On the monogenity of pure quartic relative extensions of Q(i)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {Q}}}(i)$$\end{document}

We consider pure quartic relative extensions of the number field Q(i)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {Q}}}(i)$$\end{document} of type K=Q(a+bi4)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K={{\mathbb {Q}}}(\root 4 \of {a+bi})$$\end{document}, where a,b∈Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a,b\in {{\mathbb {Z}}}$$\end{document} and b≠0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$b\ne 0$$\end{document}, such that a+bi∈Z[i]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a+bi\in {{\mathbb {Z}}}[i]$$\end{document} is square-free. We describe integral bases of these fields. The index form equation is reduced to a relative cubic Thue equation over Q(i)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {Q}}}(i)$$\end{document} and some corresponding quadratic form equations. We consider monogenity of K and relative monogenity of K over Q(i)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb {Q}}}(i)$$\end{document}. We shall show how our former method based on the factors of the index form can be used in the relative case to exclude relative monogenity in some cases.


Monogenity: basic concepts
Let ϑ be an algebraic integer of degree n, set K = Q(ϑ).Denote by Z K the ring of integers of K and by D K the discriminant of K.The number field K is called monogenic if there exists α ∈ Z K such that (1, α, . . ., α n−1 ) is an integral basis, called power integral basis (for a detailed discussion of monogenity and power integral bases of number fields see [3]).
The index of a primitive element α ∈ Z K is defined as α generates a power integral basis if and only if I(α) = 1.Denoting by α (j) (j = 1, . . ., n) the conjugates of α, we have I. Gaál and L. Remete for a primitive element α ∈ Z K .If α, β ∈ Z K and β = ±α + a with a ∈ Z, then α and β are called equivalent.Equivalent algebraic integers have the same indices.
If K contains a proper subfield M , then for a primitive algebraic integer α in K we have where is called the relative index of α over M (cf.[3]).α, β ∈ Z K are called relative equivalent if α = εβ + a where ε is a unit in M and a ∈ Z M .Relative equivalent algebraic integers have the same relative indices.
To determine all inequivalent generators of power integral bases is a complicated problem, leading to the resolution of index form equations.
There are efficient methods for the resolution of index form equations in cubic and quartic number fields.However, this problem is only partially solved for higher degree fields, since in those cases the index form equation becomes much more complicated (cf.[3]).

The purposes of the present paper
In the present paper we consider pure quartic relative extensions of the number field M = Q(i).These fields are of type K = Q( 4√ a + bi) where a, b ∈ Z and b = 0, such that a + bi is not a square in Z M .We describe the integral bases of these fields and characterize their absolute and relative monogenity for the case that a + bi is square-free.
Remark that formerly in [7] we considered monogenity of number fields K = Q(i, 4  √ m).Those are also octic fields containig Q(i) as a subfield, but those fields are composites of Q(i) and Q( 4√ m) allowing a much easier investigation.In the present paper we study quartic relative extensions of Q(i), which are not composites of Q(i) with a quartic field.
To construct an integral basis in K = Q( 4 √ a + bi) we shall apply an extension of Lemma 2.17 of Narkiewicz [8] stating that if an algebraic integer is defined by a minimal polynomial which is p-Eisenstein with respect to the prime p, then its index is not divisible by p.
We shall use the result of Gaál and Pohst [4], see also Chapter 14.1 of Gaál [3], reducing the index form equations in relative quartic extensions to a cubic relative Thue equation and to some corresponding relative quadratic form equations, similarly as in [5] in the absolute case.
Further, in our proofs we shall give a relative analogue of another method, based on the factors of the index form.If the index form as a polynomial with integer coefficients in x 2 , . . ., x n has factors F 1 , F 2 (also polynomials with integer coefficients in x 2 , . . ., x n ) and there exists an element of index 1 (a generator of power integral basis), then there exist y 2 , . . ., y n ∈ Z such that However, in some cases it can be shown that e.g.F 1 − aF 2  2 is divisible by some constant b (a, b ∈ Z\{0}).If b does not divide ±1−a with neither of the possible choices of the sign, then it contradicts the existence of a generator of a power integral basis with coordinates y 2 , . . ., y n ∈ Z in the integral basis.This method was used e.g. in [6,7] to exclude monogenity of the corresponding fields in the absolute case.In our present paper we show how a relative analogue of this method can be used to exclude the existence of relative power integral bases in relative extensions.This version of the method is used here for the first time.We believe it will have several further applications in the study of monogenity of higher degree number fields.

Pure quartic extensions of M = Q(i): preliminaries
We shall consider number fields of type K = Q(α) with α = 4 √ a + bi, where a, b ∈ Z, b = 0 and a + bi is square-free in the ring of Gaussian integers Z and the conjugates of α are In the proof of our theorem on the integral basis of the number field K we need the following Lemma.This is a slight extension of Lemma 2.17 [8] to the relative case, when the ground field is M instead of Q.Our proof strictly follows the arguments of Lemma 2.17 [8].The same arguments seem to remain valid also for any ground field having unique factorization in its ring of integers.Lemma 1.Let M = Q(i) and K = M (α), where α is an algebraic integer.If the minimal polynomial over M of α is π-Eisenstein for a prime π ∈ Z M , i.e. it has the form with a 0 , a 1 , . . ., a n−1 divisible by π and π 2 a 0 , then there is no algebraic integer Proof.The π-Eisenstein property of the minimal polynomial of α over M implies that α n /π is an algebraic integer, and π 2 N K/M (α) = a 0 .Assume that there exists an algebraic integer ξ Let j be the minimal index with π β j .Then Therefore, This implies We shall also use the result of Gaál and Pohst [4], see also Theorem 14.1 of Gaál [3].We briefly detail the lemma here, because it is an essential tool for the proof of Theorem 4. Let K be a quartic extension of M , generated by an α with relative minimal polynomial f Denote by n the relative index of α over M .
We can represent any ϑ ∈ Z K in the form where A, X, Y, Z ∈ Z M and d ∈ Z is a common denominator.Then are ternary quadratic forms over Z M .
Lemma 2 (Gaál and Pohst [4]).If ϑ of (2) generates a power integral basis of K over M , then there is a unit In our case a 1 = a 2 = a 3 = 0 and a 4 = −(a + bi).

Integral bases
Let K = Q(α) where α = 4 √ a + bi, b = 0 and a + bi is square-free in Z M , as above.

Theorem 3. An integral basis of K is given by the following table:
Integral basis where D K/M is the relative discriminant of K over M (see [8] Chapter IV., Proposition 4.15).The minimal polynomial of α over M = Q(i) is X 4 − (a + bi) with discriminant 4 4 (a + bi) 3 .Since we assumed a + bi ∈ Z[i] to be square-free, for any prime π of Z M dividing a + bi the minimal polynomial of α is π-Eisenstein.By Lemma 1, this implies that for any prime π | a + bi, the basis (1, α, α 2 , α 3 ) of K over M is π-maximal, i.e. (a + bi) 3 so the only prime which may divide the index of α is 2.
If both a and b are odd, then 1 + i | a + bi, so by Lemma 1, the basis (1, α, α 2 , α 3 ) of K over M is 1 + i-maximal and 2 does not divide the relative index of α over Q(i).In this case (1, α, α 2 , α 3 ) is a relative power integral basis of K over M and therefore If a is even, and b is odd, then the minimal polynomial of α − 1, . We obtain again that (1, α, α 2 , α 3 ) is a relative power integral basis of K over M and therefore 1, α, α 2 , α 3 , i, iα, iα 2 , iα 3   is an integral basis of K over Q.
The interesting case is when a is odd and b is even.(They can not be both even, because a + bi is square-free.)This case splits into 6 subcases.The integral bases are listed in the table of the Theorem, as cases 2-7.
In these cases we showed, that the elements listed in the corresponding bases are always algebraic integers.These cases are very similar, we detail only one of them.
If a = 8k + 1 and b = 8l + 4, then 4 is an algebraic integer, because it is a root of the integer polynomial Furthermore, we checked that the given bases are 2-maximal by calculating the minimal polynomial of each element of the form where B is the basis given in the table, λ i ∈ {0, 1}, (i = 1, . . ., 8), a = 8k + r, b = 8l + s, (k, l = 0, . . ., 255) with the corresponding remainders r, s modulo 8.For example, if a = 8k+1 and b = 8l+4, then by the theorem on symmetric polynomials, the minimal polynomial of is a polynomial of the form , We did not find any algebraic integers in this form, which proves the 2-maximality.
Remark.We do not use the assumption b = 0 in the proof of Theorem 3, so the given integral bases are valid also for the case b = 0 and for K = Q(α, i).Cases 2. and 7. extends the results of [7] to the case m ≡ 1 (mod 4).

Monogenity, relative monogenity
We investigate the relative and absolute monogenity of the field K in the seven distinct cases corresponding to the table of Theorem 3.

Remarks:
• In the cases ( 6) and ( 7) we conjecture that K is not absolutely monogenic.• The absolute monogenity of K implies the relative monogenity of K over M , but not conversely.Therefore, the cases (2), ( 3), ( 4) and ( 5) are interesting in the sense that in these cases K is absolutely monogenic if and only if it is relatively monogenic over M .
Proof of Theorem 4. In all cases we will use Lemma 2 with a 1 = a 2 = a 3 = 0 and a 4 = −(a + bi).So to check whether some algebraic integer ϑ ∈ K of the form generates a power integral basis, we have to solve the equations where ε is a unit in Z[i], and γ ∈ Z[i] is of norm d 12 /I K/M (α).We intend to use also the representation of ϑ ∈ Z K of the form where (1, B 2 , B 3 , B 4 , B 5 , B 6 , B 7 , B 8 ) is the integral basis given in Theorem 3. Considering these integral bases, we can see that for case (1) we have d = 1, for cases (3),( 5) and ( 6) we have d = 2 and for cases (2),( 4) and (7) we have d = 4. Furthermore, if S is the transition matrix form the basis (1, α, α 2 , α 3 , i, iα, iα 2 , iα 3 ) to the basis (1, B 2 , B 3 , B 4 , B 5 , B 6 , B 7 , B 8 ), then , and therefore the square of the relative index of α is where 8 is the discriminant of the minimal polynomial f (x) of α and we applied the well-known fact D(f ) = I(α) 2 • D K .Since the discriminant of the basis (1, α, α 2 , α 3 , i, iα, iα 2 , iα 3 ) is 2 24 (a 2 + b 2 ), which is equal to D(f )/b 8 , and then we have .
Proof of (1).In the first case (1, α, α 2 , α 3 ) is a relative power integral basis of K over M .The relative index of α is I K/M (α) = 1 and we can choose d = 1, so γ is also a unit and the right hand side of equation ( 6) is ε.This equation implies N M/Q (U ) = 1, V = 0. So, by Eq. ( 7), X and Z are coprime Gaussian integers, and therefore by Eq. ( 8) either X • Z = 0 or both X and Z are associates of squares in Z[i].
If X = 0, then both a + bi and Z are units, i.e. a = 0, b = ±1.In this case α generates an absolute power integral basis, not only a relative power integral basis in K.We note that this field K is the algebraic number field generated by the 16th roots of unity.
If Z = 0, then Y = 0 and X is a unit ε X .This yields the solutions ϑ = A + ε X α, which are relative equivalent to α, so they also generates a relative power integral basis in K over M .This ϑ also generates an absolute power integral basis if b = ±1, but otherwise we conjecture that the absolute index of ϑ is greater than 1. If is a unit (with either + or −).By considering the real and the imaginary parts of this expression, we obtain the system of equations where either j = ±1, k = 0 or j = 0, k = ±1.If s 1 , s 2 , t 1 , t 2 are rational integers, such that a and b are rational integer solutions of the system of linear equations above, then ϑ generates a relative power integral basis in K corresponding to the current parameters a and b.
For certain parameters a, b there exist trivial solutions.For example, if Z is a unit, i.e. t 2 1 + t 2 2 = 1, then then the algebraic integer generates a relative power integral basis in K over M .Moreover, there exist infinitely many parameters a, b for which the field K is monogenic.The simplest example is when we set b = ±1, and then the absolute index of α is also 1, because the discriminant 2 24 (a 2 + 1) of f is equal to the discriminant of K.
We can go further, there exist infinitely many paramters a, b for which K admits at least 2 inequivalent generators of power integral bases.Let q be an arbitrary rational integer, a = q 4 and b = −1, then the minimal polynomial of with discriminant 2 24 (q 8 + 1) = 2 24 (a 2 + b 2 ) = D K , so β also generates a power integral basis in K.
In order to find all monogenic fields K, and all generators of power integral bases, we have to find all solutions of ( 14) or (15), and then determine the imaginary part of A, such that the discriminant of ϑ (which is a polynomial in (A)) is equal to the discriminant of K.
Proof of (2).In the second case the relative index of α is I K/M (α) = 2 8 and we can use the common denominator d = 4, so γ is an algebraic integer of norm 4 12 /2 8 = 2 16 .Up to associates, we can choose γ = 256 and then equation ( 6) is of the form On the other hand, Eqs.(10)-(13) give Substituting these expressions into the Eqs.( 6), ( 7) and ( 8), we can see that u = U/4, v = V /2 and (u 2 + (a + bi)v 2 )/4 are Gaussian integers, so Eq. ( 6) can be written as and F 2 = u.Since F 1 • F 2 = ε, then both factors have to be a unit, such that a + bi is a divisor of 4F 1 − F 2 2 .To sum up, if the relative index of ϑ is 1, then a + bi divides 3, 5, 1 − 4i or 1 + 4i.With a ≡ 1 (mod 8) and b ≡ 0 (mod 8), b = 0 it is not possible, so there is no such algebraic integer ϑ, and therefore K is not relatively monogenic over M , and also not monogenic over Q.
Proof of (3).In this case the relative index of α is I K/M (α) = 2 4 and we can use the common denominator d = 2, so γ is an algebraic integer of norm 2 12 /2 4 = 2 8 .Up to associates, the only Gaussian integer of norm 256 is γ = 16, so Eq. ( 6) is of the form Substituting these expressions into the Eqs.( 6), ( 7) and ( 8), we can see that u = U/2, v = V and (u 2 + (a + bi)v 2 )/2 are Gaussian integers, so Eq. ( 6) is equivalent to , where F 1 and F 2 are units.With a ≡ 1, 5 (mod 8) and b ≡ 2, 6 (mod 8) it is possible only if F 1 = i, a = 1 and b = ±2.In these cases K is indeed monogenic.For example In both cases, the discriminant of the polynomials are 2 16 • 5 3 = D K , so β generates a power integral basis in K.
For other values of a and b, K is not relatively monogenic over M , and therefore, K is not monogenic over Q.
Proof of (4).It is almost the same as Case (2), except the expressions for A, X, Y and Z.The relative index of α over M is I K/M (α) = 2 8 and d = 4. Equation ( 6) is of the form Substituting expressions into the Eqs.( 6), ( 7) and ( 8) we obtain the same equations and conditions as in Case (2).If the relative index of ϑ is 1, then a + bi divides 3, 5, 1 + 4i or 1 − 4i.With a ≡ 1 (mod 8) and b ≡ 4 (mod 8), it is possible only if a = 1 and b = ±4.In these cases K is indeed monogenic.with the same discriminant, so K is monogenic in this case too.
For other values of a and b, K is not relatively monogenic over M , and therefore, K is not monogenic over Q.
Proof of (5).This case is similar to Case 3. The relative index of α is I K/M (α) = 2 4 and we can use the common denominator d = 2, so Eq. ( 6) is of the form Substituting these expressions into the Eqs.( 6), (7) and (8) we obtain that u = U/4 and v = V are Gaussian integers.Then Eq. ( 6) is equivalent to Thus a + bi divides F 1 − 4F 2 2 , i.e. a + bi divides 3, 5, 1 − 4i or 1 + 4i.With a ≡ 3, 7 (mod 8), b ≡ 0, 4 (mod 8) and b = 0, it is possible only if a = −1 and b = ±4, and therefore F 1 = ±i.However, if b = ±4, then the imaginary part of F 1 = 4u 2 + (a + bi)v 2 is even, so there is no solution of the equation above, K is not relatively monogenic over M , and therefore, K is not monogenic over Q.
Proof of (7).This is a slightly more interesting case.The relative index of α is I K/M (α) = 2 3 and we can use the common denominator d = 2, so Eq. ( 6) is of the form