[DF04] David Steven Dummit and Richard M Foote. Abstract algebra, vol- ume 1984. Wiley Hoboken, 2004. [Eis13] David Eisenbud. Commutative Algebra: with ...
Algebraic Independence of Polynomials Student: Akash Jena National Institute of Science Education and Research, Bhubaneswar
Guide: Dr. Nitin Saxena Associate Professor IIT, Kanpur
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Contents 0.1
Transcendence Degree . . . . . . . . . . . . . . . . . . . . . . .
3
0.2
Annihilating Polynomial . . . . . . . . . . . . . . . . . . . . . .
5
0.3
Connection with Krull Dimension . . . . . . . . . . . . . . . . .
8
2
0.1
Transcendence Degree
Definition. Let E/F be an extension of fields. Let S be a maximal algebraically independent subset of E with respect to inclusion. Then S is called a transcendence basis of E over F . Theorem 1. Let E/F be a transcendental extension with transcendence basis S. Then E is algebraic over F (S). Proof. Let a ∈ E be arbitrary. If a ∈ S, then a is a root of the polynomial f (X) = X − a ∈ F (S)[X]; hence is algebraic over F (S). If a 6∈ S, then by the maximality of S, the set S ∪ {a} is algebraically dependent. Therefore, a will satisfy some polynomial over F (S). Theorem 2. Let E/F be extension of fields. If S ⊆ E is such that • S is algebraically independent over F , and • E is algebraic over F (S), then S is a transcendence basis of E over F . Proof. We have to show that S is the maximal algebraically independent subset of E. Since S is already given to be algebraically independent, all it remains to prove is maximality. Assume that there exists an algebraically independent subset S 0 of E such that S ( S 0 . Consider θ ∈ S 0 − S. As E is algebraic over F (S), the element θ will be the root of some polynomial in F (S)[X]. Therefore an θn + an−1 θn−1 · · · + a1 θ + a0 = 0
(1)
where ai ∈ F (S). F (S) is the fraction field of F [S]. Clearing out the denominators in ai , we get dependence relation for θ in F [S], which contradicts the algebraic independence of S 0 . Theorem 3. E/F is a transcendental extension and S ⊆ E. Then the following are equivalent • S is a transcendence basis of E over F . • S is the maximal algebraically independent subset of E with respect to inclusion. • S is algebraically independent and E is algebraic over F (S). Proof. Follows directly from the previous two theorems.
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Definition. We say that the field extension E/F is purely transcendental if E = F (S) for some algebraically independcnt subset S of E over F . Remark. If E/F is a purely transcendental extension with E = F (S), where S is algebraically independent over F , then S is a transcendence basis of E over F . Example. If S1 and S2 are two transcendence bases of E/F , it is not necessary that F (S1 ) = F (S2 ). Consider F = Q and E = Q(t), where t is an indeterminate. Consider S1 = {t} and S2 = {t2 }. It is clear that S1 is a transcendence basis for E over F . We claim that S2 is also a transcendence basis. The element t is a root of X 2 − t2 ∈ Q(t2 )[X], hence is algebraic over Q(t2 ). Therefore, E = Q(t) is algebraic over F = Q(t2 ). Also, S2 = {t2 } is algebraically independent over F = Q. Therefore, S2 is also a transcendence basis. But F (S2 ) = Q(t2 ) ( Q(t) = F (S1 ). Theorem 4. Let E/F be an arbitrary field extension. The field E has a transcendency basis over F and any two such bases have the same cardinality. Also, any algebraically independent set over F can be extended to a transcendency basis of E over F . Proof. [Kar89] Corollary 2.4, Chapter 3. The uniqueness of the cardinality of transcendence bases allows us to define the following concept Definition. Let E/F be any field extension. The transcendence degree of E over F is defined as the cardinality of any transcendence basis of E over F . We denote it by tr.deg(E/F ). Example. We have already seen that {t2 } is a transcendense basis for Q(t)/Q. We can use the previous theorem to prove it. We know that {t} is a transcendency basis. Then the previous theorem tells that any other transcendence basis will also have cardinality one. We know that {t2 } is algebraically independent over Q, hence can be extended to a transcendency basis. But it cannot be properly contained in an algebraically independent set (otherwise we will obtain a transcendency basis with cardinality more than one). Therefore, {t2 } is a maximal algebraically independent set, and hence, a transcendency basis. .
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Example. Let F be any field. Consider E = F (X1 , X2 . . . Xn ). The set {X1 , X2 . . . Xn } is algebraically independent over F and E is algebraic over F (X1 , X2 . . . Xn ). Therefore, {X1 , X2 . . . Xn } is a transcendence basis. Consider f1 , f2 . . . fk ∈ F [X1 , X2 . . . Xn ] ⊆ F (X1 , X2 . . . Xn ). If f1 , f2 . . . fk are algebraically independent, then we claim that k ≤ n. Since f1 , f2 . . . fk are algebraically independent, we can extend {f1 , f2 . . . fk } to a transcendency basis. If k > n, then we will obtain a transcendence basis of cardinality more than n, a contradiction. Theorem 5 (Jacobian Criterion). Let F be a field of with char(F ) = 0. Consider f1 , f2 . . . fn ∈ F [X1 , X2 . . . Xn ] ⊆ F (X1 , X2 . . . Xn ). Then f1 , f2 . . . fn are algebraically independent iff their jacobian is zero. Proof. [Hum92] Section 3.10.
0.2
Annihilating Polynomial
Definition. Let f1 , . . . fk ∈ K[X1 , . . . , Xn ] be algebraically dependent. Any polynomial A ∈ K[y1 , . . . , yk ] which satisfies A(f1 , . . . fk ) is called an annihilating polynomial of f1 , . . . fk . We have already seen that any n + 1 polynomials in n variables are algebraically dependent with the help of field theoretic arguments. The following theorem is an elementary proof of the same. It also gives a procedure to construct an annihilating polynomial. For that we need the following lemma. Lemma. Let n, d ∈ N. Then, n n+1 Y Y n+1 (n + 1) [(n + 1)d + j] < [(n + 1)dn + j] j=1
j=1
Proof. We have (n + 1)
n Y
[(n + 1)d
n+1
+ j] =
n X
j=1
ck dk(n+1)
k=0
n where ck = (n + 1)k+1 Sn−k . Also, n+1 Y
n+1 X
j=1
k=0
[(n + 1)dn + j] =
5
ek dkn
n+1 where ek = (n + 1)k Sn+1−k . Then,
n+1 Y
n Y [(n + 1)d + j] − (n + 1) [(n + 1)dn+1 + j] n
j=1
j=1
=
n+1 X
ek dkn −
k=0
= e0 +
n X k=0
n+1 X
ek dkn −
k=1
= e0 +
ck dk(n+1)
n X
n X k=0
ek+1 d
(k+1)n
k=0
= e0 + = e0 +
n X k=0 n X
ck dk(n+1)
−
n X
ck dk(n+1)
k=0
[ek+1 d(k+1)n − ck dk(n+1) ] n+1 n [(n + 1)k+1 · Sn−k · d(k+1)n − (n + 1)k+1 · Sn−k · dk(n+1) ]
k=0
= e0 +
n X
n+1 n [Sn−k · d(k+1)n − Sn−k · dk(n+1) ] · (n + 1)k+1 .
k=0
We have Skn+1 ≥ Skn and d(k+1)n ≥ dk(n+1) for 0 ≤ k ≤ n. Therefore, the RHS of the above equation is greater than zero and we have our result. Theorem 6. Let K be a field. Let f1 , . . . , fn+1 ∈ K[X1 , . . . Xn ]. Then there exists A(y1 , . . . , yn+1 ) ∈ K[y1 , . . . , yn+1 ] such that A(f1 , . . . , fn+1 ) = 0. Furthermore, deg(A) ≤ (n + 1)dn , where d = max deg(fi ). i
Proof. Consider an arbitrary n + 1 variate polynomial A ∈ K[y1 , . . . yn+1 ] with deg(A) = λ. A is of the form X in+1 A(y1 , . . . , yn+1 ) = ci1 ,i2 ,...,in+1 y1i1 y2i2 . . . , yn+1 (2) We consider the ci1 ,i2 ,...,in+1 ’s as unknowns in the field K, which we will find out by solving a system of linear equations. As deg(A) = λ, we have i1 + i2 + · · · + in+1 ≤ λ. So the number of monomials of A is λ+n+1 . After we plug in λ f1 , . . . , fn+1 in A we get X in+1 A(f1 , . . . , fn+1 ) = ci1 ,i2 ,...,in+1 f1i1 f2i2 . . . , fn+1 X = bj1 ,j2 ,...,jn X1j1 X2j2 . . . , Xnjn 6
where the bj1 ,j2 ,...,jn ’s are K-linear combinations of the ci1 ,i2 ,...,in+1 ’s appearing in (2). As deg(fi ) ≤ d, the highest power of any monomial of A(f1 , . . . , fn+1 ) is bounded by i1 d+i2 d+· · ·+in+1 d = (i1 +i2 +· · ·+i n+1 )d ≤ λd. So the number of λd+n monomials of A(f1 , . . . , fn+1 ) is at most λd . If we set A(f1 , . . . , fn+1 ) = 0, (equivalently, equating the coefficients of all monomials of A(f1 , . . . , fn+1 ) to zero) we get a system of K-linear equations in the unknowns ci1 ,i2 ,...,in+1 ’s. The number of equations is same as the number of monomials in A(f1 , . . . , fn+1 ), which is at most λd+n . The number of unknowns is same as the number λd of monomials in A(y1 , . . . , yn+1 ), which is λ+n+1 . If the number of variables λ is more than the number of unknowns, then there will be a solution for the λ+n+1 system. Hence, a sufficient condition for solution to exist is λd+n < λd λ or equivalently, n n+1 Y Y (n + 1) (λd + j) < (λ + j) (3) j=1
j=1
From the previous lemma it is clear that λ = (n + 1)dn will satisfy the above inequality. For our next result we need to define the notion of resultant of two polynomials. Let f (X) = a0 X l + · · · + a0 and g(X) = b0 X m + . . . bm be two polynomials in K[X]. The following (l + m) × (l + m) matrix is called the Sylvester matrix of f and g. a0 a1 a0 a a ... 2 1 . .. · · · a0 . . Syl(f, g, x) = . a1 al .. al . . .. al {z } |
b0 b1
m columns
b2 .. .
b0 b1
... ···
.. .
b0 b1
bm
.. .
bm ...
bm |
{z
l columns
}
The resultant of f and g are defined as the determinant of the Sylvester matrix. We denote it by Res(f, g, x). Lemma. Given two polynomials f, g ∈ K[X] of positive degree, there exist polynomials A, B in K[X] such that Af + Bg = Res(f, g, x) 7
Moreover, f and g have a common factor in K[x] if and only if Res(f, g, x) = 0 Proof. [CLO07] Chapter 3, §5, Proposition 8 and Proposition 9. Theorem 7. Consider f1 , . . . , fm ∈ K[X1 , . . . , Xn ]. Suppose that out of these m polynomials, m − 1 are algebraically independent. If I ⊆ K[y1 , . . . , ym ] is the set of annihilating polynomials of f1 , . . . fm , then I is a principal ideal. Proof. Without loss of generality we can assume that f1 , . . . , fm−1 are algebraically independent. Let F1 and F2 be two irreducible polynomials in I. To prove our theorem it is sufficient to prove that F2 = cF1 for some c ∈ K ∗ . We will consider F1 , F2 as elements of R[ym ] where R = K[y1 , . . . , ym−1 ]. Let g ∈ K[y1 , . . . , ym−1 ] be the resultant of F1 and F2 with respect to the variable in the ring R. By the previous lemma there exist g1 , g2 ∈ R[ym ] such that g = g1 F1 + g2 F2 . We have g(f1 , . . . , fm−1 ) = g1 (f1 , . . . , fm )F1 (f1 , . . . , fm )+g2 (f1 , . . . , fm )F2 (f1 , . . . , fm ) = 0 Since f1 , . . . , fm−1 are algebraically independent, it follows that g = 0. Hence, by the previous lemma F1 and F2 have a common factor. But we have chosen F1 and F2 to be irreducible. Hence F2 = cF1 for some c ∈ K ∗ .
0.3
Connection with Krull Dimension
Definition. Let R be a commutative ring. Let P0 , . . . Pn be prime ideals of R such that P0 ( P 1 ( · · · ( Pn . Then we say that P0 , . . . Pn form a chain of length n. The Krull dimension of R is defined as the supremum of the length of all such chains. For our next result we need the following lemma. Lemma. Let F be a field. Let B be an integral domain containing F . If every element of B is algebraic over F , then B is a field. Theorem 8. Let K be a field. Let A = K[S] be an affine K-domain. Then dim(A) ≤ max { |T | | T ⊆ S is algebraically independent} Proof. Define n to be the maximum of the right hand side of the claimed inequality. We will prove the result by induction on n. When n = 0, then every element of S is algebraic over K. Therefore, A = K[S] is algebraic over 8
K. Hence, by the previous lemma A is a field. Therefore, dim(A) = 0, which proves the claim for n = 0. Now consider the case n > 0. Let P 0 ( P1 ( · · · ( Pm be a chain of prime ideals of length m > 0 in A. Our aim is to show that m ≤ n. Quotienting the above chain by P1 we get a chain of length m − 1 in A/P1 . If we can show that all algebraically independent subsets T ⊆ {a + P1 |a ∈ S} ⊆ A/P1 have size |T | < n, then we can use induction on n and conclude that dim(A/P1 ) < n, so m − 1 < n, which proves our claim. Assume to the contrary that there are a1 , . . . an ∈ A such that a1 +P1 , . . . at + P1 are algebraically independent. Then {a1 , a2 , . . . an } ⊆ A is algebraically indef dependent. By definition of n, all a ∈ S are algebraic over L = K(a1 , . . . an ). Therefore K(S) is also algebraic over L. There exists a nonzero element a ∈ P1 . Consider the minimal polynomial of a over L G(x) =
k X
gi xi
i=0
where gi ∈ K(a1 , . . . an ). Since a 6= 0 and G is the minimal polynomial of a, we have g0 6= 0. Since gi ∈ K(a1 , . . . an ), we can clear out the denominators to obtain a new polynomial k X H(x) = hi xi i=0
with the coefficients hi ∈ K[a1 , . . . an ] and H(a) = 0. Also, the constant term h0 6= 0. Since H(a) = 0, we have h0 = −
k X
hi ai ∈ P1
(4)
i=1
Reducing the above equation modulo P1 , we obtain a polynomial dependence for a1 + P1 , . . . , an + P1 , namely, h0 (a1 + P1 , . . . , an + P1 ) = 0. This contradicts the algebraic independence of {a1 + P1 , . . . , an + P1 }. Lemma. Let K be a field. Let A be an affine K-domain. If A is a field, then A is algebraic over K. Proof. [Kem11] Lemma 1.1b.
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Theorem 9. Let K be a field. Let A = K[S] be an affine K-domain. Then dim(A) ≥ tr.degK (A) Proof. Let n ≥ 0 be an integer. We will use induction to prove that if tr.degK (A) ≥ n, then dim(A) ≥ n. Consider the case n = 0. The ideal {0} is prime and hence dim(A) ≥ 0. Now we will consider the case n > 0. As tr.degK (A) ≥ n we can find n def algebraically independent elements a1 , a2 , . . . , an in A. Let L = K(a1 ). Define def A0 to be the L-algebra generated by S, that is A0 = L[S]. A0 is an affine L-domain and {a2 , a3 , . . . , an } is algebraically independent over L. Hence, tr.degL (A0 ) ≥ n − 1. So the induction hypothesis implies that dim(A0 ) ≥ n − 1 So we have a chain of prime ideals P0 ( P1 ( · · · ( Pn−1 def
in A0 . We have A ,→ A0 , as rings. Define Pi as Pi = Pic , the contraction of Pi in A. Then we have Pi−1 ⊆ Pi . We claim that this containment is proper. This is because I ce = I for any ideal I of A0 (for any ideal J of A we denote J e as the extension of J to A0 ). So we get the chain P0 ( P1 ( · · · ( Pn−1 in spec(A). Also, L ∩ Pn−1 = {0}, otherwise there will be an invertible element in Pn−1 , which will imply Pn−1 = A0 . It follows that a1 + Pn−1 ∈ A/Pn−1 is not algebraic over K. Hence, by the previous lemma we get that A/Pn−1 is not a field. Therefore, Pn−1 is not a maximal ideal. So we can extend our previous chain in A to P0 ( P1 ( · · · ( Pn−1 ( Pn where Pn is a maximal ideal containing Pn−1 . This completes the induction process.
Theorem 10. Let K be a field. Let A = K[S] be an affine K-domain. Then dim(A) = tr.degK (A) = tr.degK (S) Proof. It follows directly from the previous two theorems Example. Let K be a field. Consider R = K[X1 , . . . , Xn ]. The elements X1 , . . . , Xn are algebraically independent over K. Hence, by the previous theorem we have dim(R) = n.
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References [AM69] Michael Francis Atiyah and Ian Grant Macdonald. Introduction to commutative algebra, volume 2. Addison-Wesley Reading, 1969. [BMS13] Malte Beecken, Johannes Mittmann, and Nitin Saxena. Algebraic independence and blackbox identity testing. Information and Computation, 222:2–19, 2013. [CLO07] David Cox, John Little, and Donal OShea. Ideals, varieties, and algorithms, 3rd edn. undergraduate texts in mathematics, 2007. [DF04]
David Steven Dummit and Richard M Foote. Abstract algebra, volume 1984. Wiley Hoboken, 2004.
[Eis13]
David Eisenbud. Commutative Algebra: with a view toward algebraic geometry, volume 150. Springer Science & Business Media, 2013.
[Hum92] James E Humphreys. Reflection groups and Coxeter groups, volume 29. Cambridge university press, 1992. [Kar89] Gregory Karpilovsky. Topics in field theory, volume 155. Elsevier, 1989. [Kem11] Gregor Kemper. A course in commutative algebra, volume 256 of graduate texts in mathematics, 2011.
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