. Since g ∈ I was arbitrary, this shows that I ⊆ hf i, and thus I = hf i. A.0.8 Proposition. If n > 1, k[x1 , . . . , xn ] is not principal. Proof: Suppose that I is principal. Let I = hx1 , . . . , xn i. Then I = hpi for some p ∈ k[x1 , . . . , xn ]. Hence p|q for every q ∈ I. In particular, q|xi for 1 ≤ i ≤ n. Since the only elements in k[x1 , . . . , xn ] that divide every indeterminate are the non-zero scalars, p must be a scalar. However, this a contradiction, as there are no non-zero scalars in I. Therefore, our assumption that I is principal is false, and k[x1 , . . . , xn ] is not principal. A.0.9 Definition. We say that a ring R is Noetherian if every ideal of R is finitely generated. A.0.10 Proposition. Let R be a ring. Then the following are equivalent: 11

(i) R is Noetherian, (ii) R satisfies the ascending chain condition on ideals, i.e. if I0 ⊆ I1 ⊆ · · · ⊆ In ⊆ · · · is a chain of ideals of R, there exists a k ∈ N such that Ik = Ik+1 = · · · = Ik+n = · · · . Proof: Suppose R is Noetherian, and let I0 ⊆ I1 ⊆ · · · ⊆ In ⊆ · · · be a chain of ideals of R. Let I=

[

Ik .

k∈N

In general, the union of ideals is not an ideal, but the union of an increasing chain of ideals can easily be seen to be an ideal. Thus I is an ideal. Since R is Noetherian, I is finitely generated, i.e. there exist a1 , . . . , am ∈ I such that I = ha1 , . . . , am i. Let k ∈ N be such that a1 , . . . , am ∈ Ik . Then I = Ik = Ik+1 = · · · = Ik+n = · · · . Conversely, suppose R satisfies the ascending chain condition but is not Noetherian, and let I be an ideal of R that is not finitely generated. Pick a0 ∈ I, and let I0 = ha0 i. Since I is not finitely generated, I0 6= I. Pick a1 ∈ I \ I0 , and let I1 = ha0 , a1 i. Since I is not finitely generated, I0 ( I1 6= I. Continuing by induction, we get an increasing chain of ideals I0 ( I1 ( · · · ( In ( · · · , in contradiction to the ascending condition on R. Therefore, our assumption that R is not Noetherian is false. We now establish that polynomial rings over an arbitrary Noetherian ring are Noetherian. A.0.11 Theorem (Hilbert Basis Theorem). If R is a Noetherian ring, then R[x] is Noetherian. Proof: Suppose R[x] is not Noetherian, and let I is an ideal of R[x] that is not finitely generated. Let f0 be a polynomial of minimum degree in I. Continuing by induction, let fk+1 be a polynomial of minimum degree in I \ hf0 , . . . , fk i. For every k ∈ N, let dk = deg(fk ), and let ak be the leading coefficient of fk , and let J = h{ak : k ∈ N}i. Since R is Noetherian and ha0 i ⊆ ha0 , a1 i ⊆ · · · ha0 , . . . , an i ⊆ · · · 12

is an increasing chain of ideals whose union is J, there exists an n ∈ N such that J = ha0 , . . . , an i. Let I0 = hf0 , . . . , fn i. By construction, fn+1 ∈ / I0 . Since J = ha0 , . . . , an i and an+1 ∈ J, there exist b0 , . . . , bn ∈ R such that an+1 = b0 a0 + · · · + bn an . Then, as fn+1 ∈ I \ I0 , we have g = mn+1 − xdn+1 −d0 b0 f0 − · · · − xdn+1 −dn bn fn ∈ I, so deg(g) < deg(fn+1 ). However, g ∈ / I0 , as fn+1 ∈ / I0 , contradicting the minimality of deg(fn+1 ). Therefore, our assumption that R[x] is not Noetherian is false. A.0.12 Corollary. If R is a Noetherian ring, then R[x1 , . . . , xn ] is Noetherian. Proof: Since R[x1 , . . . , xn+1 ] ∼ = R[x1 , . . . , xn ][xn+1 ], the result follows by induction from the Hilbert Basis Theorem. A.0.13 Definition. Let R be a ring, and I an ideal in R. The radical of I is the ideal √ I = {a ∈ R | an ∈ I for some n > 0}. √ If I = I, we say that I is radical . A.0.14 Proposition. Let R be a ring, and I an ideal of R. Then ideal of R. Proof: If a ∈ R and b ∈

√

√

I is an

I, then bn ∈ I for some n > 0, so (ab)n = an bn ∈ I,

√ √ and ab ∈ I. If a, b ∈ I, am ∈ I and bn ∈ I for some m, n > 0. Therefore, by the Binomial Theorem, m+n+1 X m + n − 1 ak bm+n−1−k . (a + b)m+n+1 = k k=0

For every k ∈ N, either k ≥ m, or m−1 ≥ k and m+n−1−k ≥ n. This implies that for any k ∈ N, either ak ∈ I or bm+n−1−k ∈ I. Therefore, every term of m+n+1 the series expansion of (a + is in I, showing that (a + b)m+n+1 ∈ I, √ √ b) or a + b ∈ I. Therefore, I is an ideal. A.0.15 Proposition. Let R be a ring, and I an ideal of R. Then I is radical if and only if an ∈ I implies that a ∈ I for all a ∈ R and n > 0. √ Proof: Suppose I is radical and an ∈ I. Then a ∈ I = I. Conversely, n suppose √ that a ∈ I implies that a ∈√I for all a ∈ √R and n n> 0. Clearly, I ⊆ I, so we only need to show √ that I ⊆ I. If a ∈ I then a ∈ I for some n > 0. Thus a ∈ I, showing that I ⊆ I and that I is radical. 13

A.0.16 Proposition. Let R be a ring, and I a prime ideal of R. Then I is radical. Proof: Given a ∈ R and n > 0 such that an ∈ I, we will show that a ∈ I by induction on the n such that an ∈ I. If n = 1 and an ∈ I, then clearly a ∈ I. Suppose that bn ∈ I implies b ∈ I, and that an+1 ∈ I. Since I is prime, either a ∈ I or an ∈ I, in which case we also have a ∈ I by our induction hypothesis. Therefore, by Proposition A.0.15, I is radical.

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