# Show that $K$ can be generated over $F$ by the root of an irreducible quartic polynomial of the form $x^4+bx^2+c$.

I am attempting to solve Artin problem 16.6.3. The chapter is on Galois extensions, so I assume the tools of Galois theory will be key to the solution.

Let $K\supset L \supset F$ be a chain of extension fields of degree 2. Assume $char(F)=0$. Show that $K$ can be generated over $F$ by the root of an irreducible quartic polynomial of the form $x^4+bx^2+c$.

I take it that $[K:L]=[L:F]=2$, implying that $[K:F]=4$. Since $[K:F]$ is finite and $char(F)=0$, we have by Primitive Element Theorem that $K=F(α)$ for some $α\in F$.

Therefore, $α$ is the root of some monic, degree 4 polynomial $f(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0 \in F[x]$.

If I can show that $a_3=a_1=0$, I will be done, but how can I show this? Or if it is impossible to show this, is there another way I can solve the problem?

Let $\alpha$ be a primitive element for $K/L$ with the property that $\alpha^2 \in L$, and let $\beta$ be a primitive element for $L/F$ with the property that $\beta^2 \in F$. Then there exist $a,b \in F$ such that $\alpha^2 = a + b \beta$.
First, suppose $b \neq 0$. Then simply adjoining $\alpha$ to $F$ gives the full extension $K/F$. The minimal polynomial of $\alpha$ is $(x^2 - a)^2 - (b\beta)^2$, which has the desired form.
Next, suppose $b = 0$. The extension $K/F$ is generated by $\alpha + t \beta$ for some $t \in F$ by the proof of the Primitive Element Theorem. The minimal polynomial of $\alpha + t\beta$ over $F$ is $(x^2 - \alpha^2 - (t\beta)^2)^2 - 4t^2 \alpha^2\beta^2$, which has the desired form.
• Does every primitive element for $K/L$ have the property that its square is in $L$? May 16 '18 at 20:15
• No; for example, $\mathbb{Q}(1 + \sqrt{2}) = \mathbb{Q}(\sqrt{2})$, but $(1+ \sqrt{2})^2 \not\in \mathbb{Q}$. But in the characteristic $0$ setting, you can always find such a primitive element in a quadratic extension. (I think this actually works as long as the characteristic is not $2$.) May 16 '18 at 21:03
• How do you find such a primitive element? I mean, say $K=L(\alpha_1)$ but $\alpha_1^2$ is not necessarily in $L$. From this $\alpha_1$, can I construct an $\alpha$ such that $K=L(\alpha)$ AND $\alpha^2 \in L$? May 16 '18 at 22:05
• The minimal polynomial of $\alpha_1$ is something of the form $x^2 + cx + d$ for $c,d \in L$. Then $\alpha_1 = \frac{-c \pm \sqrt{c^2 - 4d}}{2}$ by the quadratic formula. So $\sqrt{c^2 - 4d}$ is a primitive element. May 16 '18 at 22:25
• In the case where the minimal polynomial of $\alpha_1$ over $L$ is $x^2 + cx + d$, then $\alpha_1 - c/2$ will also be an element that generates $K$ whose square is in $L$. May 11 '20 at 16:38