3.1. $\mathbf {C}_{\mathbf {2}}\times \mathbf {C}_{\mathbf {2}}$

Since $f_t(x\kern1.4pt{-}\kern1.4pt1\kern-0.4pt)\kern1.4pt{=}\kern1.4pt x^4\kern1.4pt{-}\kern1.4pt 4x^3\kern1.4pt{+}\kern1.4pt(4t\kern1.4pt{+}\kern1.4pt 6\kern-0.1pt)x^2\kern1.4pt{-}\kern1.4pt(8t\kern1.4pt{+}\kern1.4pt 4\kern-0.1pt)x\kern1.4pt{+}\kern1.4pt 4t\kern1.4pt{+}\kern1.4pt2$ is 2-Eisenstein, we conclude that $f_t(x)=x^4+4tx^2+1$ is irreducible over ${\mathbb Q}$ . The cubic resolvent of $f_t(x)$ is

$$ \begin{align*} r_t(x)=x^3-4tx^2-4x+16t=(x-2)(x+2)(x-4t). \end{align*} $$

By Theorem 2.1, it follows that $\mathrm {Gal}(f_t)\simeq C_2\times C_2$ , which proves item (1a).

For item (1b), a straightforward calculation reveals that

(3.1) $$ \begin{align} \Delta(f_t)=2^8(4t^2-1)^2. \end{align} $$

To establish the monogenicity of $f_t(x)$ , we use Theorem 2.3 with $T(x):=f_t(x)$ and we examine the prime divisors q of $\Delta (f_t)$ .

First, let $q=2$ . Then $\overline {T}(x)=(x+1)^4$ and we can let $h_1(x)=x+1$ and ${h_2(x)=(x+1)^3}$ . Then,

$$ \begin{align*} F(x)=\dfrac{(x+1)^4-f_t(x)}{2}=2x^3-(2t-3)x^2+2x \equiv x^2 \pmod{2}. \end{align*} $$

Thus, $\gcd (\overline {h_1},\overline {h_2},\overline {F})=1$ and $[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$ is not divisible by $q=2$ .

Next, we give details only for the case when a prime q divides $2t-1$ since the case when q divides $2t+1$ is similar. Since $t\equiv 1/2 \pmod {q}$ , it follows that

$$ \begin{align*}\overline{T}(x)=\left\{\begin{array}{@{}ll} (x^2+1)^2 &\mbox{ if } q\equiv 3 \pmod{4},\\ (x-y)^2(x+y)^2 & \mbox{ if } q\equiv 1 \pmod{4}, \end{array} \right.\end{align*} $$

where $y\in {\mathbb Z}$ such that $y^2\equiv -1\pmod {q}$ . If $q\equiv 3 \pmod {4}$ , we can let $h_1(x)=h_2(x)=x^2+1$ . Then,

$$ \begin{align*} \overline{F}(x)=\overline{-2\bigg(\dfrac{2t-1}{q}\bigg)}x^2 \end{align*} $$

so that $\gcd (\overline {h_1},\overline {h_2},\overline {F})=1$ if and only if $[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$ is not divisible by q if and only if $q^2\nmid (2t-1)$ . If $q\equiv 1 \pmod {4}$ , then $y^2=zq-2t$ for some integer z since ${-2t\equiv -1 \pmod {q}}$ . Then,

$$ \begin{align*} \overline{F}(x)=\overline{-2z}(x^2+1)+\overline{\bigg(\dfrac{4t^2-1}{q}\bigg)}, \end{align*} $$

from which we see that

$$ \begin{align*} \overline{F}(\pm y)=\overline{\bigg(\dfrac{4t^2-1}{q}\bigg)}\ne 0 \quad \mbox{if and only if}\quad q^2\nmid (2t-1). \end{align*} $$

Thus, we conclude that $f_t(x)$ is monogenic if and only if $4t^2-1$ is squarefree, completing the proof of item (1b).

For item (1c), we note first that ${\mathcal F}_2$ is an infinite set since there exist infinitely many integers t such that $4t^2-1$ is squarefree [Reference Booker and Browning1]. To see that the fields generated by the elements $f_t(x)$ of ${\mathcal F}_2$ are all distinct, we proceed by way of contradiction. We assume for integers $t_1\ne t_2$ , that $K_1=K_2$ , where $K_1={\mathbb Q}(\alpha _1)$ and $K_2={\mathbb Q}(\alpha _2)$ with $f_{t_1}(\alpha _1)=0=f_{t_2}(\alpha _2)$ . Since both $f_{t_1}(x)$ and $f_{t_2}(x)$ are monogenic by item (1b), it follows that $\Delta (f_{t_1})=\Delta (f_{t_2})$ . Consequently, from (3.1),

$$ \begin{align*} 8(t_1-t_2)(t_1+t_2)(2t_1^2+2t_2^2-1)=0. \end{align*} $$

Hence, $t_1=-t_2$ . Without loss of generality, we may assume that $t_1>0$ so that $t_2<0$ . Since the zeros of $f_t(x)$ are

$$ \begin{align*} \sqrt{-2t+\sqrt{4t^2-1}},\quad \sqrt{-2t-\sqrt{4t^2-1}},\quad -\sqrt{-2t+\sqrt{4t^2-1}},\quad -\sqrt{-2t-\sqrt{4t^2-1}},\end{align*} $$

it follows that all zeros of $f_t(x)$ are real if $t<0$ , while all zeros of $f_t(x)$ are nonreal if $t\ge 0$ , which contradicts the assumption that $K_1=K_2$ .

3.2. $\mathbf {D}_{\mathbf {4}}$

Note that $f_0(x-1)=x^4-4x^3+10x^2-8x+2$ is 2-Eisenstein. Hence, $f_0(x)$ is irreducible over ${\mathbb Q}$ . Suppose next that $t\ne 0$ . Then, $|{t}|>10/12$ , from which we deduce that

$$ \begin{align*}|{24t}|=|{12t}|+12|{t}|>|{12t}|+10\ge |{12t+4}|+6.\end{align*} $$

Thus, $f_t(x)=x^4+24tx^3+(12t+4)x^2+4x+1$ is irreducible over ${\mathbb Q}$ when $t\ne 0$ by Perron’s irreducibility criterion [Reference Perron16].

To determine $\mathrm {Gal}(f_t)$ , we use Theorem 2.1. The cubic resolvent of $f_t(x)$ is

$$ \begin{align*} r_t(x)&=x^3-(12t+4)x^2+(96t-4)x-(576t^2-48t)\\ &=(x-12t)(x^2-4x+(48t-4)). \end{align*} $$

Since

$$ \begin{align*} \Delta(x^2-4x+(48t-4))=-192t+32=16(-12t+2), \end{align*} $$

and $-12t+2 \equiv 2 \pmod {4}$ , we conclude that $r_t(x)$ has exactly the one rational zero $x=12t$ . Consequently, the splitting field of $r_t(x)$ is $L={\mathbb Q}(\sqrt {-2(6t-1)})$ . Then, in Theorem 2.1, we have $g(x)=g_1(x)g_2(x)$ , where

$$ \begin{align*}g_1(x)=x^2-12tx+1 \quad\mbox{and}\quad g_2(x)=x^2+24tx+4.\end{align*} $$

Since $\Delta (g_1)=4(6t-1)(6t+1)$ and $\Delta (g_2)=16(6t-1)(6t+1)$ , it follows that both $g_1(x)$ and $g_2(x)$ are irreducible over L. Hence, $\mathrm {Gal}(f_t)\simeq D_4$ by Theorem 2.1, which proves item (2a).

To establish item (2b), we use Theorem 2.3 with $T(x):=f_t(x)$ . A straightforward calculation yields

$$ \begin{align*}\Delta(f_t)=-2^9(6t-1)^3(6t+1)^2.\end{align*} $$

First let $q=2$ . Then, $\overline {T}(x)=(x+1)^4$ , and we can select

(3.2) $$ \begin{align} h_1(x)=x+1 \quad\mbox{and}\quad h_2=(x+1)^3. \end{align} $$

Thus,

$$ \begin{align*}F(x)=\dfrac{(x+1)^4-f_t(x)}{2}=(2-12t)x^3+(1-6t)x^2 \equiv x^2 \pmod{2}.\end{align*} $$

Therefore, $\gcd (\overline {h_1},\overline {F})=1$ so that $[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$ is not divisible by $q=2$ .

When q is a prime divisor of $6t-1$ , we also get

$$ \begin{align*}\overline{T}(x)=x^4+4x^3+6x^2+4x+1=(x+1)^4,\end{align*} $$

with $h_i(x)$ as in (3.2). Then,

$$ \begin{align*}F(x)=\dfrac{(x+1)^4-f_t(x)}{2}=\dfrac{-2(6t-1)}{q}(2x+1)x^2,\end{align*} $$

and it follows that

$$ \begin{align*}\gcd(\overline{h_1},\overline{F})=1 \quad \mbox{ if and only if}\quad q^2\nmid (6t-1).\end{align*} $$

Thus, $q\nmid [{\mathbb Z}_K:{\mathbb Z}[\theta ]]$ if and only if $q^2\nmid (6t-1)$ .

Suppose next that q is a prime divisor of $6t+1$ . Then, $t\equiv -1/6 \pmod {q}$ and

$$ \begin{align*}T(x)\equiv (x^2-2x-1)^2 \pmod{q}.\end{align*} $$

If $x^2-2x-1$ is irreducible over ${\mathbb F}_q$ , then we can let $h_1(x)=h_2(x)=x^2-2x-1$ . In this case, we get

$$ \begin{align*} F(x)=\dfrac{(x^2-2x-1)^2-f_t(x)}{2}=-2\bigg(\dfrac{6t+1}{q}\bigg)x^2(2x+1),\end{align*} $$

so that

$$ \begin{align*}\gcd(\overline{h_1},\overline{F})=1 \quad\mbox{if and only if}\quad q^2\nmid (6t+1).\end{align*} $$

If $x^2-2x-1$ is reducible over ${\mathbb F}_q$ , then

$$ \begin{align*}x^2-2x-1=(x-(1+y))(x-(1-y)),\end{align*} $$

where $y\in {\mathbb Z}$ with $y^2\equiv 2\pmod {q}$ . Then, we can select

$$ \begin{align*}h_1(x)=h_2(x)=(x-(1+y))(x-(1-y)),\end{align*} $$

so that

$$ \begin{align*}F(x)=-4\bigg(\dfrac{6t+1}{q}\bigg)x^3-2\bigg(\dfrac{6t+1}{q}+\dfrac{y^2-2}{q}\bigg)x^2+4\bigg(\dfrac{y^2-2}{q}\bigg)x+y^2\bigg(\dfrac{y^2-2}{q}\bigg).\end{align*} $$

Then, computer calculations reveal that

$$ \begin{align*}\overline{F}(1\pm y)=2\bigg(\dfrac{6t+1}{q}\bigg)(\mp 12y-17).\end{align*} $$

If $\mp 12y-17\equiv 0 \pmod {q}$ , then, since $y^2\equiv 2\pmod {q}$ ,

$$ \begin{align*}288\equiv 144y^2\equiv (\mp 12y)^2\equiv 17^2\equiv 289 \pmod{q},\end{align*} $$

which yields the contradiction that $0\equiv 1\pmod {q}$ . Hence,

$$ \begin{align*}\gcd(\overline{h_1},\overline{F})=1 \quad\mbox{if and only if}\quad q^2\nmid (6t+1),\end{align*} $$

completing the proof that $f_t(x)$ is monogenic if and only if $36t^2-1$ is squarefree.

For item (2c), we proceed as in the case of $C_2\times C_2$ . The set ${\mathcal F}_3$ is infinite since there exist infinitely many values of t such that $36t^2-1$ is squarefree [Reference Booker and Browning1]. We assume for integers $t_1\ne t_2$ that $K_1=K_2$ , where $K_1={\mathbb Q}(\alpha _1)$ and $K_2={\mathbb Q}(\alpha _2)$ with $f_{t_1}(\alpha _1)=0=f_{t_2}(\alpha _2)$ . Since both $f_{t_1}(x)$ and $f_{t_2}(x)$ are monogenic by item (2b), it follows that $\Delta (f_{t_1})=\Delta (f_{t_2})$ . Solving this discriminant equation using Maple shows that $6t_1$ must be a zero of the polynomial

$$ \begin{align*} {\mathcal G}(X)&:=X^4 +(6t_2-1)X^3+(36t_2^2-6t_2-2)X^2+(216t_2^3-36t_2^2-12t_2+2)X \\ &\quad +1296t_2^4-216t_2^3+12t_2-72t_2^2+1, \end{align*} $$

which is impossible since ${\mathcal G}(0)\not \equiv 0 \pmod {6}$ . This contradiction completes the proof of item (2c) for the case of $D_4$ .

3.3. $\mathbf {A}_{\mathbf {4}}$

Since $f_t(x)$ is 2-Eisenstein, $f_t(x)$ is irreducible over ${\mathbb Q}$ . Straightforward calculations using Maple give

$$ \begin{align*} \Delta(f_t)=2^6(4t-1)^2(108t^2-54t+7)^2 \end{align*} $$

and the cubic resolvent of $f_t(x)$ as

$$ \begin{align*} r_t(x)=x^3-2x^2-(144t^2-72t+8)x+128t^2-64t+8. \end{align*} $$

If $r_t(x)$ is reducible over ${\mathbb Q}$ , then

$$ \begin{align*} r_t(x)=(x+z)(x^2+Ax+B)=x^3+(A+z)x^2+(Az+B)x+Bz, \end{align*} $$

for some integers $A,B,z$ . Hence, by equating coefficients, we see that

$$ \begin{align*} A+z&=-2\\ Az+B&=-(144t^2-72t+8)\\ Bz&=128t^2-64t+8. \end{align*} $$

Solving this system in Maple yields

$$ \begin{align*}B=(8z^2+16z+8)/(9z+8).\end{align*} $$

Let $d:=\gcd (8z^2+16z+8,9z+8)$ . Then, easy greatest common divisor (gcd) calculations show that $d\mid 8$ . Since B is an integer, $d=9z+8$ . Hence, the only possibilities for z are $z=0$ and $z=-1$ . However, checking the values of $-z$ in $r_t(x)$ gives $r(0)=8(4t-1)^2\ne 0$ and $r(1)=-(4t-1)^2\ne 0$ , since $t\in {\mathbb Z}$ . Hence, $r_t(x)$ is irreducible over ${\mathbb Q}$ , and it follows from Theorem 2.1 that $\mathrm {Gal}(f_t)\simeq A_4$ , which proves item (3a).

For item (3b), we use Theorem 2.3 with $T(x):=f_t(x)$ to check the prime divisors q of $\Delta (f_t)$ . Suppose first that $q=2$ . Then, $\overline {T}(x)=x^4$ , so that we can let $h_1(x)=x$ and $h_2(x)=x^3$ , and we get

$$ \begin{align*}F(x)=-x^3-x^2-2tx-18t^2+8t-1\equiv x^3+x^2+1 \pmod{2}.\end{align*} $$

Hence, we see quite easily that $\gcd (\overline {h_1},\overline {F})=1$ , from which we conclude that $[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$ is not divisible by $q=2$ .

Next, let q be a prime divisor of $4t-1$ . Then, $t\equiv 1/4\pmod {q}$ and

$$ \begin{align*}T(x)\equiv (x^2+x+1/2)^2 \pmod{q}.\end{align*} $$

Suppose first that $x^2+x+1/2$ is irreducible over ${\mathbb F}_q$ . Note that in this situation, we must have $q\equiv 3 \pmod {4}$ since $-1$ is not a square. Then we can let

$$ \begin{align*}h_1(x)=h_2(x)=x^2+x+(q+1)/2,\end{align*} $$

which yields

$$ \begin{align*} \overline{F}(x)&=x^2-\overline{\bigg(\dfrac{4t-1}{q}-1\bigg)}x-\overline{\bigg(\dfrac{(4t-1)(36t-7)-q^2-2q}{4q}\bigg)}\\ &=x^2-\overline{\bigg(\dfrac{4t-1}{q}-1\bigg)}x-\overline{\bigg(\dfrac{(4t-1)(36t-7)}{4q}\bigg)}+\overline{\dfrac{1}{2}}. \end{align*} $$

Note that $36t-7 \equiv 2\pmod {q}$ so that $q\nmid (36t-7)$ . Hence, it follows that ${\gcd (\overline {h_1},\overline {F})\ne 1}$ if and only if $\overline {F}(x)=\overline {h_1}(x)$ , which is true if and only if $q^2\mid (4t-1)$ .

Suppose next that $x^2+x+1/2$ is reducible over ${\mathbb F}_q$ . Then, $q\equiv 1 \pmod {4}$ and

$$ \begin{align*}x^2+x+1/2\equiv\bigg(x-\bigg(\dfrac{-1+y}{2}\bigg)\bigg)\bigg(x-\bigg(\dfrac{-1-y}{2}\bigg)\bigg)\pmod{q},\end{align*} $$

where $y^2\equiv -1\pmod {q}$ . Choosing $y\equiv 1 \pmod {2}$ , we can let

$$ \begin{align*}h_1(x)=h_2(x)=\bigg(x-\bigg(\dfrac{-1+y}{2}\bigg)\bigg)\bigg(x-\bigg(\dfrac{-1-y}{2}\bigg)\bigg).\end{align*} $$

Then, computer calculations produce

$$ \begin{align*}\overline{F}(x)=\overline{\bigg(\dfrac{-y^2-1}{2q}\bigg)}x^2+\overline{\bigg(\dfrac{1-y^2-8t}{2q}\bigg)}x +\overline{\bigg(\dfrac{y^4-2y^2-576t^2+256t-31}{16q}\bigg)},\end{align*} $$

so that

$$ \begin{align*}\overline{F}((-1\pm y)/2)=\overline{\dfrac{-(4t-1)(9(4t-1)\pm 2y)}{q}+\dfrac{(y^2+1)^2}{4q}}=\overline{\mp 2y\bigg(\dfrac{4t-1}{q}\bigg)}.\end{align*} $$

Hence, we see that $\overline {F}((-1\pm y)/2)=0$ if and only if $q^2\mid (4t-1)$ , which completes the proof when $q\mid (4t-1)$ .

Now, suppose that q is a prime divisor of $108t^2-54t+7$ . Note that $q\not \in \{2,3\}$ . Then, it follows that

$$ \begin{align*}\overline{T}(x)=(x-(18t-5))(x-(1-6t))^3,\end{align*} $$

and therefore, we can let

(3.3) $$ \begin{align} h_1(x)=(x-(18t-5))(x-(1-6t)) \quad\mbox{and}\quad h_2(x)=(x-(1-6t))^2. \end{align} $$

Observe from (3.3) that we only need to check $\overline {F}(1-6t)$ to determine $\gcd (\overline {h_1},\overline {h_2},\overline {F})$ . More precisely,

(3.4) $$ \begin{align} [{\mathbb Z}_K:{\mathbb Z}[\theta]]\equiv 0 \ \mathrm{mod}\ q \quad \mbox{if and only if}\quad \overline{F}(1-6t)=0. \end{align} $$

Straightforward calculations yield

$$ \begin{align*}F(x)=\bigg(\dfrac{108t^2-54t+7}{q}\bigg)(-2x^2-2(8t-1)x-(36t^2-10t+1)),\end{align*} $$

so that

$$ \begin{align*}\overline{F}(1-6t)=-\overline{\bigg(\dfrac{108t^2-54t+7}{q}\bigg)}(12t^2-6t+1)=\overline{-\dfrac{2}{9}\bigg(\dfrac{108t^2-54t+7}{q}\bigg)},\end{align*} $$

since

$$ \begin{align*}9(12t^2-6t+1)-2=108^2-54t+7\equiv 0 \pmod{q}.\end{align*} $$

Hence,

$$ \begin{align*}\overline{F}(1-6t)=0 \quad \mbox{if and only if}\quad 108t^2-54t+7\equiv 0\ \mathrm{mod}\ q^2.\end{align*} $$

Consequently, from (3.4),

$$ \begin{align*}[{\mathbb Z}_K:{\mathbb Z}[\theta]]\equiv 0 \ \mathrm{mod}\ q \quad \mbox{if and only if}\quad 108t^2-54t+7\equiv 0\ \mathrm{mod}\ q^2.\end{align*} $$

Since $\gcd (4t-1,108t^2-54t+7)=1$ , the proof of item (3b) for $A_4$ is complete.

For item (3c), we proceed as in the previous cases. The set ${\mathcal F}_4$ is infinite since there exist infinitely many values of t such that $(4t-1)(108t^2-54t+7)$ is squarefree [Reference Booker and Browning1]. We assume for integers $t_1\ne t_2$ that $K_1=K_2$ , where $K_1={\mathbb Q}(\alpha _1)$ and $K_2={\mathbb Q}(\alpha _2)$ with $f_{t_1}(\alpha _1)=0=f_{t_2}(\alpha _2)$ . Since both $f_{t_1}(x)$ and $f_{t_2}(x)$ are monogenic by item (3b), it follows that $\Delta (f_{t_1})=\Delta (f_{t_2})$ . Using Maple to solve this equation, we get six possible solutions. One solution has $t_1=t_2$ , which we are not considering. A second solution has $t_1=1/2-t_2$ , which is impossible in integers $t_1$ and $t_2$ . The remaining four solutions contain the expression $\sqrt {-255+1944t_2-3888t_2^2}$ . Thus, for one of these solutions to be viable, $-255+1944t_2-3888t_2^2$ must be a perfect square. However, it is easy to see that $-3888x^2+1944x-255<0$ for all $x\in {\mathbb R}$ , and this fact completes the proof of item (3c) for the case of $A_4$ .

3.4. $\mathbf {S}_{\mathbf {4}}$

For item (4a), $f_t(x)$ is irreducible over ${\mathbb Q}$ for all $t\in {\mathbb Z}$ since $f_t(x)$ is 2-Eisenstein. Using Maple, we calculate

$$ \begin{align*} \Delta(f_t)=16(4t+1)(4t-7)(64t+13). \end{align*} $$

Since $\gcd (4t+1,4t-7)=1$ with $4t+1$ and $4t-7$ squarefree, it follows that $(4t+1)(4t-7)$ divides $64t+13$ if $\Delta (f_t)$ is a square. Observe then that $64t+13=16(4t+1)-3$ implies that $4t+1$ divides 3, from which we conclude that $t\in \{-1,0\}$ . However, $64t+13=16(4t-7)+5^3$ implies that $4t-7$ divides $5$ , which in turn implies that $t\in \{2,3\}$ . This impossibility shows that $\Delta (f_t)$ is not a square. The cubic resolvent of $f_t(x)$ is

$$ \begin{align*}r_t(x)=x^3+2x^2-4(4t+1)x-12(4t+1).\end{align*} $$

If $r_t(x)$ is reducible over ${\mathbb Q}$ , then

$$ \begin{align*}r_t(x)=(x+z)(x^2+Ax+B)=x^3+(A+z)x^2+(Az+B)x+Bz\end{align*} $$

for some integers $A,B,z$ . Equating coefficients,

$$ \begin{align*}A+z=2, \quad Az+B=-16t-4 \quad \mbox{and} \quad Bz=-48t-12.\end{align*} $$

Thus,

$$ \begin{align*}B:=-\dfrac{3z(z-2)}{z-3}=-3(z+1)-\dfrac{9}{z-3}\in {\mathbb Z},\end{align*} $$

which implies that $z-3$ divides 9, and hence, $-z\in \{-12,-6,-4,-2,0,6\}$ . However, solving $r_t(-z)=0$ for t for each of these values of $-z$ yields

$$ \begin{align*}t\in \{39/4,\ 11/4,\ 7/4,\ -1/4\},\end{align*} $$

which contradicts the fact that $t\in {\mathbb Z}$ . Therefore, $r_t(x)$ is irreducible for all $t\in {\mathbb Z}$ , and we deduce from Theorem 2.1 that $\mathrm {Gal}(f_t)\simeq S_4$ .

To establish item (4b), we use Theorem 2.3 with $T(x):=f_t(x)$ to check the prime divisors q of $\Delta (f_t)$ . Since the details are similar for any prime divisor q of $\Delta (f_t)$ , we only give details in the case when q divides $4t-7$ . Then, $t\equiv 7/4 \pmod {q}$ and

$$ \begin{align*}T(x)\equiv (x^2-4x+5)(x+1)^2 \pmod{q}.\end{align*} $$

Since $x=-1$ is a zero of $x^2-4x+5$ if and only if $q=5$ , we have three cases to consider:

(3.5) $$ \begin{align} \overline{T}(x)=\left\{\begin{array}{@{}ll} x(x+1)^3 & \mbox{ if } q=5,\\ (x^2-4x+5)(x+1)^2 & \mbox{ if } x^2-4x+5 \text{ is irreducible over } {\mathbb F}_q,\\ (x-A)(x-B)(x+1)^2 & \mbox{ if } x^2-4x+5 \text{ is reducible over } {\mathbb F}_q, \end{array} \right. \end{align} $$

where A and B are integers with $A,B\not \equiv -1 \pmod {q}$ . We see in all of these cases that $h_2(x)=x+1$ so that $\gcd (\overline {h_1},\overline {h_2},\overline {F})$ can be determined by calculating $\overline {F}(-1)$ . Thus, more precisely,

$$ \begin{align*} \gcd(\overline{h_1},\overline{h_2},\overline{F})=1\Longleftrightarrow \overline{F}(-1)\ne 0 \Longleftrightarrow [{\mathbb Z}_K:{\mathbb Z}[\theta]] \mbox{ is not divisible by } q. \end{align*} $$

Straightforward calculations reveal in all cases of (3.5) that

$$ \begin{align*} \overline{F}(-1)=\overline{-\bigg(\dfrac{4t-7}{q}\bigg)}. \end{align*} $$

Consequently, we deduce that

$$ \begin{align*} [{\mathbb Z}_K:{\mathbb Z}[\theta]]\not \equiv 0 \pmod{q} \quad \mbox{if and only if}\quad 4t-7 \mbox{ is squarefree}. \end{align*} $$

For item (4c), we first note that there exist infinitely many integers t such that

$$ \begin{align*}(4t+1)(4t-7)(64t+13)\end{align*} $$

is squarefree [Reference Booker and Browning1]. Hence, there are infinitely many integers t such that $4t+1$ , $4t-7$ and $65t+13$ are simultaneously squarefree. Thus, the set ${\mathcal F}_5$ is infinite. To see that each such value of t generates a distinct field, we proceed as in previous cases, and let Maple solve the equation $\Delta (f_{t_1})=\Delta (f_{t_2})$ . In all solutions given by Maple, other than $t_1=t_2$ , we see that $W(t_2):=-12288t_2^2+10624t_2+19049$ must be a perfect square. It is easy to verify that $W(t_2)<0$ for all integers $t_2\not \in \{0,1\}$ . Since neither $W(0)=19\,049$ nor $W(1)=17\,385$ is a square, the proofs of item (4c) and the theorem are complete.

4.1. $\mathbf {C}_{\mathbf {2}}\times \mathbf {C}_{\mathbf {2}}$

The literature on quartic polynomials with Galois group $C_2\times C_2$ is fairly extensive, with some authors addressing the issue of monogenicity [Reference Chang2, Reference Gras and Tanoé7, Reference Jones9, Reference Nyul14, Reference Williams20]. However, we found only one explicit two-parameter family of monogenic $C_2\times C_2$ -quartic polynomials [Reference Jones9], given by

$$ \begin{align*} {\mathcal F}(x)=x^4+(36rp-1)x^2+1, \end{align*} $$

where $r\ge 3$ and p are primes, such that r is a primitive root modulo 9 and

$$ \begin{align*}(12rp-1)(12rp+1)(36rp-1)(36rp+1)\end{align*} $$

is squarefree. Suppose that ${\mathcal F}(\alpha )=0$ and $f_t(\theta )=0$ . To see that $K\ne L$ , where $K:={\mathbb Q}(\alpha )$ and $L:={\mathbb Q}(\theta )$ , suppose to the contrary that $K=L$ . Since both ${\mathcal F}(x)$ and $f_t(x)$ are monogenic, it follows that

$$ \begin{align*}144(36rp+1)^2(12rp-1)^2=\Delta({\mathcal F})=\Delta(K)=\Delta(f_t)=256(2t-1)^2(2t+1)^2,\end{align*} $$

which is clearly seen to be impossible by examining the power of 2 dividing each side. Hence, $K\ne L$ .

4.2. $\mathbf {D}_{\mathbf {4}}$

Several authors have investigated integral bases for $D_4$ -quartic number fields [Reference Huard, Spearman and Williams8–Reference Kable10, Reference Ledermann and van der Ploeg12], but we found only two (two-parameter) families of monogenic quartic $D_4$ -polynomials [Reference Jones9], given by

$$ \begin{align*} {\mathcal F}^+&(x)=x^4+x^3+(100rp+1)x^2+x+1\quad \mbox{and} \\ &{\mathcal F}^-(x)=x^4-x^3+(100rp+1)x^2-x+1, \end{align*} $$

where $r\ge 3$ and p are primes, such that r is a primitive root modulo 25 and

$$ \begin{align*} (20rp+1)(100rp+1)(80rp-1) \end{align*} $$

is squarefree. Note that, for fixed values of r and p, ${\mathcal F}^+(x)$ and ${\mathcal F}^-(x)$ generate the same field. Since

$$ \begin{align*} \Delta({\mathcal F}^{\pm})=5^3(20rp+1)(100rp+1)(80rp-1)^2 \quad\mbox{and}\quad \Delta(f_t)=-2^9(6t-1)^3(6t+1)^2, \end{align*} $$

an argument similar to that given for $C_2\times C_2$ easily shows that the family ${\mathcal F}^+(x)$ differs from the family given in Theorem 1.1 for $D_4$ .

4.3. $\mathbf {A}_{\mathbf {4}}$

The only monogenic $A_4$ -quartic family of polynomials we could find in the literature is the single-parameter family

$$ \begin{align*} {\mathcal F}_m(x)=x^4+18x^2-4mx+m^2+81, \end{align*} $$

where $m(m^2+108)$ is squarefree [Reference Spearman18]. Note that m is odd. To see that the family ${\mathcal F}_m(x)$ has no overlap with the family presented in this article, we assume for some integers m and t for which ${\mathcal F}_m(x)$ and $f_t(x)$ are respectively monogenic, that $K=L$ , where $K:={\mathbb Q}(\alpha )$ and $L:={\mathbb Q}(\theta )$ with ${\mathcal F}_m(\alpha )=0$ and $f_t(\theta )=0$ . Then,

$$ \begin{align*} 2^8m^2(m^2+108)^2=\Delta({\mathcal F})=\Delta(K)=\Delta(f_t)=2^6(4t-1)^2(108t^2-54t+7)^2. \end{align*} $$

Using Maple to solve this equation, we find that any solution must have

(4.1) $$ \begin{align} z^2=11664m^6+2519424m^4+136048896m^2+1 \end{align} $$

for some integer z. Multiplying both sides of (4.1) by 4 shows that $(x,y)=(36m^2,2z)$ is an integral point on the elliptic curve

(4.2) $$ \begin{align} y^2=x^3+7776x^2+15116544x+4. \end{align} $$

Using Sage to find all integral points $(x,y)$ (with $y\ge 0$ ) on (4.2), we find

$$ \begin{align*}(x,y)\in \{(0,2),(-3888,2),(14281868898720,53973124902433105922)\}.\end{align*} $$

Since $x=36m^2$ with m odd, it is easy to see that none of these points yields a valid solution to (4.1). That is, there is no overlap with these two families.

4.4. $\mathbf {S}_{\mathbf {4}}$

We found two families of monogenic $S_4$ -quartic trinomials in [Reference Smith17] and one family in [Reference Gassert, Smith and Stange5]. The first family in [Reference Smith17] is

(4.3) $$ \begin{align} {\mathcal F}_{b}(x)=x^4+bx+b\in {\mathbb Z}[x] \quad \mbox{with } \Delta({\mathcal F}_b)=(256-27b)b^3, \end{align} $$

where $b\not \in \{3,5\}$ , and both b and $256-27b$ are squarefree, and the second is

(4.4) $$ \begin{align} {\mathcal G}_{d}(x)=x^4+x^3+d\in {\mathbb Z}[x] \quad \mbox{with } \Delta({\mathcal G}_d)=(256d-27)d^2, \end{align} $$

where $d\ne -2$ , and both d and $256d-27$ are squarefree. The family in [Reference Gassert, Smith and Stange5] is

(4.5) $$ \begin{align} {\mathcal H}_m(x)=x^4-6x^2-mx-3\in {\mathbb Z}[x] \quad \mbox{with } \Delta({\mathcal H}_m)=-27(m-8)^2(m+8)^2. \end{align} $$

Recall that $f_t(x):=x^4-2x^3-2x^2+6x+4t-2$ , with

$$ \begin{align*} \Delta (f_t)=16(4t+1)(4t-7)(64t+13).\end{align*} $$

The question is whether $f_t(x)$ and ${\mathcal F}_{b}(x)$ (or ${\mathcal G}_d(x)$ or ${\mathcal H}_m(x)$ ) generate the same quartic field for some integers t and b (or d or m). If so, then $K=L$ , where $K={\mathbb Q}(\alpha )$ , $L={\mathbb Q}(\theta )$ , $f_t(\theta )=0$ and ${\mathcal F}_{b}(\alpha )=0$ (or ${\mathcal G}_{d}(\alpha )=0$ or ${\mathcal H}_m(\alpha )=0$ ). More importantly in this situation, since all of these polynomials are monogenic, their respective discriminants must be equal. In each case, we assume that these polynomial discriminants are equal and proceed towards a contradiction.

We begin with ${\mathcal F}_b(x)$ in (4.3). Here we are assuming that

$$ \begin{align*} (256-27b)b^3=16(4t+1)(4t-7)(64t+13) \end{align*} $$

for some integers b and t. Since $4t+1$ , $4t-7$ and $64t+13$ are all odd and squarefree, it follows that $b=2$ . Then, Maple tells us that $P(t)=0$ , where

$$ \begin{align*} P(x):=128x^3-166x^2-95x-24. \end{align*} $$

Since $\Delta (P)<0$ , we know that $P(x)$ has exactly one real zero, and since $P(1)<0$ while $P(2)>0$ , it follows that $P(t)$ has no integer zero, which contradicts the fact that $t\in {\mathbb Z}$ .

For ${\mathcal G}_d(x)$ in (4.4), a similar argument shows that $d=4$ and $P(t)=0$ , where

$$ \begin{align*} P(x):=128x^3-166x^2-95x-136. \end{align*} $$

As before, $P(x)$ has exactly one real noninteger zero between 1 and 2, which contradicts the fact that $t\in {\mathbb Z}$ .

Finally, we address ${\mathcal H}_m(x)$ in (4.5). We assume that

(4.6) $$ \begin{align} -27(m-8)^2(m+8)^2=16(4t+1)(4t-7)(64t+13) \end{align} $$

for some integers m and t. An examination of each of the factors $A:=4t+1$ , $B:=4t-7$ and $C:=64t+13$ modulo 3 produces

$$ \begin{align*} [A \ \mathrm{mod}\ 3,B \ \mathrm{mod}\ 3,C\ \mathrm{mod}\ 3]=\left\{\begin{array}{@{}cl} [1,2,1] & \mbox{if}\ t\equiv 0\ \mathrm{mod}\ 3,\\{}[2,0,2] & \mbox{if}\ t\equiv 1\ \mathrm{mod}\ 3,\\{}[0,1,0] & \mbox{if}\ t\equiv 2\ \mathrm{mod}\ 3. \end{array} \right. \end{align*} $$

Thus, since A, B and C are squarefree, it follows that $ABC\not \equiv 0 \pmod {27}$ , and so there are no solutions to (4.6).