1. Introduction and statement of results
An algebraic number field K having ring of integers
$\mathbb {Z}_K$
is said to be monogenic if there exists an algebraic integer
$\beta \in K$
such that
$\mathbb {Z}_K=\mathbb {Z}[\beta ]$
. Monogenic number fields play a significant role in the study of algebraic number fields due to their simplified structure and computational advantages. One interesting problem in this direction is to study the monogenity of polynomials. A polynomial
$f(x)\in \mathbb {Z}[x]$
having a root
$\theta $
is said to be monogenic if it is irreducible over
$\mathbb {Q}$
and the ring of algebraic integers of
$\mathbb {Q}(\theta )$
is
$\mathbb {Z}[\theta ]$
. Some known families of monogenic polynomials are
$x^2-a\in \mathbb {Z}[x],$
where
$a\equiv 2,3\ \mod 4$
,
$x^3-b,$
where
$b\ne \pm 1$
is a squarefree integer, and cyclotomic polynomials. Determining infinite families of higher degree monogenic polynomials can be difficult and much research has been done to identify such families [Reference Jhorar and Khanduja4–Reference Kaur and Kumar7]. Despite the difficulty in determining higher degree monogenic polynomials, we are inspired by [Reference Gaál2, Reference Harrington and Jones3, Reference Kaur, Kumar and Remete8, Reference Sharma, Sharma and Laishram11] concerning the monogenity of the composition of polynomials.
Let
$f(x)$
and
$g(x)$
be two monic polynomials belonging to
$\mathbb {Z}[x]$
such that
$f(g(x))$
is irreducible over
$\mathbb {Q}$
. Let
$\alpha $
be a root of
$f(g(x))$
. Then, we have a tower of fields
$$ \begin{align*}\mathbb{Q}\rightarrow \mathbb{Q}(\beta)\rightarrow \mathbb{Q}(\alpha),\end{align*} $$
where
$\beta =g(\alpha )$
. We focus on the monogenity of the number fields of this tower. We give necessary and sufficient conditions for the composition
$f(g(x))$
to be monogenic and a further sufficient condition. Finally, we construct some towers of monogenic number fields.
Throughout the paper,
$D(f)$
and
$(f\circ g)(x)$
will denote the discriminant of
$f(x)\in \mathbb {Z}[x]$
and the composition of
$f(x)$
and
$g(x)$
, respectively. The discriminant and the ring of integers of an algebraic number field K will be denoted by
$d_K$
and
$\mathbb {Z}_K$
, respectively. For a field extension
$K/F$
and
$\gamma \in K$
, the symbol
$N_{K/F}(\gamma )$
denotes the norm of
$\gamma $
over F. The index of a monic irreducible polynomial
$f(x)\in \mathbb {Z}[x]$
having root
$\theta $
is the group index
$[\mathbb {Z}_{\mathbb {Q}(\theta )}:\mathbb {Z}[\theta ]]$
.
Theorem 1.1. Let
$f(x)$
and
$g(x)$
belonging to
$\mathbb {Z}[x]$
be two monic polynomials and
$\alpha $
be a root of
$(f\circ g) (x)$
. Then,
$(f\circ g)(x)$
is monogenic if and only if
-
(1)
$f(x)$
is monogenic and -
(2)
$\mathbb {Z}_L=\mathbb {Z}_K[\alpha ]$
, where
$L=\mathbb {Q}(\alpha )$
and
$K=\mathbb {Q}(g(\alpha ))$
.
Clearly, the theorem states that if
$L= \mathbb {Q}(\alpha )$
and
$ K=\mathbb {Q}(\beta )$
with
$\beta =g(\alpha )$
, then
$[\mathbb {Z}_L:\mathbb {Z}[\alpha ]]=1$
if and only if
$[\mathbb {Z}_K:\mathbb {Z}[\beta ]]=1$
and
$[\mathbb {Z}_L:\mathbb {Z}_K[\alpha ]]=1.$
For the monogenity of
$f\circ g$
, the necessity of condition
$(2)$
is also mentioned in [Reference Gaál1, Corollary 1.7].
The following result provides a sufficient condition for the composition to be nonmonogenic.
Corollary 1.2. Let
$f(x)\in \mathbb {Z}[x]$
be a monic polynomial. If there exists a prime p and an integer a such that
$p\mid f'(a)$
and
$p^2\mid f(a)$
, then
$(f\circ g)(x)$
is nonmonogenic for each monic polynomial
$g(x)\in \mathbb {Z}[x]$
.
Note that the prime factors of
$f(0)$
and
$f(a)$
can be different. So the primes different from the divisors of
$f(0)$
satisfying the conditions of the above corollary cannot divide the index of
$(f\circ g)(x)$
. The particular case of the corollary with
$a=0$
and
$f=g$
is proved in [Reference Sharma, Sharma and Laishram11].
Applying Theorem 1.1 with mathematical induction gives the following result.
Corollary 1.3. Let
$F_n(x)=(f_1\circ f_2\circ \cdots \circ f_n)(x) $
be the composition of n monic polynomials
$f_1,f_2,\ldots ,f_n$
belonging to
$\mathbb {Z}[x]$
. If
$F_n(x)$
is monogenic, then
$F_i(x)$
is monogenic for all
$i=1,2\ldots ,n$
.
In [Reference Harrington and Jones3], Harrington and Jones characterised those binomials
$f(x), g(x)$
which satisfy the condition that both
$f(x)$
and
$f(g(x))$
are monogenic. In general, it is not always true that
$f(x)$
and
$f(g(x))$
are both monogenic. An example of such polynomials is
$$ \begin{align*}f(x)=x^2+x+4 \quad \text{and} \quad g(x)=x^3.\end{align*} $$
Here,
$f(x)$
is monogenic whereas
$f(x^3)$
is nonmonogenic, because
$2$
divides the index of
$f(x^3)$
. We provide a sufficient condition under which both
$f(x)$
and
$(f\circ g)(x)$
are monogenic. First, we prove the following theorem that yields the prime divisors of the index of
$(f\circ g)(x)$
when the composition is of a certain type.
Theorem 1.4. Let
$f(x)$
and
$ g(x)$
be two monic polynomials in
$\mathbb {Z}[x]$
of degree m and n, respectively, such that
$(f\circ g)(x)$
is irreducible. Suppose
$\alpha $
is a root of
$(f\circ g)(x)$
. If
-
(1)
$f(x)$
is monogenic and -
(2)
$N_{\mathbb {Q}(\alpha )/\mathbb {Q}}\big (g'(\alpha )\big )/n^{mn}$
is a squarefree integer,
then every prime divisor of the index of
$(f\circ g)(x)$
will be a divisor of n.
From the theorem, we obtain the next two corollaries.
Corollary 1.5. Let
$f(x)$
and
$ g(x)$
be two monic polynomials belonging to
$\mathbb {Z}[x]$
of degree m and n, respectively. Suppose
$(f\circ g)(x)$
is an Eisenstein polynomial with respect to each prime divisor of n and has root
$\alpha $
. If
$N_{\mathbb {Q}(\alpha )/\mathbb {Q}}\big (g'(\alpha )\big )/n^{mn}$
is a squarefree integer, then
$f(x)$
is monogenic if and only if
$(f\circ g)(x)$
is monogenic.
The following corollary is useful in the construction of infinite towers of monogenic number fields. Some examples are given at the end of the paper.
Corollary 1.6. Let
$\{f_n\}$
be a sequence of monic polynomials belonging to
$\mathbb {Z}[x]$
and let
$f_n$
have degree
$t_n$
. For each
$n>1$
, assume that the polynomial
$F_n(x)=(F_{n-1}\circ f_n)(x)$
is Eisenstein with respect to each prime divisor of
$t_n$
. Let
$\alpha _n$
be a root of
$F_n$
and
$K_n=\mathbb {Q}(\alpha _n)$
. If
$N_{K_n/\mathbb {Q}}\big (f_{n}'(\alpha _n)\big )/t_n^{t_n\deg F_{n-1}}$
is a squarefree integer for each
$n>1$
, then
$f_1$
is monogenic if and only if
$F_n$
is monogenic.
2. Preliminaries
Theorem 2.1 (Capelli; see [Reference Tschebotaröw12, page 288]).
Let F be a field. Let
$f(x)$
and
$g(x)$
be two monic polynomials belonging to
$F[x]$
and
$\beta $
be a root of
$f(x)$
in the algebraic closure of F. Then, the following statements are equivalent:
-
(i)
$(f\circ g)(x)$
is irreducible over F; -
(ii)
$f(x)$
is irreducible over F and
$g(x)-\beta $
is irreducible over
$F(\beta ).$
The following two results proved in [Reference Uchida13] will be used in the proof of Lemma 2.4.
Lemma 2.2. An ideal
$\mathfrak M$
of
$\mathbb {Z}[x]$
containing a monic polynomial is maximal if and only if
$\mathfrak M = \langle p, g(x) \rangle $
for some prime number p and a monic polynomial
$g(x)$
belonging to
$\mathbb Z [x]$
which is irreducible modulo p.
Theorem 2.3. Let
$K=\mathbb Q(\theta )$
with
$\theta \in \mathbb {Z}_K$
having minimal polynomial
$f(x)$
over
$\mathbb Q$
. Then,
$\mathbb {Z}_K=\mathbb {Z}[\theta ]$
if and only if
$f(x)$
does not belong to
$\mathfrak M ^2$
for any maximal ideal
$\mathfrak M$
of the polynomial ring
$\mathbb Z[x]$
.
The following lemma will be used in the proof of Theorem 1.1.
Lemma 2.4. Let
$f(x)$
and
$g(x)$
be two monic polynomials belonging to
$\mathbb {Z}[x]$
such that the composition
$(f\circ g)(x)$
is irreducible over
$\mathbb {Q}$
. If
$f(x)$
is nonmonogenic, then so is
$(f\circ g)(x)$
.
Proof. Suppose that
$f(x)$
is reducible over
$\mathbb {Q}$
, so that
$f(x)=a(x)b(x)$
for some nonconstant polynomials
$a(x),b(x) \in \mathbb {Z}[x]$
. Clearly,
$f(g(x))=a(g(x))b(g(x))\in \mathbb {Z}[g(x)]$
. As
$\mathbb {Z}[g(x)]\subset \mathbb {Z}[x]$
, we see that
$f(g(x))$
is reducible over
$\mathbb {Q}$
. Thus, our supposition is wrong and
$f(x)$
is irreducible over
$\mathbb {Q}$
. As
$f(x)$
is nonmonogenic, by Theorem 2.3, there exists a maximal ideal
$\mathfrak {M}$
of
$\mathbb {Z}[x]$
such that
$f(x)\in \mathfrak {M}^2$
. By Lemma 2.2, there is a prime p and a monic polynomial
$r(x)\in \mathbb {Z}[x]$
which is irreducible modulo p such that
$\mathfrak {M}=\langle p, r(x)\rangle .$
Since
$f(x)\in \langle p, r(x)\rangle ^2,$
we can write
$$ \begin{align*}f(x)=p^2u(x)+p\cdot r(x)\cdot v(x)+r^2(x)\cdot w(x),\end{align*} $$
for suitable
$u(x), v(x)$
and
$w(x) \in \mathbb {Z}[x].$
By replacing x with
$g(x)$
in the above equation, we see that
$$ \begin{align} f(g(x))=p^2u(g(x))+p\cdot r(g(x))\cdot v(g(x))+r(g(x))^2\cdot w(g(x)).\end{align} $$
If
$r(g(x))\in \mathbb {Z}[x]$
is irreducible modulo p, then by Lemma 2.3,
$f(g(x))\in \langle p, r(g(x))\rangle ^2.$
Let
$s(x)\in \mathbb {Z}[x]$
be a monic factor of
$r(g(x))$
which is irreducible modulo p. Set
$r(g(x))=s(x)\alpha (x)+p\beta (x),$
for some
$\alpha (x)$
and
$\beta (x)$
belonging to
$\mathbb {Z}[x].$
Substituting
$r(g(x))=s(x)\alpha (x)+p\beta (x)$
into (2.1) and simplifying the formula,
$$ \begin{align*}f(g(x))=p^2c(x)+p\cdot s(x)\cdot d(x)+s(x)^2\cdot t(x)\end{align*} $$
for some
$c(x),d(x)$
and
$t(x) \in \mathbb {Z}[x]$
. Thus, by Lemma 2.2,
$f(g(x))$
belongs to the square of the maximal ideal
$\langle p, s(x)\rangle $
. Hence, by Theorem 2.3,
$f(g(x))$
is nonmonogenic.
Theorem 2.5 [Reference Khanduja9, Theorem 1.19].
If
$K/F$
is an extension of number fields of degree n and
$\tau _1,\ldots ,\tau _n$
are all F-isomorphisms of K into a normal extension of F containing
$K,$
then for every
$\alpha \in K,$
we have
$N_{K/F}(\alpha )=\prod _{i=1}^{n}\tau _i(\alpha ).$
For the discriminant formula of the composition of two nonzero polynomials belonging to
$\mathbb {Z}[x]$
, we need various results concerning the resultant of two polynomials. Let F be a field. Given two polynomials
$f(x)$
and
$g(x)$
belonging to the ring
$F[x]$
with leading coefficients a and b, respectively, the resultant
$R(f,g)$
of
$f(x)$
and
$g(x)$
is
$$ \begin{align} R(f,g)=a^{\deg g}b^{\deg f}\displaystyle\prod_{\substack{1\le i\le \deg f\\ 1\le j \le \deg g}}(\lambda_i-\mu_j),\end{align} $$
where
$\lambda _1,\ldots ,\lambda _{\deg f}$
and
$\mu _1,\ldots ,\mu _{\deg g}$
are the roots of f and g, respectively.
In the next lemma, we list some well-known properties of the resultant of two polynomials.
Lemma 2.6. Let
$f(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots +a_0$
and
$g(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots +b_0$
be two polynomials over a field F with
$a_mb_n\ne 0$
. If
$\lambda _1,\ldots ,\lambda _m$
and
$\mu _1\ldots ,\mu _n$
are the roots of
$f(x)$
and
$g(x)$
, respectively, then the following statements hold:
-
(1)
$R(f,g)=a_m^n\prod _{i=1}^m g(\lambda _i)$
and
$R(f,g)=(-1)^{mn}b_n^m\prod _{j=1}^n f(\mu _j)$
; -
(2)
$R(f,gh)=R(f,g)R(f,h)$
for any
$h(x)\in \mathbb {Z}[x]$
.
Proof. (1) From (2.2),
$$ \begin{align*}R(f,g)=a_m^nb_n^m\displaystyle\prod_{\substack{1\le i\le m\\ 1\le j \le n}}(\lambda_i-\mu_j)=a_m^n\displaystyle\prod_{i=1}^m b_n^m\displaystyle\prod_{1\le j \le n}(\lambda_i-\mu_j).\end{align*} $$
Therefore,
$R(f,g)=a_m^n\prod _{i=1}^m g(\lambda _i).$
The other identity can be proved similarly.
(2) Let
$\deg h(x)=k$
,
$\gamma _1,\ldots ,\gamma _k$
be the roots of
$h(x)$
and
$c_k$
be the leading coefficient of
$h(x)$
. Then, by (2.2),
$$ \begin{align*}R(f,gh)=a_m^{n+k}(b_nc_k)^m\displaystyle\prod_{\substack{1\le i\le m\\ 1\le j \le n+k}}(\lambda_i-\alpha_j),\end{align*} $$
where
$\alpha _j=\mu _j$
for
$1\le j\le m$
and
$\alpha _j=\gamma _{j-m}$
for
$m+1\le j\le m+k$
. This implies
$$ \begin{align*}R(f,gh)=a_m^{n}b_n^m\displaystyle\prod_{\substack{1\le i\le m\\ 1\le j \le n}}(\lambda_i-\mu_j)a_m^kc_k^m\displaystyle\prod_{\substack{1\le i\le m\\ 1\le j \le k}}(\lambda_i-\gamma_j).\end{align*} $$
Hence,
$R(f,gh)=R(f,g)R(f,h).$
Let F be a field and
$f(x)\in F[x]$
be a monic irreducible polynomial of degree m. Suppose
$f(x)$
has a root
$\alpha $
. From Lemma 2.6(1),
$$ \begin{align*}R(f,f')=\displaystyle\prod_{i=1}^{m}f'(\lambda_i),\end{align*} $$
where
$f(\lambda _i)=0$
for
$1\le i\le m.$
Hence, using Theorem 2.5 and the fact that
$D(f)=(-1)^{{m(m-1)}/{2}}\prod _{i=1}^{m}f'(\lambda _i)$
, the discriminant of
$f(x)$
is
$$ \begin{align} D(f)=(-1)^{{m(m-1)}/{2}}R(f,f')=(-1)^{{m(m-1)}/{2}}N_{L/F}(f'(\alpha)), \end{align} $$
where
$L=F(\alpha )$
.
The following lemma provides a formula for the discriminant of the composition of two monic polynomials belonging to
$\mathbb {Z}[x]$
.
Lemma 2.7. Let
$f(x)$
and
$g(x)$
be two nonzero monic polynomials belonging to
$\mathbb {Z}[x]$
of degree m and n, respectively. If
$(f \circ g)(x)$
is irreducible, then the discriminant of
$(f\circ g)(x)$
is given by
$$ \begin{align*}D(f\circ g)=(-1)^{{mn}[m(n+1)-2]/2} (D(f))^{n}N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(g'(\alpha)),\end{align*} $$
where
$\alpha $
is a root of
$(f\circ g)(x).$
Proof. Let
$\alpha $
be a root of
$F(x):=(f\circ g)(x).$
Then, using (2.3),
$$ \begin{align*}D(F)= (-1)^{{mn(mn-1)}/{2}}R(F, F') = (-1)^{{mn(mn-1)}/{2}}R(F(x),(f'\circ g)(x) g'(x)).\end{align*} $$
Applying Lemma 2.6(2),
$$ \begin{align*}D(F)= (-1)^{{mn(mn-1)}/{2}}R(F(x),(f'\circ g)(x))R(F(x),g'(x)).\end{align*} $$
If
$\delta _1,\ldots , \delta _{mn}$
and
$\mu _1,\ldots , \mu _m$
are the roots of
$F(x)$
and
$f(x)$
, respectively, then by Lemma 2.6(1),
$$ \begin{align*}D(F)= (-1)^{{mn(mn-1)}/{2}}\displaystyle\prod_{i=1}^{mn}f'(g(\delta_i))R(F(x),g'(x)).\end{align*} $$
For each j with
$1\le j\le n$
, we can assume that
$g(\delta _{i+j-1})=\mu_j$
for
$1\le i\le m$
, so that
$$ \begin{align*}D(F)= (-1)^{{mn(mn-1)}/{2}}\displaystyle\prod_{j=1}^{m}(f'(\mu_j))^nR(F(x),g'(x)).\end{align*} $$
Thus, by (2.3) and Lemma 2.6(1),
$$ \begin{align*}D(F)= (-1)^{{mn}[m(n+1)-2]/2}(D(f(x)))^nR(F(x),g'(x)).\end{align*} $$
Finally, Lemma 2.6(1) implies that
$$ \begin{align*} D(F)=(-1)^{{mn}[m(n+1)-2]/2} (D(f))^{n}N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(g'(\alpha)).\\[-34pt] \end{align*} $$
Remark 2.8. With the notation of the above lemma, if
$g(x)=x^n+a\in \mathbb {Z}[x],$
then
$$ \begin{align*} D(f\circ g)=(-1)^{{mn}[m(n+1)-2]/2} (D(f))^{n}n^{mn}f(a)^{n-1}.\end{align*} $$
The following lemmas are useful in the construction of a tower of monogenic number fields.
Lemma 2.9 [Reference Narkiewicz10, Lemma 2.17].
Let
$\alpha $
be an algebraic integer and let
$L=\mathbb {Q}(\alpha )$
be the field generated by it. If the minimal polynomial
$f(x)$
of
$\alpha $
over
$\mathbb {Q}$
is an Eisenstein polynomial with respect to a prime p, that is, it has the form
$x^n+a_{n-1}x^{n-1}+\cdots +a_0,$
with
$a_0,\ldots ,a_{n-1}$
divisible by p and
$p^2\nmid a_0,$
then the index of
$f(x)$
is not divisible by
$p.$
Lemma 2.10 [Reference Khanduja9, Corollary 7.10].
If
$\mathbb {Q}\subset K\subset L$
is a tower of number fields, then
$$ \begin{align*}d_L=(d_K)^{[L:K]}N_{K/\mathbb{Q}}(d_{L/K}),\end{align*} $$
where
$d_{L/K}$
denotes the discriminant of the extension
$L/K.$
3. Proofs of Theorem 1.1, Corollary 1.2 and Theorem 1.4
Proof of Theorem 1.1.
Here,
$L=\mathbb {Q}(\alpha )$
and
$K=\mathbb {Q}(\beta )$
, where
$\beta =g(\alpha )$
. First, assume that
$(f\circ g)(x)$
is monogenic. Then, by Lemma 2.4, the polynomial
$f(x)$
is monogenic. By Theorem 2.1, the polynomial
$h(x)=g(x)-\beta $
belonging to
$\mathbb {Z}_K[x]$
is irreducible over
$K.$
Now, we show that
$\mathbb {Z}_K[\alpha ]=\mathbb {Z}_L.$
As both
$f(x)$
and
$(f\circ g)(x)$
are monogenic, we have
$\mathbb {Z}_K=\mathbb {Z}[\beta ]$
and
$\mathbb {Z}_L=\mathbb {Z}[\alpha ]$
. Clearly,
$$ \begin{align*}\mathbb{Z}_K[\alpha]=(\mathbb{Z}[\beta])[\alpha]=\mathbb{Z}[\beta,\alpha]\subset \mathbb{Z}[\alpha]=\mathbb{Z}_L.\end{align*} $$
Also,
$\mathbb {Z}\subset \mathbb {Z}_K$
and so
$\mathbb {Z}[\alpha ]\subset \mathbb {Z}_K[\alpha ],$
that is,
$\mathbb {Z}_L\subset \mathbb {Z}_K[\alpha ]$
. Hence,
$\mathbb {Z}_L=\mathbb {Z}_K[\alpha ].$
Conversely, assume that
$\mathbb {Z}_K=\mathbb {Z}[\beta ]$
and
$\mathbb {Z}_L=\mathbb {Z}_K[\alpha ]$
. Using Theorem 2.1, the irreducibility of
$g(x)-\beta $
over K and irreducibility of
$f(x)$
over
$\mathbb {Q}$
imply that
$(f\circ g) (x)$
is irreducible over
$\mathbb {Q}$
. To show that
$(f\circ g)(x)$
is monogenic, we need to show that
$\mathbb {Z}_L=\mathbb {Z}[\alpha ].$
Clearly,
$\mathbb {Z}_L=\mathbb {Z}_K[\alpha ]=\mathbb {Z}[\beta ,\alpha ]\subset \mathbb {Z}[\alpha ].$
Since
$\mathbb {Z}[\alpha ]\subset \mathbb {Z}_L,$
we have
$\mathbb {Z}_L=\mathbb {Z}[\alpha ]$
. Hence, the proof is complete.
Proof of Corollary 1.2.
If
$f(x)$
or
$(f\circ g)(x)$
is reducible over
$\mathbb {Q}$
, then the result is obvious. Suppose that both
$f(x)$
and
$(f\circ g)(x)$
are irreducible over
$\mathbb {Q}$
. In view of Theorem 1.1, it is enough to prove that the polynomial
$f(x)$
is nonmonogenic. Set
$f(a)=p^2b$
and
$f'(a)=pc$
for some integers b and
$c.$
Write
$$ \begin{align*} f(x)=f(x-a+a)=\displaystyle\sum_{i=2}^{n}\frac{f^{(n+2-i)}(a)}{(n+2-i)!}(x-a)^{n+2-i}+f'(a)(x-a)+f(a),\end{align*} $$
where
$f^{(i)}$
denotes the ith derivative of
$f.$
Consequently,
$$ \begin{align*}f(x)&=(x-a)^2h(x)+pc(x-a)+p^2b,\\ &\quad\mbox{where}\ h(x)=\displaystyle\sum_{i=2}^{n}\frac{f^{(n+2-i)}(a)}{(n+2-i)!}(x-a)^{n-i}\in\mathbb{Z}[x].\end{align*} $$
This implies that
$f(x)\in \langle x-a, p\rangle ^2.$
Since
$x-a$
is irreducible modulo p, by Lemma 2.2 and Theorem 2.3, we see that
$f(x)$
is nonmonogenic.
Proof of Theorem 1.4.
Set
$L=\mathbb {Q}(\alpha )$
and
$K=\mathbb {Q}(\beta )$
, where
$\beta =g(\alpha )$
. By Lemma 2.7, the discriminant of
$f\circ g$
is
$$ \begin{align} D(f\circ g)=\pm(D(f))^nN_{L/\mathbb{Q}}(g'(\alpha)). \end{align} $$
Since
$f(x)$
is monogenic, we have
$d_K=D(f).$
Substituting the value of
$D(f)$
into (3.1), we obtain
$D(f\circ g)=\pm (d_K)^nN_{L/\mathbb {Q}}(g'(\alpha )).$
If I denotes the index
$[\mathbb {Z}_L:\mathbb {Z}[\alpha ]]$
, then using (1.4) and the well-known identity
$D(f\circ g)=I^2d_L$
, we see that
$I^2d_L=d_K^nN_{L/\mathbb {Q}}(g'(\alpha )),$
which gives
$$ \begin{align*}I^2\frac{d_L}{d_K^n}=n^{mn}\frac{1}{n^{mn}}N_{L/\mathbb{Q}}(g'(\alpha)).\end{align*} $$
Since
${d_L}/{d_K^n}\in \mathbb {Z}$
and
$N_{L/\mathbb {Q}}(g'(\alpha ))/{n^{mn}}$
is a squarefree integer, we can easily see that the possible prime divisors of I are prime divisors of
$n.$
4. Infinite examples
In the following proposition, we construct an infinite tower of monogenic number fields. This construction follows from Corollary 1.6.
Proposition 4.1. If
$m\equiv 2\ \mod 4$
is a squarefree integer and n is a positive integer, then
$$ \begin{align*}\mathbb{Q}\rightarrow \mathbb{Q}(\sqrt{m})\rightarrow \mathbb{Q}(\kern-2pt\sqrt[2^2]{m})\rightarrow\cdots\rightarrow \mathbb{Q}(\kern-2pt\sqrt[2^n]{m})\rightarrow \cdots\end{align*} $$
is an infinite tower of monogenic number fields.
Proof. Clearly,
$x^{2^n}-m$
is an Eisenstein polynomial with respect to
$2,$
so it is irreducible. Thus, we have a tower of number fields
$$ \begin{align*}\mathbb{Q}\rightarrow L_1\rightarrow L_2\rightarrow\cdots\rightarrow L_n\rightarrow L_{n+1}\rightarrow \cdots,\end{align*} $$
where
$L_n=\mathbb {Q}(\kern -2pt\sqrt [2^n]{m})$
. Clearly,
$f_1(x)=x^2-m$
is monogenic, that is,
$L_1=\mathbb {Q}(\kern -2pt\sqrt [2]{m})$
is monogenic. To prove the monogenity of
$L_n$
, we shall use induction on
$n.$
Let
$f_2(x)=x^2$
and
$L_2=\mathbb {Q}(\kern -2pt\sqrt [2^2]{m})$
, then
$$ \begin{align*}N_{L_2/\mathbb{Q}}(2\kern-2pt\sqrt[2^2]{m})=\pm 2^{2^2}m.\end{align*} $$
Applying Theorem 1.4, we observe that the only prime that can divide the group index
$[\mathbb {Z}_{L_2}:\mathbb {Z}[\kern -2pt\sqrt [2^2]{m}]]$
is
$2.$
However,
$(f_1\circ f_2)(x)=x^{2^2}-m$
is an Eisenstein polynomial with respect to
$2$
, so using Lemma 2.9, we see that the field
$L_2=\mathbb {Q}(\kern -2pt\sqrt [2^2]{m})$
is monogenic. Suppose that
$f_{n-1}(x)=x^{2^{n-1}}-m$
is monogenic and
$g(x)=x^2$
. Then,
$L_n=\mathbb {Q}(\kern -2pt\sqrt [2^{n}]{m})$
, which is generated by a root
$\kern -2pt\sqrt [2^{2^n}]{m}$
of
$(f_{n-1}\circ g)(x).$
As
$ N_{L_n/\mathbb {Q}}(2\kern -2pt\sqrt [2^{n}]{m})/{2^{2^n}}=\pm m$
is a squarefree integer, by Theorem 1.4,
$[\mathbb {Z}_{L_n}:\mathbb {Z}[\kern -2pt\sqrt [2^n]{m}]]=2^s$
for some nonnegative integer
$s.$
Furthermore, by Lemma 2.9,
$\mathbb {Z}_{L_{n}}=\mathbb {Z}[\kern -2pt\sqrt [2^{n}]{m}]]$
, that is,
$L_n$
is monogenic.
If
$m\ne 0,1$
is a squarefree integer, then the polynomial
$f(x)=x^m-m$
is irreducible and has discriminant
$D(f)=m^{2m-1}.$
Let
$K=\mathbb {Q}(\theta ),$
where
$f(\theta )=0.$
Keeping in mind Lemma 2.9 and the fact that
$D(f)=[\mathbb {Z}_K:\mathbb {Z}[\theta ]]^2d_K$
, we observe that the polynomial is monogenic. Using Remark 2.8 together with Lemma 2.9, we easily get the infinite tower of monogenic number fields
$$ \begin{align*}\mathbb{Q}\rightarrow\mathbb{Q}(\kern-2pt\sqrt[m]{m})\rightarrow\mathbb{Q}(\kern-2pt\sqrt[m^2]{m})\rightarrow\cdots\rightarrow\mathbb{Q}(\kern-2pt\sqrt[m^n]{m})\rightarrow\cdots.\end{align*} $$
Example 4.2. Let a be an even squarefree integer. Let
$f(x)=x^n-a^{n-1}x+a+a^n$
and
$g(x)=x^2+a$
be polynomials belonging to
$\mathbb {Z}[x].$
Here,
$$ \begin{align*} (f\circ g) (x)=(x^2+a)^n-a^{n-1}x^2+a. \end{align*} $$
Clearly,
$(f\circ g)(x)$
is an Eisenstein polynomial with respect to
$2.$
So
$(f\circ g)(x)$
is an irreducible polynomial. If
$\alpha $
is a root of
$(f\circ g)(x)$
, then
$$ \begin{align*}N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(g'(\alpha))=\pm2^{2n}N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)=\pm2^{2n}a.\end{align*} $$
This implies that
$N_{\mathbb {Q}(\alpha )/\mathbb {Q}}(g'(\alpha ))/4^n=\pm a$
is squarefree. Since
$f(g(x))$
is an Eisenstein polynomial with respect to
$2$
, by Corollary 1.5, it follows that
$f(x)$
is monogenic if and only if
$f(g(x))$
is monogenic.
Example 4.3. Let
$L=\mathbb {Q}(\alpha )$
be an algebraic number field, where
$\alpha $
is a root of
$$ \begin{align*}F(x)&=x^{27}-18x^{24}+144x^{21}-672x^{18}+2016x^{15}-4032x^{12}\\&\quad+5376x^9+4608x^6+2304x^3-1.\end{align*} $$
Note that
$F(x+1)$
is an Eisenstein polynomial with respect to the prime
$3;$
therefore,
$F(x)$
is an irreducible polynomial. Take
$f(x)=x^9+511$
and
$g(x)=x^3-2,$
so that
$F(x)=(f\circ g)(x)$
. Now,
$$ \begin{align*}N_{L/\mathbb{Q}}(g'(\alpha))=3^{27}N_{L/\mathbb{Q}}(\alpha)^2=3^{27}.\end{align*} $$
Clearly,
$N_{L/\mathbb {Q}}(g'(\alpha )/3^{27}$
is squarefree. If
$f(x)$
is monogenic, then using Theorem 1.4, we can see that
$[\mathbb {Z}_L:\mathbb {Z}[\alpha ]]=3^t$
for some nonnegative integer
$t.$
As
$F(x+1)$
is an Eisenstein polynomial with respect to
$3,$
it follows that
$t=0$
in view of Lemma 2.9. Hence,
$F(x)$
is monogenic. Thus, we have the
$3$
-tower of monogenic number fields
$$ \begin{align*}\mathbb{Q}\subset \mathbb{Q}(\alpha^3)\subset \mathbb{Q}(\alpha).\end{align*} $$
Now, we prove that
$f(x)=x^9+511$
is monogenic. Here,
$D(f)=\pm 9^97^873^8.$
For each prime divisor p of
$511$
, the polynomial
$f(x)$
is Eisenstein. So using Lemma 2.9, p does not divide the index of
$f(x)$
. Also,
$f(x-1)$
is an Eisenstein polynomial with respect to
$3$
, so
$3$
does not divide the index of
$f(x)$
. In view of the identity
$D(f)=[\mathbb {Z}_L:\mathbb {Z}[\alpha ]]^2d_L$
, we see that the polynomial
$f(x)$
is monogenic.
