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

2. Let V be a variety. If fm 2 I(V), then f 2 I(V).
-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 fm 2 I(V), then f 2 I(V).

3. Proof:
Let x 2 V. If fm 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.

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

5. 1. I ½ , since f 2 I means f1 2 I and therefore f 2 by definition.
Properties of :
1. I ½ , since f 2 I means f1 2 I and therefore f 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, y3 i½ k[x, y].
Neither x nor y lie in J; but x and y

6. by the Binomial Theorem.
Also, (x ¢ y)2 = x2y2 2 J, because x2 2 J.
Then x ¢ y Finally, x + y
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 x2 or y3, (x + y)4 2 J,
and therefore x + y

7. Now consider the binomial expansion of (f + g)m+l-1 .
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 ½ has already been shown. We want to prove
is an ideal. Let f, g ; then by definition there exist
m, l 2 Z+ so that fm, gl 2 I.
Now consider the binomial expansion of (f + g)m+l-1 .

8. Finally, suppose f 2 and h 2 k[x1,…,xn].
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 2 I or
gj 2 I. This implies that figj 2 I, so every term of the
expansion lies in I.
Therefore, (f + g)m+l-1 2 I, so f + g
Finally, suppose f and h 2 k[x1,…,xn].
This means that fm 2 I for some integer m ¸ l.

9. Theorem 6 (The Strong Nullstellensatz):
Therefore, hmfm 2 I, or (h¢f)m 2 I, so hf
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. This means that f 2 . And because f was arbitrary,
Proof:
Consider any f Then by definition fm 2 I for some m ¸ l. Therefore fm 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 fm 2 I.
This means that f And because f was arbitrary,

11. 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!

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 I1 ½ I2 are ideals, then
V(I1) ¾ V(I2) and, similarly, if V1 ½ V2 are
varieties, then I(V1) ¾ I(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
radical ideals ===> affine varieties
are inclusion-reversing bijections which are inverses
of each other.

14. By definition, every f 2 I(V) vanishes on V, so
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 2 I(V) vanishes on V, so
V ½ V(I(V)).

15. definition of I. Therefore, h f1,…,fs i ½ I(V).
On the other hand, we have f1,…,fs 2 I(V) from
definition of I. Therefore, h f1,…,fs i ½ I(V).
V is inclusion reversing, hence
V(I(V)) ½ V(h f1,…,fs i) = V.
Before we had V ½ V(I(V)), and now we
showed V(I(V)) ½ V.

16. We also know that V(I(V)) = V from Part 1.
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)) = 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.

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,…,fs i :
1. Radical Generators: Is there an algorithm to produce a set
{g1,…,gm} so that = h g1,…,gm i ?
2. Radical Ideal: Is there an algorithm to determine if I is radical?
3. Radical Membership: Given f 2 k[x1,…,xn], is there an algorithm to determine if f ?

20. Proposition 8 (Radical Membership):
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,…,fs i ½ k[x1,…,xn]
be an ideal. Then f 2 if and only if the constant
polynomial 1 belongs to the ideal
= h f1,…,fs, 1 – yf i ½ k[x1,…,xn,y].

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

22. Now we multiply both sides by fm : fm =  Ai fi ,
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 :
fm =  Ai fi ,
for some polynomials Ai 2 k[x1,…,xn].

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

24. so the proof is complete.
Hence, f implies that
And before we had that implies f 2
so the proof is complete.
Radical Membership Algorithm:
-To determine if
f ½ k[x1,…,xn]
we first compute a reduced Groebner basis for:

25. h f1,…,fs, 1 – yf i ½ k[x1,…,xn,y]. -If the result is {1}, then f 2 .
-Example:
Consider the ideal I = h xy2 + 2y2, x4 – 2x2 + 1 i
in k[x, y].
We want to determine if f = y – x lies in
Using lex order on k[x, y, z], we compute a
reduced Groebner basis of the following ideal:

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

27. Proposition 9: Definition 10:
Let f 2 k[x1,…,xn] and I = h f i.
If is the factorization of f into a
product of distinct irreducible polynomials, then
= h f1f2 ··· fr i
Definition 10:
If f 2 k[x1,…,xn] is a polynomial, we define the reduction of f, denoted fred , to be the polynomial such
that h fred i = , where I = 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. Stay tuned for the next lecture,
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!