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Galois closures of quartic subfields of rational function fields
Finite Fields and Their Applications 26 (2014) 100–103
Contents lists available at ScienceDirect
Finite Fields and Their Applications www.elsevier.com/locate/ffa
Galois closures of quartic subfields of rational function fields Robert C. Valentini Department of Mathematics and Statistics, California State University, Long Beach, Long Beach, CA 90840, USA
a r t i c l e
i n f o
Article history: Received 21 August 2013 Accepted 21 November 2013 Available online 14 December 2013 Communicated by H. Stichtenoth MSC: 11R32 12F10 11T55
a b s t r a c t Let k be a finite field with q elements. Let f (x) ∈ k[x] be a monic quartic polynomial. Then k(x)/k( f (x)) is a field extension of degree 4. If the extension is separable, then the Galois group of the Galois closure is isomorphic to a transitive subgroup of the symmetric group on 4 letters. We determine the number of f (x)’s having a given subgroup as Galois group. © 2013 Elsevier Inc. All rights reserved.
Keywords: Galois group Quartic polynomial Rational function field
1. Introduction Let p be a prime. Let k be the finite field with q = p e elements. For a monic polynomial f (x) = x4 + ax3 + bx2 + cx + d ∈ k[x], set y = f (x). Then the field extension k(x)/k( y ) is of degree 4 and the minimal polynomial of x over k( y ) is F ( X ) = X 4 + a X 3 + b X 2 + c X + (d − y ). When F ( X ) is separable, then the Galois group G f of the Galois closure of k(x)/k( y ) is a transitive subgroup of S 4 (the symmetric group on 4 letters). Hence G f is isomorphic to S 4 , A 4 (the alternating group on 4 letters), D 8 (the dihedral group of order 8), V (the Klein 4 group), or C 4 (the cyclic group of order 4). In this paper we determine the number of f (x)’s producing each of these possibilities. It is classical that the Galois group of a quartic polynomial can largely be determined from the degree of the splitting field of a resolvent cubic. A succinct statement of the situation can be found in [1, pp. 51–52].
E-mail address: [email protected]. 1071-5797/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.ffa.2013.11.005
R.C. Valentini / Finite Fields and Their Applications 26 (2014) 100–103
101
QG Theorem. Let g be a separable irreducible quartic polynomial over a field K . Let m be the degree over K of the splitting field of a resolvent cubic of g. Let G be the Galois group of g over K . Then (1) (2) (3) (4)
If m = 6, G If m = 3, G If m = 1, G If m = 2, G
is isomorphic to is isomorphic to is isomorphic to is isomorphic to
S4. A4 . V. D 8 or C 4 .
We apply this theorem to analyze the situation for F ( X ) over k( y ). 2. p = 2 First we deal with the situation that p = 2. Then it is easily seen that F ( X ) is separable and we may apply the QG Theorem. We start by looking at those f (x)’s with a = d = 0. So
F ( X ) = X 4 + b X 2 + c X − y.
(1)
We employ the resolvent cubic as developed in [2, pp. 613–614], which for (1) is
H ( X ) = X 3 − 2b X 2 + b2 + 4 y X + c 2 .
(2)
The discriminant D of H ( X ) as given in [2, p. 614] is
D = − 4b3 + 27c 2 c 2 − 16b b3 + 9c 2 y − 128b2 y 2 − 256 y 3 and is the same as the discriminant of F ( X ). As a polynomial in y, D has degree 3, so cannot be a square in k( y ). Hence G f cannot be a subgroup of A 4 . So the only possibilities for G f are S 4 , D 8 , and C 4 . From the QG Theorem we see that we will get S 4 if H ( X ) is irreducible, which occurs precisely when it has no roots in k( y ). Applying the rational root theorem, we have H ( X ) is irreducible when c = 0. So there are q(q − 1) F ( X )’s of the type (1) for which H ( X ) is irreducible and G f is isomorphic to S 4 . When c = 0, F ( X ) = X 4 + b X 2 − y. If G f is isomorphic to C 4 , then G f is a group of automorphisms of the rational function field k(x). By Theorem 1 of [3], this is only possible for a strict set of arithmetic conditions. The factorization y = x4 + bx2 = x2 (x2 + b), shows that both the zero and the pole of y are ramified in the extension k(x)/k( y ). By Theorem 1 of [3] this can occur for C 4 only when both the zero and the pole of y are completely ramified. But this means b = 0. Then F ( X ) = X 4 − y. The roots of this equation are ±x, ±ux, where u is a primitive fourth root of unity. Thus if k(x)/k( y ) is to be Galois, k must contain the fourth roots of unity and hence 4 | (q − 1). Thus if q ≡ 1 (mod 4), there is exactly one F ( X ) of type (1) for which G f is isomorphic to C 4 and none if q ≡ 3 (mod 4). Since the only remaining possibility for G f is D 8 , by the process of elimination, if q ≡ 1 (mod 4), then there are q − 1F ( X )’s of type (1) for which G f is isomorphic to D 8 and q such F ( X )’s when q ≡ 3 (mod 4). Now for an arbitrary f (x) = x4 + ax3 + bx2 + cx + d the automorphism of k(x) sending x to x − a/4 sends f (x) to f (x − a/4) = x4 + β x2 + γ x + δ , where β = (−3a2 + b)/8, γ = (a3 − 4ab + 8c )/8, and δ = (−3a4 + 16a2 b − 64ac + 256d)/256. So k(x)/k( f (x)) is isomorphic to k(x)/k( f (x − a/4) − δ) and the Galois closures are isomorphic also. But f (x − a/4) − δ gives an F ( X ) of type (1). From the equations for β and γ , we see that for a fixed pair (a, d) as (b, c ) runs over all elements of k2 , (β, γ ) also runs over all elements of k2 . So for a fixed quartic with a = d = 0, there are q2 arbitrary f (x)’s such that f (x − a/4) − δ is equal to the fixed quartic. So we have the following theorem.
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Theorem 1. Let p = 2 be a prime. Let q = p e , where e is a positive integer. Let k be the finite field with q elements. Let M = { f (x) ∈ k[x] | f (x) is monic of degree 4}. For f (x) ∈ M, let G f be the Galois group of the Galois closure of the extension k(x)/k( f (x)). Then the number of f (x)’s in M for which G f is isomorphic to S 4 is precisely q4 − q3 . If q ≡ 3 (mod 4), then the remaining q3 f (x)’s in M have G f isomorphic to D 8 . If q ≡ 1 (mod 4), then q3 − q2 f (x)’s have G f isomorphic to D 8 and q2 f (x)’s have G f isomorphic to C 4 . 3. p = 2 Now we consider the case when p = 2. Since k( f (x) + d) = k( f (x)), if we set y = f (x) + d, then the minimal polynomial of x over k( y ) is
F ( X ) = X 4 + a X 3 + b X 2 + c X + y.
(3)
This polynomial is irreducible over k( y ), but may not be separable. Indeed by inspection we see it is separable only if a = 0 or c = 0. So in this case we only want to consider the q4 − q2 quartic polynomials f (x) for which this is true. Using the resolvent cubic as developed in [1, pp. 51–52], we find that for (3) we get
H ( X ) = X 3 + b X 2 + ac X + a2 y + c 2 .
(4)
We first consider the situation when a = 0. Then the constant term of H ( X ) is a prime element of k[ y ]. So, again by the rational root theorem we can easily conclude that H ( X ) is irreducible. So in the QG Theorem either m = 3 or m = 6. Now to have m = 3 means that if z is a root of H ( X ), then k( z)/k( y ) is a cyclic Galois extension of degree 3. Again by Theorem 1 of [3], this can only occur for a strict set of arithmetic conditions. Analysis of the equation z3 + bz2 + acz = (a2 y + c 2 ) shows that the pole of z is fully ramified in the extension k( z)/k( y ). From Theorem 2 of [3], this can only occur if 3 | (q − 1). Furthermore, by Theorem 1 of [3], a finite degree one prime of k( z) must also be fully ramified. This means for some r of k, there exist s and t = 0 of k so that we have a2 y + c 2 + r = t ( z + s)3 . But then we would have t ( z + s)3 = z3 + bz2 + acz + r. Equating coefficients we see this requires t = 1, s = b, s2 = ac and s3 = r. For any b = 0, this is possible as long as c = b2 a−1 . If c = b2 a−1 , then H ( X + b) = X 3 + (a2 y + c 2 + b3 ) would have z + b, u ( z + b) and u 2 ( z + b) as roots, with u a primitive third root of unity in k, making k( z)/k( y ) a cyclic Galois extension. Thus if q ≡ 1 (mod 3), here we get q(q − 1)2 f (x)’s with G f isomorphic to A 4 and q(q − 1)(q2 − q + 1) with G f isomorphic to S 4 . And, if q ≡ 2 (mod 3), G f cannot be isomorphic to A 4 and we q3 (q − 1) f (x)’s with G f isomorphic to S 4 . Now we consider those f (x)’s with a = 0. This means that c = 0 and we have for (3)
F (X) = X4 + b X2 + c X + y
(5)
H ( X ) = X 3 + b X 2 + c2 .
(6)
and (4) becomes
The coefficients of (6) are in k, so its splitting field has a Galois group which is cyclic with order m = 1, 2 or 3. Furthermore, the splitting field of H ( X ) will be the same as the splitting field of H 1 ( X ) = H ( X + b) = X 3 + b2 X + c 2 . Now m = 1 occurs when H 1 ( X ) factors completely over k. Since c = 0, none of the roots is 0 and since H 1 ( X ) is separable the roots must be distinct. Furthermore, the sum of the roots must be 0. Since b2 can be any element of k and c 2 can be any element of k∗ , any 3 distinct elements of k∗ which sum to 0 will be the roots of some polynomial of type H 1 ( X ). Thus here we will get q(q − 1)(q − 2)/6 f (x)’s for which m = 1 and G f is isomorphic to V .
R.C. Valentini / Finite Fields and Their Applications 26 (2014) 100–103
103
For m = 2, H 1 ( X ) factors as a linear times an irreducible quadratic. Since the quadratic term of H 1 ( X ) is 0, this means H 1 ( X ) = ( X + r )( X 2 + r X + s). Irreducibility of X 2 + r X + s guarantees r = 0 and s = 0, so that rs = 0. Again, since b2 and c 2 are arbitrary (with c = 0), any irreducible monic quadratic in k[ X ] will produce an H 1 ( X ) of this type. Since there are q(q − 1)/2 monic irreducible quadratics over k, we get q2 (q − 1)/2 f (x)’s with G f isomorphic to D 8 . (We note that for p = 2, by Theorem 1 of [3], C 4 cannot occur as a group of automorphisms of a rational function field and hence G f cannot be isomorphic to C 4 .) Finally, the remaining H 1 ( X )’s must be irreducible and give m = 3. Since there are q(q − 1) polynomials of type (6), by the process of elimination there are (q2 − 1)/3 irreducible H 1 ( X )’s. Hence, q(q2 − 1)/3 f (x)’s with G f isomorphic to A 4 here. So we have established the following theorem. Theorem 2. Let q = 2e , where e is a positive integer. Let k be the finite field with q elements. Let M s = { f (x) ∈ k[x] | f (x) = x4 + ax3 + bx2 + cx + d, with a = 0 or c = 0}. For f (x) ∈ M s , let G f be the Galois group of the Galois closure of the extension k(x)/k( f (x)). Then there are no f (x)’s in M s with G f isomorphic to C 4 . There are q(q − 1)(q − 2)/6 f (x)’s in M s with G f isomorphic to V and q2 (q − 1)/2 with G f isomorphic to D 8 . If q ≡ 1 (mod 3), then there are 2q(q − 1)(2q − 1)/3 f (x)’s in M s with G f isomorphic to A 4 and q(q − 1)(q2 − q + 1) with G f isomorphic to S 4 . If q ≡ 2 (mod 3), then there are q(q2 − 1)/3 f (x)’s with G f isomorphic to A 4 and q3 (q − 1) with G f isomorphic to S 4 . References [1] Irving Kaplansky, Fields and Rings, 2nd ed., The University of Chicago Press, Chicago, 1972. [2] David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd ed., John Wiley and Sons, Hoboken, NJ, 2004. [3] Robert C. Valentini, Manohar L. Madan, A Haupsatz of L.E. Dickson and Artin–Schreier extensions, J. Reine Angew. Math. 318 (1980) 156–177.
Contents lists available at ScienceDirect
Finite Fields and Their Applications www.elsevier.com/locate/ffa
Galois closures of quartic subfields of rational function fields Robert C. Valentini Department of Mathematics and Statistics, California State University, Long Beach, Long Beach, CA 90840, USA
a r t i c l e
i n f o
Article history: Received 21 August 2013 Accepted 21 November 2013 Available online 14 December 2013 Communicated by H. Stichtenoth MSC: 11R32 12F10 11T55
a b s t r a c t Let k be a finite field with q elements. Let f (x) ∈ k[x] be a monic quartic polynomial. Then k(x)/k( f (x)) is a field extension of degree 4. If the extension is separable, then the Galois group of the Galois closure is isomorphic to a transitive subgroup of the symmetric group on 4 letters. We determine the number of f (x)’s having a given subgroup as Galois group. © 2013 Elsevier Inc. All rights reserved.
Keywords: Galois group Quartic polynomial Rational function field
1. Introduction Let p be a prime. Let k be the finite field with q = p e elements. For a monic polynomial f (x) = x4 + ax3 + bx2 + cx + d ∈ k[x], set y = f (x). Then the field extension k(x)/k( y ) is of degree 4 and the minimal polynomial of x over k( y ) is F ( X ) = X 4 + a X 3 + b X 2 + c X + (d − y ). When F ( X ) is separable, then the Galois group G f of the Galois closure of k(x)/k( y ) is a transitive subgroup of S 4 (the symmetric group on 4 letters). Hence G f is isomorphic to S 4 , A 4 (the alternating group on 4 letters), D 8 (the dihedral group of order 8), V (the Klein 4 group), or C 4 (the cyclic group of order 4). In this paper we determine the number of f (x)’s producing each of these possibilities. It is classical that the Galois group of a quartic polynomial can largely be determined from the degree of the splitting field of a resolvent cubic. A succinct statement of the situation can be found in [1, pp. 51–52].
E-mail address: [email protected]. 1071-5797/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.ffa.2013.11.005
R.C. Valentini / Finite Fields and Their Applications 26 (2014) 100–103
101
QG Theorem. Let g be a separable irreducible quartic polynomial over a field K . Let m be the degree over K of the splitting field of a resolvent cubic of g. Let G be the Galois group of g over K . Then (1) (2) (3) (4)
If m = 6, G If m = 3, G If m = 1, G If m = 2, G
is isomorphic to is isomorphic to is isomorphic to is isomorphic to
S4. A4 . V. D 8 or C 4 .
We apply this theorem to analyze the situation for F ( X ) over k( y ). 2. p = 2 First we deal with the situation that p = 2. Then it is easily seen that F ( X ) is separable and we may apply the QG Theorem. We start by looking at those f (x)’s with a = d = 0. So
F ( X ) = X 4 + b X 2 + c X − y.
(1)
We employ the resolvent cubic as developed in [2, pp. 613–614], which for (1) is
H ( X ) = X 3 − 2b X 2 + b2 + 4 y X + c 2 .
(2)
The discriminant D of H ( X ) as given in [2, p. 614] is
D = − 4b3 + 27c 2 c 2 − 16b b3 + 9c 2 y − 128b2 y 2 − 256 y 3 and is the same as the discriminant of F ( X ). As a polynomial in y, D has degree 3, so cannot be a square in k( y ). Hence G f cannot be a subgroup of A 4 . So the only possibilities for G f are S 4 , D 8 , and C 4 . From the QG Theorem we see that we will get S 4 if H ( X ) is irreducible, which occurs precisely when it has no roots in k( y ). Applying the rational root theorem, we have H ( X ) is irreducible when c = 0. So there are q(q − 1) F ( X )’s of the type (1) for which H ( X ) is irreducible and G f is isomorphic to S 4 . When c = 0, F ( X ) = X 4 + b X 2 − y. If G f is isomorphic to C 4 , then G f is a group of automorphisms of the rational function field k(x). By Theorem 1 of [3], this is only possible for a strict set of arithmetic conditions. The factorization y = x4 + bx2 = x2 (x2 + b), shows that both the zero and the pole of y are ramified in the extension k(x)/k( y ). By Theorem 1 of [3] this can occur for C 4 only when both the zero and the pole of y are completely ramified. But this means b = 0. Then F ( X ) = X 4 − y. The roots of this equation are ±x, ±ux, where u is a primitive fourth root of unity. Thus if k(x)/k( y ) is to be Galois, k must contain the fourth roots of unity and hence 4 | (q − 1). Thus if q ≡ 1 (mod 4), there is exactly one F ( X ) of type (1) for which G f is isomorphic to C 4 and none if q ≡ 3 (mod 4). Since the only remaining possibility for G f is D 8 , by the process of elimination, if q ≡ 1 (mod 4), then there are q − 1F ( X )’s of type (1) for which G f is isomorphic to D 8 and q such F ( X )’s when q ≡ 3 (mod 4). Now for an arbitrary f (x) = x4 + ax3 + bx2 + cx + d the automorphism of k(x) sending x to x − a/4 sends f (x) to f (x − a/4) = x4 + β x2 + γ x + δ , where β = (−3a2 + b)/8, γ = (a3 − 4ab + 8c )/8, and δ = (−3a4 + 16a2 b − 64ac + 256d)/256. So k(x)/k( f (x)) is isomorphic to k(x)/k( f (x − a/4) − δ) and the Galois closures are isomorphic also. But f (x − a/4) − δ gives an F ( X ) of type (1). From the equations for β and γ , we see that for a fixed pair (a, d) as (b, c ) runs over all elements of k2 , (β, γ ) also runs over all elements of k2 . So for a fixed quartic with a = d = 0, there are q2 arbitrary f (x)’s such that f (x − a/4) − δ is equal to the fixed quartic. So we have the following theorem.
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Theorem 1. Let p = 2 be a prime. Let q = p e , where e is a positive integer. Let k be the finite field with q elements. Let M = { f (x) ∈ k[x] | f (x) is monic of degree 4}. For f (x) ∈ M, let G f be the Galois group of the Galois closure of the extension k(x)/k( f (x)). Then the number of f (x)’s in M for which G f is isomorphic to S 4 is precisely q4 − q3 . If q ≡ 3 (mod 4), then the remaining q3 f (x)’s in M have G f isomorphic to D 8 . If q ≡ 1 (mod 4), then q3 − q2 f (x)’s have G f isomorphic to D 8 and q2 f (x)’s have G f isomorphic to C 4 . 3. p = 2 Now we consider the case when p = 2. Since k( f (x) + d) = k( f (x)), if we set y = f (x) + d, then the minimal polynomial of x over k( y ) is
F ( X ) = X 4 + a X 3 + b X 2 + c X + y.
(3)
This polynomial is irreducible over k( y ), but may not be separable. Indeed by inspection we see it is separable only if a = 0 or c = 0. So in this case we only want to consider the q4 − q2 quartic polynomials f (x) for which this is true. Using the resolvent cubic as developed in [1, pp. 51–52], we find that for (3) we get
H ( X ) = X 3 + b X 2 + ac X + a2 y + c 2 .
(4)
We first consider the situation when a = 0. Then the constant term of H ( X ) is a prime element of k[ y ]. So, again by the rational root theorem we can easily conclude that H ( X ) is irreducible. So in the QG Theorem either m = 3 or m = 6. Now to have m = 3 means that if z is a root of H ( X ), then k( z)/k( y ) is a cyclic Galois extension of degree 3. Again by Theorem 1 of [3], this can only occur for a strict set of arithmetic conditions. Analysis of the equation z3 + bz2 + acz = (a2 y + c 2 ) shows that the pole of z is fully ramified in the extension k( z)/k( y ). From Theorem 2 of [3], this can only occur if 3 | (q − 1). Furthermore, by Theorem 1 of [3], a finite degree one prime of k( z) must also be fully ramified. This means for some r of k, there exist s and t = 0 of k so that we have a2 y + c 2 + r = t ( z + s)3 . But then we would have t ( z + s)3 = z3 + bz2 + acz + r. Equating coefficients we see this requires t = 1, s = b, s2 = ac and s3 = r. For any b = 0, this is possible as long as c = b2 a−1 . If c = b2 a−1 , then H ( X + b) = X 3 + (a2 y + c 2 + b3 ) would have z + b, u ( z + b) and u 2 ( z + b) as roots, with u a primitive third root of unity in k, making k( z)/k( y ) a cyclic Galois extension. Thus if q ≡ 1 (mod 3), here we get q(q − 1)2 f (x)’s with G f isomorphic to A 4 and q(q − 1)(q2 − q + 1) with G f isomorphic to S 4 . And, if q ≡ 2 (mod 3), G f cannot be isomorphic to A 4 and we q3 (q − 1) f (x)’s with G f isomorphic to S 4 . Now we consider those f (x)’s with a = 0. This means that c = 0 and we have for (3)
F (X) = X4 + b X2 + c X + y
(5)
H ( X ) = X 3 + b X 2 + c2 .
(6)
and (4) becomes
The coefficients of (6) are in k, so its splitting field has a Galois group which is cyclic with order m = 1, 2 or 3. Furthermore, the splitting field of H ( X ) will be the same as the splitting field of H 1 ( X ) = H ( X + b) = X 3 + b2 X + c 2 . Now m = 1 occurs when H 1 ( X ) factors completely over k. Since c = 0, none of the roots is 0 and since H 1 ( X ) is separable the roots must be distinct. Furthermore, the sum of the roots must be 0. Since b2 can be any element of k and c 2 can be any element of k∗ , any 3 distinct elements of k∗ which sum to 0 will be the roots of some polynomial of type H 1 ( X ). Thus here we will get q(q − 1)(q − 2)/6 f (x)’s for which m = 1 and G f is isomorphic to V .
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For m = 2, H 1 ( X ) factors as a linear times an irreducible quadratic. Since the quadratic term of H 1 ( X ) is 0, this means H 1 ( X ) = ( X + r )( X 2 + r X + s). Irreducibility of X 2 + r X + s guarantees r = 0 and s = 0, so that rs = 0. Again, since b2 and c 2 are arbitrary (with c = 0), any irreducible monic quadratic in k[ X ] will produce an H 1 ( X ) of this type. Since there are q(q − 1)/2 monic irreducible quadratics over k, we get q2 (q − 1)/2 f (x)’s with G f isomorphic to D 8 . (We note that for p = 2, by Theorem 1 of [3], C 4 cannot occur as a group of automorphisms of a rational function field and hence G f cannot be isomorphic to C 4 .) Finally, the remaining H 1 ( X )’s must be irreducible and give m = 3. Since there are q(q − 1) polynomials of type (6), by the process of elimination there are (q2 − 1)/3 irreducible H 1 ( X )’s. Hence, q(q2 − 1)/3 f (x)’s with G f isomorphic to A 4 here. So we have established the following theorem. Theorem 2. Let q = 2e , where e is a positive integer. Let k be the finite field with q elements. Let M s = { f (x) ∈ k[x] | f (x) = x4 + ax3 + bx2 + cx + d, with a = 0 or c = 0}. For f (x) ∈ M s , let G f be the Galois group of the Galois closure of the extension k(x)/k( f (x)). Then there are no f (x)’s in M s with G f isomorphic to C 4 . There are q(q − 1)(q − 2)/6 f (x)’s in M s with G f isomorphic to V and q2 (q − 1)/2 with G f isomorphic to D 8 . If q ≡ 1 (mod 3), then there are 2q(q − 1)(2q − 1)/3 f (x)’s in M s with G f isomorphic to A 4 and q(q − 1)(q2 − q + 1) with G f isomorphic to S 4 . If q ≡ 2 (mod 3), then there are q(q2 − 1)/3 f (x)’s with G f isomorphic to A 4 and q3 (q − 1) with G f isomorphic to S 4 . References [1] Irving Kaplansky, Fields and Rings, 2nd ed., The University of Chicago Press, Chicago, 1972. [2] David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd ed., John Wiley and Sons, Hoboken, NJ, 2004. [3] Robert C. Valentini, Manohar L. Madan, A Haupsatz of L.E. Dickson and Artin–Schreier extensions, J. Reine Angew. Math. 318 (1980) 156–177.