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Section 4-2 Radical Ideals and the Ideal-Variety Correspondence by Pablo Spivakovsky-Gonzalez

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-We ended Section 4-1 with Hilbert’s Nullstellensatz -In this section we will look at Hilbert’s Nullstellensatz from a different perspective -If we have some variety V, can we identify those ideals that consist of all polynomials which vanish on that variety? Lemma 1: Let V be a variety. If f m 2 I(V), then f 2 I(V).

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Proof: Let x 2 V. If f m 2 I(V ), then (f(x)) m = 0. This can only be true if f(x) = 0. This reasoning applies to any x 2 V, so we conclude that f 2 I(V). -Therefore, I(V) has the property that if some power of a polynomial is in the ideal, then that polynomial itself must also belong to I(V). -This leads to the definition of radical ideal.

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Definition 2: An ideal I is radical if f m 2 I for some integer m ¸ 1 implies that f 2 I. -We can now rephrase Lemma 1 using radical ideals. Corollary 3: I(V) is a radical ideal. Definition 4: Let I ½ k[x 1,…,x n ] be an ideal. The radical of I, denoted, is the set {f : f m 2 I for some integer m ¸ 1 }.

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Properties of : 1.I ½, since f 2 I means f 1 2 I and therefore f 2 by definition. 2.An ideal I is radical if and only if I =. 3.For any ideal I, is always an ideal. Example: Consider the ideal J = h x 2, y 3 i½ k[x, y]. Neither x nor y lie in J; but x 2 and y 2.

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Also, (x ¢ y) 2 = x 2 y 2 2 J, because x 2 2 J. Then x ¢ y 2. Finally, x + y 2. To see this, we note that (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4, by the Binomial Theorem. Since each term above is a multiple of either x 2 or y 3, (x + y) 4 2 J, and therefore x + y 2.

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Lemma 5: If I is an ideal in k[x 1,…,x n ] then is an ideal in k[x 1,…,x n ] containing I. Furthermore, is a radical ideal. Proof: I ½ has already been shown. We want to prove is an ideal. Let f, g 2 ; then by definition there exist m, l 2 Z + so that f m, g l 2 I. Now consider the binomial expansion of (f + g) m+l-1.

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Every term in the expansion has a factor f i g j with i + j = m + l – 1. Therefore, either i ¸ m or j ¸ l, so either f i 2 I or g j 2 I. This implies that f i g j 2 I, so every term of the expansion lies in I. Therefore, (f + g) m+l-1 2 I, so f + g 2. Finally, suppose f 2 and h 2 k[x 1,…,x n ]. This means that f m 2 I for some integer m ¸ l.

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Therefore, h m f m 2 I, or (h ¢ f) m 2 I, so hf 2. This completes the proof that is an ideal. The book leaves the proof that is a radical ideal as an exercise at the end of the section. Theorem 6 (The Strong Nullstellensatz): Let k be an algebraically closed field. If I is an ideal in k[x 1,…,x n ], then I(V(I)) =

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Proof: Consider any f 2. Then by definition f m 2 I for some m ¸ l. Therefore f m vanishes on V(I), so clearly f must vanish on V(I) also. It follows that f 2 I(V(I)), so we have ½ I(V(I)). Conversely, suppose that f 2 I(V(I)). Then f vanishes on V(I). Now, by Hilbert’s Nullstellensatz, 9 m ¸ l such that f m 2 I. This means that f 2. And because f was arbitrary,

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I(V(I)) ½. Before we had ½ I(V(I)), so clearly I(V(I)) = This concludes our proof. - Note: From now on, Theorem 6 will be referred to simply as “the Nullstellensatz”. - The Nullstellensatz allows us to set up a “dictionary” between algebra and geometry => very important!

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Theorem 7 (The Ideal-Variety Correspondence): Let k be an arbitrary field. 1. The maps I affine varieties ===> ideals V ideals ===> affine varieties are inclusion-reversing. If I 1 ½ I 2 are ideals, then V( I 1 ) ¾ V( I 2 ) and, similarly, if V 1 ½ V 2 are varieties, then I( V 1 ) ¾ I( V 2 ).

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In addition, for any variety V, we have V(I(V)) = V, so that I is always one-to-one. 2. If k is algebraically closed, and we restrict ourselves to radical ideals, then the maps I affine varieties ===> radical ideals V radical ideals ===> affine varieties are inclusion-reversing bijections which are inverses of each other.

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Proof: 1. The proof that I and V are inclusion reversing is given as an exercise at the end of the section. We will now prove that V(I(V)) = V, when V = V(f 1,…,f s ) is a subvariety of k n. By definition, every f 2 I(V) vanishes on V, so V ½ V(I(V)).

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On the other hand, we have f 1,…,f s 2 I(V) from definition of I. Therefore, h f 1,…,f s i ½ I(V). V is inclusion reversing, hence V(I(V)) ½ V( h f 1,…,f s i ) = V. Before we had V ½ V(I(V)), and now we showed V(I(V)) ½ V.

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Therefore V(I(V)) = V, and I is one-to-one because it has a left inverse. This completes the proof of Part 1 of Thm. 7. 2. By Corollary 3, I(V) is a radical ideal. We also know that V(I(V)) = V from Part 1. The next step is to prove I(V(I)) = I whenever I is a radical ideal.

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The Nullstellensatz tells us I(V(I)) =. Also, if I is radical, I = (Exercise 4). Therefore, I(V(I)) = I whenever I is a radical ideal. We see that V and I are inverses of each other. V and I define bijections between the set of radical ideals and affine varieties. This completes the proof.

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Consequences of Theorem 7: - Allows us to consider a question about varieties (geometry) as an algebraic question about radical ideals, and viceversa. - We can move between algebra and geometry => powerful tool for solving many problems! - Note that the field we are working over must be algebraically closed in order to apply Theorem 7.

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Questions about Radical Ideals: Consider an ideal I = h f 1,…,f s i : 1. Radical Generators: Is there an algorithm to produce a set {g 1,…,g m } so that = h g 1,…,g m i ? 2. Radical Ideal: Is there an algorithm to determine if I is radical? 3. Radical Membership: Given f 2 k[x 1,…,x n ], is there an algorithm to determine if f 2 ?

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Answer: -Yes, algorithms exist for all 3 problems. -We will focus on the easiest question, #3, the Radical Membership Problem. Proposition 8 (Radical Membership): Let k be an arbitrary field and let I = h f 1,…,f s i ½ k[x 1,…,x n ] be an ideal. Then f 2 if and only if the constant polynomial 1 belongs to the ideal = h f 1,…,f s, 1 – yf i ½ k[x 1,…,x n,y].

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In other words, f 2 if and only if = k[x 1,…,x n,y]. Proof: Suppose 1 2. Then we can write 1 as: s p i (x 1,…,x n, y) f i + q (x 1,…,x n, y) (1 – yf), i for some p i, q 2 k[x 1,…,x n, y].

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We set y = 1 / f(x 1,…,x n ). Then our expression becomes s p i (x 1,…,x n, 1/f ) f i, i Now we multiply both sides by f m : s f m A i f i, i for some polynomials A i 2 k[x 1,…,x n ].

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Therefore, f m 2 I, and so f 2. Going the other way, suppose that f 2. Then f m 2 I ½ for some m. At the same time, 1 – yf 2. Then, 1 = y m f m + (1 – y m f m ) = = y m f m + (1 – yf)(1 + yf +…+ y m-1 f m-1 ) 2

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Hence, f 2 implies that 1 2. And before we had that 1 2 implies f 2 so the proof is complete. Radical Membership Algorithm: - To determine if f 2 ½ k[x 1,…,x n ] we first compute a reduced Groebner basis for:

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h f 1,…,f s, 1 – yf i ½ k[x 1,…,x n,y]. -If the result is {1}, then f 2. -Example: Consider the ideal I = h xy 2 + 2y 2, x 4 – 2x 2 + 1 i in k[x, y]. We want to determine if f = y – x 2 + 1 lies in Using lex order on k[x, y, z], we compute a reduced Groebner basis of the following ideal:

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= h xy 2 + 2y 2, x 4 – 2x 2 + 1, 1 – z (y – x 2 + 1) i The basis we obtain is {1}, so by Proposition 8 f 2. - In fact, (y – x 2 + 1) 3 2 I, but no lower power of f is in I. Principal Ideals: -If I = h f i, we can compute the radical of I as follows:

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Proposition 9: Let f 2 k[x 1,…,x n ] and I = h f i. If is the factorization of f into a product of distinct irreducible polynomials, then = h f 1 f 2 ··· f r i Definition 10: If f 2 k[x 1,…,x n ] is a polynomial, we define the reduction of f, denoted f red, to be the polynomial such that h f red i =, where I = h f i.

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-A polynomial is said to be reduced, or square-free, if f = f red. -Section 4-2 ends with a formula for computing the radical of a principal ideal => see pg. 181.

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Sources Used - Ideals, Varieties, and Algorithms, by Cox, Little, O’Shea; UTM Springer, 3 rd Ed., 2007. Thank You! Stay tuned for the next lecture, by ShinnYih Huang!

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