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

1. Section 4-2Radical Ideals and the Ideal-Variety Correspondenceby Pablo Spivakovsky-Gonzalez

2. -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 fm2I(V), then f 2I(V).

3. Proof: Let x 2 V. If fm2I(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 2I(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.

4. Definition 2: An ideal I is radical if fm2 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 ½k[x1,…,xn] be an ideal. The radical of I, denoted , is the set {f :fm2 I for some integer m ¸ 1 }.

5. Properties of : 1. I ½, since f 2 I means f12 I and therefore f 2 by definition. • An ideal I is radical if and only if I = . • For any ideal I, is always an ideal. Example: Consider the ideal J =h x2, y3i½ k[x, y]. Neither x nor y lie in J; but x 2 and y 2 .

6. Also, (x ¢ y)2= x2y22 J, because x22 J. Then x ¢ y 2 . Finally, x + y 2 . To see this, we note that (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 , by the Binomial Theorem. Since each term above is a multiple of either x2or y3, (x + y)42 J, and therefore x + y 2 .

7. Lemma 5: If I is an ideal in k[x1,…,xn] then is an ideal in k[x1,…,xn] containing I. Furthermore, is a radical ideal. Proof: I ½has already been shown. We want to prove is an ideal. Let f, g 2 ; then by definition there exist m, l 2Z+ so that fm, gl2I. Now considerthe binomial expansion of (f + g)m+l-1 .

8. Every term in the expansion has a factor figj with i + j = m + l – 1. Therefore, either i ¸ m or j ¸ l, so either fi 2I or gj 2I. This implies that figj 2I, so every term of the expansion lies in I. Therefore, (f + g)m+l-12I, so f + g 2 . Finally, suppose f 2 and h2 k[x1,…,xn]. This means that fm2 I for some integer m ¸ l.

9. Therefore, hmfm 2 I, or (h¢f)m2 I, so hf2 . 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[x1,…,xn], then I(V(I)) =

10. Proof: Consider any f2 . Then by definition fm2 I for some m ¸ l. Therefore fm vanishes on V(I), so clearly f must vanish on V(I) also. It follows that f2 I(V(I)) , so we have ½I(V(I)) . Conversely, suppose that f2 I(V(I)). Then f vanishes on V(I). Now, by Hilbert’s Nullstellensatz, 9m ¸ l such that fm2 I. This means that f 2 . And because f was arbitrary,

11. I(V(I)) ½ . Before we had ½I(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!

12. 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 I1½ I2 are ideals, then V(I1) ¾V(I2)and, similarly, if V1½ V2 are varieties, then I(V1) ¾I(V2).

13. 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 radicalideals ===> affine varieties are inclusion-reversing bijections which are inverses of each other.

14. 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(f1,…,fs) is a subvariety of kn. By definition,every f 2I(V) vanishes on V, so V ½V(I(V)).

15. On the other hand, we have f1,…,fs2I(V) from definition of I. Therefore, h f1,…,fsi½ I(V). V is inclusion reversing, hence V(I(V)) ½V(h f1,…,fsi) = V. Before we had V ½ V(I(V)), and now we showed V(I(V)) ½ V.

16. 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.

17. The Nullstellensatz tells us I(V(I)) = . Also, if I is radical, I = (Exercise 4). Therefore, I(V(I)) = Iwhenever 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.

18. 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.

19. Questions about Radical Ideals: Consider an ideal I =h f1,…,fsi: 1. Radical Generators: Is there an algorithm to produce a set {g1,…,gm} so that = hg1,…,gmi? 2. Radical Ideal: Is there an algorithm to determine if I is radical? 3. Radical Membership: Given f2 k[x1,…,xn],is there an algorithm to determine if f2 ?

20. 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 f1,…,fsi½ k[x1,…,xn] be an ideal. Then f 2if and only if the constant polynomial 1 belongs to the ideal =h f1,…,fs, 1 – yf i½ k[x1,…,xn,y].

21. In other words, f 2if and only if = k[x1,…,xn,y]. Proof: Suppose 12 . Then we can write 1 as: s 1 = pi (x1,…,xn, y)fi+ q(x1,…,xn, y)(1 – yf), i=1 for somepi , q 2 k[x1,…,xn, y].

22. We set y = 1 / f(x1,…,xn). Then our expression becomes s 1 = pi (x1,…,xn, 1/f )fi , i=1 Now we multiply both sides by fm : s fm = Ai fi , i=1 for some polynomialsAi2 k[x1,…,xn].

23. Therefore, fm2I, and so f2 . Going the other way, suppose that f 2 . Then fm2 I ½for some m. At the same time, 1 – yf2. Then, 1 = ymfm+ (1 – ymfm ) = = ymfm + (1 – yf)(1 + yf +…+ ym-1fm-1 )2

24. 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[x1,…,xn] we first compute a reduced Groebner basis for:

25. h f1,…,fs, 1 – yf i½ k[x1,…,xn,y]. -If the result is {1}, then f2 . -Example: Consider the ideal I =h xy2 + 2y2, x4– 2x2+ 1 i in k[x, y]. We want to determine if f = y – x2 + 1 lies in Using lex order on k[x, y, z], we compute a reducedGroebner basisof the followingideal:

26. =h xy2 + 2y2, x4– 2x2+ 1, 1 – z (y – x2+ 1) i The basis we obtain is {1}, so by Proposition 8 f2 . - In fact, (y – x2+ 1)32 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:

27. Proposition 9: Let f 2 k[x1,…,xn] andI =h f i. Ifis the factorization of f into a product of distinct irreducible polynomials, then =h f1f2··· fri Definition 10: If f 2 k[x1,…,xn] is a polynomial, we define the reduction of f, denoted fred, to be the polynomial such thath fred i=, whereI =h f i.

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

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