1 Introduction
In any book on field theory and Galois theory, we first encounter the primitive element theorem (see, for instance, [Reference Lang6, Theorem 4.6]) which states: ‘If
$L/K$
is a finite separable extension of fields such that K contains infinitely many elements, and
$\alpha , \beta \in L$
, then for all but finitely many
$u\in K$
, we have
$K(\alpha , \beta ) = K(\alpha +u\beta )$
. That is, for all but finitely many
$u\in K$
, the element
$\alpha +u\beta $
is a primitive element for
$K(\alpha , \beta )$
over K’.
Let K be a given field which contains infinitely many elements. If
$\alpha $
and
$\beta $
are algebraic and separable elements over K, we define
$$ \begin{align*} \xi_K(\alpha, \beta) = \left\{u\in K : K(\alpha, \beta) \ne K(\alpha+u\beta)\right\} \end{align*} $$
as the exceptional set. By the primitive element theorem,
$|\xi _K(\alpha , \beta )| < \infty $
. In the following theorem, we estimate the size of this set. We denote the degree of the extension field F over K by
$[F:K]$
.
Theorem 1.1. Let K be an arbitrary field which has infinitely many elements. Let
$\alpha $
and
$\beta $
be distinct algebraic and separable elements over K. If the degree of
$\beta $
is at least
$3$
and
${(\beta - \beta _1)}/{(\beta - \beta _2)}\not \in K$
for any two distinct Galois conjugates of
$\beta $
with
$\beta _i\ne \beta $
for
$i = 1,2$
, then
$$ \begin{align*} |\xi_K(\alpha, \beta)| \leq [K(\alpha): K]. \end{align*} $$
If
$L/K$
is a finite Galois extension of fields with Galois group G, then by the normal basis theorem, there exists
$\alpha \in L$
such that the set
$\{\sigma (\alpha ) : \sigma \in G\}$
forms a K-basis for L. Such an element
$\alpha \in L$
is called a normal basis generator for
$L/K$
. Also, for a given number field K, an algebraic element
$\beta $
over K is said to be totally real, if all its Galois conjugates are real.
Let K be a number field and
$f(x)\in K[x]$
be an irreducible polynomial of degree d. Let
$\Omega _f = \{\alpha _1, \ldots , \alpha _d\}$
be the set of all complex roots of
$f(x)$
. We say that
$\Omega _f$
is K-trivial, if
$c_1\alpha _1+\cdots +c_d\alpha _d = 0$
for some
$c_i\in K$
, implies
$c_1 = c_2 = \cdots = c_d$
. In 1977, V. A. Kurbatov [Reference Kurbatov5] proved: ‘Let
$L/K$
be a finite Galois extension of number fields of degree p, for some prime number
$p\geq 3$
with
$K\cap \mathbb {Q}(\zeta _p)=\mathbb {Q}$
. If
$L = K(\beta )$
for some
$\beta $
whose minimal polynomial is
$f(x)\in K[x]$
, then
$\Omega _f$
is K-trivial’. In 1997, Dixon [Reference Dixon1] proved: ‘Let
$L/K$
be a finite Galois extension of number fields with Galois group G and let
$L = K(\beta )$
. If
$f(x)\in K[X]$
is the minimal polynomial of
$\beta $
and G acts on
$\Omega _f\ 2$
-transitively, then
$\Omega _f$
is K-trivial’. We prove the following corollary.
Corollary 1.2. Let K be a given number field. Let
$\alpha $
and
$\beta $
be algebraic numbers over K such that
$\alpha , \beta \not \in K$
and
$\beta \not \in K(\alpha )$
. Suppose one of
$\alpha $
and
$\beta $
is of degree
$\geq 3$
.
-
(1) If
$K\subset \mathbb {R}$
, then
$|\xi _K(\alpha , \beta )| \leq \min \{[K(\alpha ):K], [K(\beta ):K]\}$
. -
(2) If
$K\not \subset \mathbb {R}$
and
$\beta $
is a totally real algebraic number of degree at least
$3$
, then
$$ \begin{align*} |\xi_K(\alpha, \beta)| \leq [K(\alpha):K]. \end{align*} $$
-
(3) If
$K\not \subset \mathbb {R}$
and
$\beta \in \bar {\mathbb {Q}}$
is such that
$\beta $
and at least two of its Galois conjugates are purely imaginary, then
$|\xi _K(\alpha , \beta )| \leq [K(\alpha ):K]$
. -
(4) If
$K(\beta )/K$
is a finite Galois extension of degree at least
$3$
and
$\beta $
is a normal basis generator of
$K(\beta )/K$
, then
$|\xi _K(\alpha , \beta )| \leq [K(\alpha ):K]$
. -
(5) If
$f(x)\in K[x]$
is the minimal polynomial of
$\beta $
and
$\Omega _f$
is K-trivial, then
$$ \begin{align*} |\xi_K(\alpha, \beta)| \leq [K(\alpha):K]. \end{align*} $$
It is easy to see that
$|\xi _K(\alpha , \beta )| \leq (d-1)^2$
, where
$d = [K(\alpha , \beta ): K]$
. In 2022, Dubickas [Reference Dubickas3] gave an argument for the upper bound
$|\xi _K(\alpha , \beta )| \leq e(e+1)/2$
, where
${e = d/p}$
and p is the least prime dividing d. For earlier results of this kind, we refer to [Reference Dubickas3]. Then, he proved
$|\xi _K(\alpha , \beta )| \leq \min \{[\mathbb {Q}(\alpha ):\mathbb {Q}], [\mathbb {Q}(\beta ):\mathbb {Q}]\}\leq d$
(as in Corollary 1.2(1), when
$K = \mathbb {Q}$
) and noted that the bound for
$|\xi _K(\alpha , \beta )|$
as a function of d is far from optimal. However, in the same paper, he also proved that the bound obtained in Corollary 1.2(1) cannot be improved, in general. This is applied to our situation in Corollaries 1.2 and 1.3.
Now, we consider K as a finite extension of
$\mathbb {Q}_p$
, the field of p-adic numbers. If
$\beta $
is an algebraic element over K of degree d, then the discriminant
$D(\beta )$
of
$\beta $
is defined by
$$ \begin{align*} D(\beta) = \prod_{1\leq i < j\leq d} (\beta_i - \beta_j)^2, \end{align*} $$
where
$\beta _1 = \beta $
and the
$\beta _j$
are the other Galois conjugates of
$\beta $
for
$j\geq 2$
. We say that the square root of
$D(\beta )$
is square-free if
$\sqrt {D(\beta )}$
does not contain a term
$(\beta _i - \beta _j)^2$
for any
$i < j$
.
Corollary 1.3. Let K be a finite extension over
$\mathbb {Q}_p$
, the field of p-adic numbers. Let
$\alpha $
and
$\beta $
be algebraic over K.
-
(1) If
$K(\beta )$
is a finite Galois extension over K of degree
$\geq 3$
and
$\beta $
is a normal basis generator of
$K(\beta )/K$
, then
$|\xi _K(\alpha , \beta )| \leq [K(\alpha ):K]$
. -
(2) If
$K(\beta )$
is a finite Galois extension over K of degree
$\geq 3$
and if the square root of the discriminant of
$\beta $
over K is square-free, then
$|\xi _K(\alpha , \beta )| \leq [K(\alpha ):K]$
. -
(3) Suppose L is the splitting field of
$\beta $
and
$L/K$
is unramified. Let
$\bar {L}$
and
$\bar {K}$
be the residue fields of L and K. If
$\beta $
is an integral element which is the pre-image of a normal basis generator of the residue field
$\bar {L}/\bar {K}$
of degree
$\geq 3$
, then
$$ \begin{align*} |\xi_K(\alpha, \beta)| \leq [K(\alpha):K]. \end{align*} $$
2 Preliminaries
We first prove some lemmas and then we prove the theorem and its corollaries. The first lemma provides a criterion for an element to be a primitive element.
Lemma 2.1 [Reference Dubickas3].
Let
$K/F$
be a finite separable extension over a field F and let L be the Galois closure of K whose Galois group is
$G = \mathrm {Gal}(L/F)$
. Let H be the subgroup of G which fixes K and let
$\{H, \sigma _1 H, \ldots , \sigma _{d-1}H\}$
be a complete set of left coset representatives of H where
$d = [K:F] = [G:H]$
. Then,
$\alpha \in K$
is a primitive element, that is,
$K = F(\alpha )$
if and only if
$\sigma _i(\alpha ) \ne \alpha $
for every
$i =1,2,\ldots , d-1$
.
Lemma 2.2. Let K be a field with infinitely many elements. Let
$\alpha _1, \alpha _2$
and
$\alpha _3$
be distinct algebraic and separable elements over K. Then, the following statements are equivalent:
-
(1)
${(\alpha _1-\alpha _2)}/{(\alpha _1 - \alpha _3)} \not \in K$
; -
(2) for any
$x_1, x_2, x_3\in K\backslash \{0\}$
with
$x_1+x_2+x_3 = 0$
, we have
$x_1\alpha _1+x_2\alpha _2+x_3\alpha _3 \ne 0$
.
Proof. Observe that
$$ \begin{align*} \frac{\alpha_1- \alpha_2}{\alpha_1-\alpha_3} = x\in K \iff (x-1)\alpha_1 +\alpha_2 - x\alpha_3 =0, \end{align*} $$
with
$x_1 = x-1$
,
$x_2 = 1$
and
$x_3 = -x$
satisfying
$x_1+x_2+x_3 =0$
. Note that
$x\ne 0$
as
${\alpha _1\ne \alpha _2}$
. Hence,
$x_1, x_2, x_3\in K\backslash \{0\}$
. Conversely, if
$x_1, x_2, x_3\in K\backslash \{0\}$
are such that
${x_1+x_2+x_3 =0}$
and
$x_1\alpha _1+x_2\alpha _2+x_3\alpha _3 = 0$
, then
$$ \begin{align*} 0 &= (-x_2-x_3)\alpha_1 +x_2\alpha_2+x_3\alpha_3\\ &= x_2(\alpha_2-\alpha_1)+x_3(\alpha_3-\alpha_1) \iff \frac{\alpha_1 - \alpha_2}{\alpha_1-\alpha_3} = -\frac{x_3}{x_2}\in K.\end{align*} $$
This proves the lemma.
The proof of the following lemma follows easily from [Reference Dubickas2, Theorem 4] (see also [Reference Dubickas3, Lemma 4]). Hence, we omit the proof.
Lemma 2.3 [Reference Dubickas2].
Let
$K\subset \mathbb {R}$
be a given number field and let
$\alpha _1, \alpha _2$
and
$\alpha _3$
be distinct algebraic numbers over K which are Galois conjugates of each other. If
$x_1, x_2$
and
$x_3$
are nonzero elements of K such that
$x_1+x_2+x_3 = 0$
, then
$x_1\alpha _1+x_2\alpha _2+x_3\alpha _3 \ne 0$
.
Lemma 2.4. Let K be a finite extension over
$\mathbb {Q}_p$
for some prime p. Let
$\alpha $
be an algebraic element over K of degree
$d\geq 3$
and let
$L/K$
be a finite Galois extension which contains
$\alpha $
. Suppose there are nonzero elements
$x_1, x_2, x_3 \in K$
with
$x_1+x_2+x_3 =0$
and distinct Galois conjugates
$\alpha _1 = \alpha , \alpha _2, \alpha _3$
of
$\alpha $
such that
$x_1\alpha _1+x_2\alpha _2+x_3\alpha _3 =0$
. Then, the square root of
$D(\alpha )$
is not square-free.
Proof. Assume that
$x_1\alpha _1+x_2\alpha _2+x_3\alpha _3 = 0$
for some nonzero elements
$x_i\in K$
and for some distinct Galois conjugates
$\alpha _1 = \alpha , \alpha _2$
and
$\alpha _3$
. We consider the valuation in L and denote it by
$v_L(\gamma )$
for
$\gamma \in L$
. Since
$x_1+x_2+x_3 =0$
, we must have
${v_L(x_i) = v_L(x_j)}$
for some
$i,j$
with
$1\leq i< j \leq 3$
. Without loss of generality, we assume that
${v_L(x_1) = v_L(x_2)}$
. Since
$$ \begin{align*} x_1\alpha_1+x_2\alpha_2+x_3\alpha_3 = 0 & \iff x_1\alpha_1 +x_2\alpha_2 + (-x_1-x_2) \alpha_3 = 0\\ & \iff \alpha_1 - \alpha_3 = -\frac{x_2}{x_1}(\alpha_2-\alpha_3), \end{align*} $$
it follows that
$v_L(\alpha _1-\alpha _3) = v_L(x_2/x_1) + v_L(\alpha _2-\alpha _3) = v_L(\alpha _2-\alpha _3)$
. Therefore,
${(\alpha _1 - \alpha _3)^2}$
divides
$\sqrt {D(\alpha )}$
. This proves the lemma.
Lemma 2.5 (Nakayama lemma, [Reference Jacobson4]).
Suppose M is a finitely generated module over a commutative ring R and
$J(R)$
is the intersection of all the maximal ideals of R. If
$M = J(R)M$
, then
$M =0$
.
Lemma 2.6. Let K be a finite extension of
$\mathbb {Q}_p$
for some prime p and let
$\beta $
be an integral element over K of degree
$\geq 3$
. Let L be the splitting field of
$\beta $
over K such that
$L/K$
is unramified. Let
$\bar {L}$
and
$\bar {K}$
be the residue fields of L and K, respectively. If
$\beta $
is a pre-image of a normal basis generator of
$\bar {L}/\bar {K}$
, then for any other distinct Galois conjugates
$\beta '$
and
$\beta "$
of
$\beta $
, if
$c_1\beta +c_2\beta '+c_3\beta " = 0$
for some
$c_1, c_2, c_3\in K$
, then
$c_1 = c_2= c_3 =0$
.
Proof. Put
$\beta = \beta _1$
and label the other given distinct conjugates
$\beta _2$
and
$\beta _3$
. Assume that
$c_1\beta _1+c_2\beta _2+c_3\beta _3 = 0$
for some
$c_1, c_2, c_3\in K$
, not all zero. Let G be the Galois group of the unramified extension
$L/K$
which is the splitting field of
$\beta = \beta _1$
. Let
$\mathcal {O}_K$
be the local ring of K and
$\mathfrak {m}_L$
be the unique maximal ideal in L. Let
$\sigma _1, \sigma _2$
and
$\sigma _3$
be the automorphisms of L such that
$\sigma _i(\beta ) = \beta _i$
for
$i = 1, 2, 3$
.
Since
$\beta = \beta _1$
is a pre-image of a normal basis generator
$\gamma $
of the residue field
$\bar {L}/\bar {K}$
, under reduction modulo
$\mathfrak {m}_L$
, we can write
$\beta _i +\mathfrak {m}_L = \bar {\beta }_i$
for
$i = 1,2,3$
. Under the reduction modulo
$\mathfrak {m}_L$
, the automorphisms
$\sigma _i$
are automorphisms
$\bar {\sigma }_i$
of
$\bar {L}$
over
$\bar {K}$
. Note that
$\mathfrak {m}_L\cap \mathcal {O}_K = \mathfrak {m}_K$
is the maximal ideal of K.
Since the
$\bar {\sigma }_i$
are automorphisms of the Galois group
$\bar {L}/\bar {K}$
and
$\overline {\sigma _1(\beta )} = \gamma $
, the normal basis generator of
$\bar {L}/\bar {K}$
along with two other Galois conjugates of
$\gamma $
are given by
$\overline {\sigma _i(\beta )} = \gamma _i$
for
$i=2,3$
. Since
$L/K$
is an unramified extension, we have
$[L:K] = [\bar {L}:\bar {K}]$
and hence
$\bar {\sigma }_2 \ne \bar {\sigma }_3$
. This in turn implies that
$\gamma _2\ne \gamma _3$
and both are distinct from
$\gamma $
. We denote the Galois group of
$\bar {L}/\bar {K}$
by
$\bar {G}$
.
Consider the
$\mathcal {O}_K$
-module
$M = \mathcal {O}_K[G] \cdot (c_1\sigma _1+c_2\sigma _2+c_3\sigma _3)$
. Then,
$$ \begin{align} M/\mathfrak{m}_LM = \left(\mathcal{O}_K/\mathfrak{m}_K\right)[\bar{G}] \cdot \left[\overline{c_1\sigma_1+ c_2\sigma_2+c_3\sigma_3}\right] = \bar{K}[\bar{G}]\cdot (\bar{c}_1\bar{\sigma}_1+\bar{c}_2\bar{\sigma}_2+\bar{c}_3\bar{\sigma}_3). \end{align} $$
Since
$c_1\beta _1+c_2\beta _2+c_3\beta _3 = 0$
, we conclude that
$$ \begin{align*} \bar{c}_1\gamma +\bar{c}_2\gamma_2+\bar{c}_3\gamma_3 = \bar{0}. \end{align*} $$
Since
$\gamma $
is a normal basis generator, we must have
$\bar {c}_i = \bar {0}$
for
$i=1,2,3$
. Together with (2.1), this implies that
$$ \begin{align*} M/\mathfrak{m}_LM = \bar{0} \iff M = \mathfrak{m}_LM. \end{align*} $$
Since
$\mathfrak {m}_L$
is the only maximal ideal in L, by Nakayama’s Lemma 2.5, we conclude that
$M =0$
. However,
$\mathcal {O}_K[G]$
is a free module over
$\mathcal {O}_K$
with generators as their elements, so we conclude that
$c_1 = c_2 = c_3 = 0$
.
3 Proof of Theorem 1.1
Let K be an infinite field. We wish to prove
$|\xi _K(\alpha , \beta )| \leq [K(\alpha ):K]$
. In fact, we prove that whenever
$u, v\in \xi _K(\alpha , \beta )$
with
$u\ne v$
in K, there are two distinct Galois conjugates of
$\alpha $
which gives the desired result.
Let
$u, v\in \xi _K(\alpha , \beta )$
with
$u\ne v$
. Since
$\alpha +u\beta $
and
$\alpha +v\beta $
are not primitive elements of
$L = K(\alpha , \beta )$
, by Lemma 2.1, there exist embeddings
$\tau _1, \tau _2 : L \to \mathbb {C}$
such that
${\tau _1(\alpha +u\beta ) = \alpha +u\beta }$
and
$\tau _2(\alpha +v\beta ) = \alpha +v\beta $
.
Note that if
$\tau _i(\beta ) = \beta $
(or
$\tau _i(\alpha ) = \alpha $
), then these equations give
$\tau _i(\alpha ) = \alpha $
(respectively,
$\tau _i(\beta ) = \beta $
) as
$u, v$
are elements of K. It follows that
$\tau _i$
must be the identity, which is in contradiction to Lemma 2.1. Therefore,
$\tau _i(\alpha ) \ne \alpha $
and
$\tau _i(\beta ) \ne \beta $
. Thus,
$$ \begin{align*} \tau_1(\alpha) + u\tau_1(\beta) = \alpha+u\beta \quad\mbox{and}\quad \tau_2(\alpha) + v\tau_2(\beta) = \alpha+v\beta, \end{align*} $$
with
$\tau _i(\alpha ) \ne \alpha $
and
$\tau _i(\beta )\ne \beta $
for
$i = 1,2$
. Since
$\tau _1(\alpha )$
and
$\tau _2(\alpha )$
are Galois conjugates distinct from
$\alpha $
, we claim that
$\tau _1(\alpha ) \ne \tau _2(\alpha )$
(and the proof of the assertion follows).
To the contrary, assume that
$\tau _1(\alpha ) = \tau _2(\alpha )$
. Since
$u\ne v$
, we must have
$\tau _1(\beta ) \ne \tau _2(\beta )$
. From these equations,
$$ \begin{align*} u(\beta - \tau_1(\beta)) = \tau_1(\alpha) - \alpha = \tau_2(\alpha) - \alpha = v(\beta - \tau_2(\beta)) \end{align*} $$
and hence
$$ \begin{align*}\frac{\beta - \tau_1(\beta)}{\beta-\tau_2(\beta)} = \frac vu\in K,\end{align*} $$
which is a contradiction. This proves the theorem.
4 Proof of Corollary 1.2
Proof of Corollary 1.2(1).
Since either
$\alpha $
or
$\beta $
is of degree
$\geq 3$
, we can assume that
$\beta $
is of degree
$\geq 3$
. Since
$K \subset \mathbb {R}$
, by Lemmas 2.3 and 2.2, we conclude that
$$ \begin{align*} \frac{\beta - \beta_1}{\beta-\beta_2}\not\in K \end{align*} $$
for any distinct conjugates
$\beta _i \ne \beta $
(
$i=1,2$
). Therefore, by Theorem 1.1,
$$ \begin{align*} |\xi_K(\alpha, \beta)| \leq [K(\alpha):K]. \end{align*} $$
If
$[K(\alpha ):K] \geq [K(\beta ):K]\geq 3$
, we swap the roles of
$\beta $
and
$\alpha $
giving
$|\xi _K(\alpha , \beta )| \leq [K(\beta ):K]$
. This proves part (1) of the corollary.
Proof of Corollary 1.2(2).
If
$K\subset \mathbb {R}$
, the result follows from Corollary 1.2(1). Hence, we assume that
$K\subset \mathbb {C}$
.
If
$x_1, x_2, x_3 \in \mathbb {C}$
, then we write
$x_j= a_{j1}+ ia_{j2}$
for
$j = 1,2,3$
and
$a_{rs}\in \mathbb {R}\cap K$
. Since
$x_1+x_2+x_3 = 0$
, we have
$$ \begin{align*} a_{11}+a_{21}+a_{31} =0 \quad\mbox{and}\quad a_{12}+a_{22}+a_{32} = 0. \end{align*} $$
Since
$\beta $
is of degree
$\geq 3$
and totally real, every Galois conjugate of
$\beta $
is real. Therefore, if
$\beta _1$
and
$\beta _2$
are any distinct Galois conjugates of
$\beta $
with
$\beta _i \ne \beta $
, then by Lemma 2.3,
$$ \begin{align*} a_{11}\beta+a_{21}\beta_1+a_{31}\beta_2 \ne 0 \quad\mbox{and}\quad a_{12}\beta+a_{22}\beta_1+a_{32}\beta_2\ne 0. \end{align*} $$
Therefore,
$x_1\beta +x_2\beta _1+x_3\beta _2\ne 0$
. Part (2) of the corollary follows by Lemma 2.2 and Theorem 1.1.
Proof of Corollary 1.2(3).
Given that
$\beta $
and two of its Galois conjugates are purely imaginary, we have distinct conjugates
$\beta = i\beta '$
,
$\beta _1= i\beta _1'$
and
$\beta _2= i\beta _2'$
with
$\beta ', \beta _1'$
and
$\beta _2'$
distinct real numbers. With the same notation in the proof of Corollary 1.2(2),
$$ \begin{align*} i(a_{11}\beta'+a_{21}\beta_1'+a_{31}\beta_2') \ne 0 \quad\mbox{and}\quad -(a_{12}\beta'+a_{22}\beta_1'+a_{32}\beta_2')\ne 0, \end{align*} $$
by Lemma 2.3. Therefore,
$$ \begin{align*} a_{11}\beta'+a_{21}\beta_1'+a_{31}\beta_2' \ne 0 \quad\mbox{and}\quad a_{12}\beta'+a_{22}\beta_1'+a_{32}\beta_2'\ne 0, \end{align*} $$
and hence
$x_1\beta +x_2\beta _1+x_3\beta _2\ne 0$
. Part (3) of the corollary follows by Lemma 2.2 and Theorem 1.1.
Proof of Corollary 1.2(4).
Let
$K(\beta )$
be a finite Galois extension over K such that
$\beta $
is a normal integral basis generator over K of degree at least
$3$
. If the Galois group of
$K(\beta )/K$
is G, then the set
$\{\sigma (\beta ) : \sigma \in G\}$
forms a K-basis and, in particular, it is K-linearly independent. Therefore, if
$\beta _1$
and
$\beta _2$
are any distinct Galois conjugates of
$\beta $
such that
$\beta _i \ne \beta $
, then
$x_1\beta +x_2\beta _1+x_3\beta _2 \ne 0$
for any nonzero tuple
$(x_1, x_2, x_3) \in K^3$
. By Lemma 2.2, we have
${(\beta - \beta _1)}/{(\beta - \beta _2)} \not \in K$
. The result follows from Theorem 1.1.
Proof of Corollary 1.2(5).
Let f be the minimal polynomial of
$\beta $
and assume that
$\Omega _f$
, the set of all roots of f over K, is K-trivial. Since
$\Omega _f$
is K-trivial and the degree of
$\beta $
is at least
$3$
, if
$c_1\beta +c_2\beta _2+c_3\beta _3 = 0$
for some
$c_i\in K$
and for any distinct conjugates
$\beta _1\ne \beta _2$
different from
$\beta $
, then
$c_1 = c_2 = c_3 =0$
. Therefore, for any nonzero
$c_1, c_2, c_3\in K$
with
$c_1+c_2+c_3 = 0$
, we must have
$c_1\beta +c_2\beta _2+c_3\beta _3 \ne 0$
. The result now follows from Theorem 1.1.
5 Proof of Corollary 1.3
Let
$K(\beta )$
be a finite Galois extension over K. Since
$\beta $
is a normal basis generator, Corollary 1.3(1) follows by the same proof as Corollary 1.2(4).
To prove Corollary 1.3(2), we can assume that
$\beta $
is not a normal basis generator of
$K(\beta )/K$
. Since the square-root of the discriminant of
$\beta $
in
$K(\beta )$
is square-free, by Lemmas 2.4 and 2.2, the element
$(\beta -\beta _1)/(\beta -\beta _2)\not \in K$
for any distinct Galois conjugates
$\beta _i \ne \beta $
. Corollary 1.3(2) now follows from Theorem 1.1.
Since
$\beta $
is a pre-image of a normal basis generator of
$\bar {L}/\bar {K}$
, by Lemma 2.6, we see that
$c_1\beta +c_2\beta _1+c_3\beta _3 \ne 0$
whenever one of the
$c_i$
is nonzero. Therefore, Corollary 1.3(3) follows from Lemma 2.2 and Theorem 1.1.
Acknowledgement
We are thankful to the referee for going through the manuscript and suggesting changes that improved the presentation. This work was finished when we were visiting the Institute of Mathematical Sciences (IMSc), Chennai. We are thankful to IMSc for the wonderful research atmosphere and their hospitality.
