There exist $y \in K$ and $b \in E \setminus F$ such that $\min(y , E) = X^3 - b$ if $\min(z , E) = X^3 - a$ for some $z \in K$ and $a \in E$

Question

Let $F \leq E \leq K$ a chain of fields, such that the characteristic of $F$ is neither $2$ nor $3$, and we suppose that we have found $z \in K$, with $K = E(z)$, and $a \in E$ such that $\min(z , E) = X^3 - a$. We suppose also that $E = F(x)$, with $[E : F] = 2$. Can we take $a$ such that $a \notin F$? In other words, can we take $y \in K$ and $b \in E \setminus F$ such that $\min(y , E) = X^3 - b$? As $a \in E$, it is not difficult take $b \in E \setminus F$: we know that there exist $\alpha , \beta \in F$ such that $a^2 = \alpha + \beta x$ because $a^2 \in E = F(x)$, being $\{1 , x\}$ a basis for $E$ as $F$-vector space. Thus $b = \frac{\beta}{2} - x$ makes sense (because the characteristic of $F$ is not $2$) and satisfies $b \notin F$. My problem now is that we do not have in general the equality $X^3 - b = \min(z , E)$, so I need to replace $z$ by a new $y \in K$, but how can I take $y \in K$ with this property and such that $K = E(y)$?

Answer 1

Yes. We can always replace $a$ with an element $b$ from $E\setminus F$ such that $K/E$ is generated by a zero of the polynomial $X^3-b$.

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Assume that $\min(z,E)=X^3-a$. For all $c\in E, c\neq0$, we have $\min(cz,E)=X^3-c^3a$ as well as $K=E(cz)$. Therefore it suffices to show that we can choose $c$ in such a way that $c^3a\notin F$.

• If $a\notin F$ there is nothing to do. Or, if you wish, we can use $c=1$.
• So we can assume that $a\in F$. Because we are not in characteristic two, there exists an element $d\in F$ such that $E=F(e)$ where $e^2=d$ (the quadratic formula!). It follows that $c=e$ does not have its cube in $F$. This is beacuse $c^3=e^3=de$ cannot be in $F$ because then $e$ would be in $F$ as well. Therefore $c^3a\notin F$ and we are done.