(b) Suppose $f(x)$ is irreducible and let $G$ be the Galois group of $f(x)$. Prove that
(i) $G≅V$, the Klein 4-group, if and only if $\sqrt{b}∈\mathbb{Q}$ if and only if $αβ∈ℚ$.
(ii) $G≅C$, the cyclic group of order $4$, if and only if $\sqrt{b(a^2-4b)}∈\mathbb{Q}$ if and only if $\mathbb{Q}(αβ)=\mathbb{Q}(α^2)$.
(iii) $G≅D_8$, the dihedral group of order $8$, if and only if $\sqrt{b},\sqrt{b(a^2-4b)}∉ℚ$ if and only if $αβ∉ℚ(α^2)$.
Proof: (a) Note that if $f(α)=0$ then $f(-α)=0$ so it makes sense to refer to the roots as above. ($⇒$) Note that $x^2-α^2~|~f(x)$ so $α^2∉ℚ$. As well, the Galois group is transitive on roots so let $φ(α)=-α$. Then we have $φ(β)=±β$ and thus $φ(α±β)=-α±β$ so either $-α+β=α+β$ in which case $α=0$, or $-α-β=α+β$ in which case $α=-β$ so $f(x)$ is inseparable and hence not irreducible, or $-α+β=α-β$ in which case $α=β$ a contradiction for the same reason, or $-α-β=α-β$ implying $α=0$. Hence $α±β∉ℚ$. ($⇐$) If $b=0$ then either $α=0$ or $α^2∈ℚ$. Note that $α^2=b/β^2$ so also $β^2∉ℚ$. Since neither of $α$ or $β$ are rational we must have $f(x)$ is either irreducible or a product of irreducible quadratics. In the latter case, write $x^4+ax^2+b=(x^2+px+q)(x^2+rx+s)$. Since $p=0$ implies $α^2$ or $β^2$ is rational, simple algebra will show $r=-p$ and $s=q$. Thus the four roots are $\dfrac{±p±\sqrt{p^2-4q}}{2}$ and $α±β$ is rational.
(b)(i) We find that the resolvent cubic is $x^3-2ax^2+(a^2-4b)x=x(x-a-2\sqrt{b})(x-a+2\sqrt{b})$ hence the equivalence is clear after seeing $αβ=\sqrt{b}$.
(ii) ($1⇒2$) Note now that $D=16b(a^2-4b)^2$. By the hypothesis and the procedure for Galois groups $f(x)$ is reducible in $ℚ(\sqrt{D})=ℚ(\sqrt{b})$. Since $f(x)$ is irreducible over $ℚ[x]$, any root generates a fourth degree extension over $ℚ$ and hence is not contained in $ℚ(\sqrt{b})$ so $f(x)$ splits into two irreducible quadratics over $ℚ(\sqrt{b})[x]$. Writing this factoring generally as a system of algebraic equations, we discover either$$x^4+ax^2+b=(x^2+px+\dfrac{p^2+a}{2})(x^2-px+\dfrac{p^2+a}{2}),~~~0≠p∈ℚ(\sqrt{b})$$or$$x^4+ax^2+b=(x^2+p)(x^2+q)$$In the latter case, we thus have $x^2+ax+b$ factoring in $ℚ(\sqrt{b})$ so since $x^2+ax+b$ doesn't factor in $ℚ[x]$ hence $\sqrt{a^2-4b}∉ℚ$ we observe $ℚ(\sqrt{b})=ℚ(\sqrt{a^2-4b})$. Therefore $b$ and $a^2-4b$ differ in a square, i.e. $bz^2=a^2-4b$ and $b(a^2-4b)$ is a square.
In the former case, we have some solution $(\dfrac{x^2+a}{2})^2=b$ in $ℚ(\sqrt{b})[x]$ so $x^2+a±2\sqrt{b}$ splits in $ℚ(\sqrt{b})[x]$. Writing the solution $x=v_1+v_2\sqrt{b}$ we see $$v_1^2+bv_2^2=-a$$$$2v_1v_2=±2$$so $v_2=±1/v_1$ and substituting into the first equation we see $f(v_1)=0$, so $f(x)$ has a linear factor, a contradiction.
($1⇐2$) Suppose $z=\sqrt{b(a^2-4b)}∈ℚ$. Then $z/\sqrt{b}=\sqrt{a^2-4b}$ so $ℚ(z)=ℚ(\sqrt{b})$. We see $x^2+ax+b$ is reducible in $ℚ(\sqrt{a^2-4b})$, so $x^4+ax^2+b$ is reducible in $ℚ(\sqrt{b})$. Since $ℚ(\sqrt{D})=ℚ(\sqrt{b})$ and $\sqrt{b}∉ℚ$ (else $\sqrt{a^2-4b}∈ℚ$ and $f(x)∈ℚ[x]$ is reducible), we have the resolvent cubic (above) splits into a linear factor and an irreducible quadratic, and $f(x)$ is reducible in $ℚ(\sqrt{D})$, so by the procedure the Galois group is $C$.
($1,2⇒3$) Since $α^2$ is a root of irreducible $x^2+ax+b$ and $αβ=\sqrt{b}$ is a root of irreducible (above) $x^2-b$, we have $α^2$ generates $\sqrt{a^2-4b}$ over $ℚ$ and $αβ$ generates $\sqrt{b}$ over $ℚ$, by (2) the implication follows.
($1,2⇐3$) Since $α^2∉ℚ$ by (a), and $ℚ(α^2)=ℚ(αβ)=ℚ(\sqrt{b})$, we have $b$ is not a square hence the resolvent cubic splits into a linear factor and irreducible quadratic, and since $ℚ(\sqrt{D})=ℚ(\sqrt{b})=ℚ(αβ)=ℚ(α^2)$ as well as $f(x)=(x^2-α^2)(x^2-β^2)$ we observe $f(x)$ is reducible in $ℚ(\sqrt{D})[x]$, so by the procedure the Galois group is $C$.
(iii) This is evident by exhaustion of the previously characterized Galois groups.$~~\square$
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