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Consider a field $F$ of characteristic zero. Let $L=F[\alpha]$ be an extension of degree $d.$ We call an element $$ x=x_0 + x_1 \alpha +\ldots+ x_{d-1}\alpha^{d-1}\in L $$ short if $x_{d-1}=0.$ Under which conditions on $\alpha$ every element in $L^\times$ is a product of short elements?

It is easy to see that every element of $L$ is a quotient of two short elements for $d\geq 3$.

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  • $\begingroup$ Oh, thanks, indeed! $\endgroup$ Jan 11, 2021 at 19:12
  • $\begingroup$ It seems that the definition of short depends on $\alpha$, not only on $F$ and $L$. $\endgroup$
    – YCor
    Jan 11, 2021 at 19:13
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    $\begingroup$ Please do a favor to an aged mathematician with failing eyesight by not using $\alpha$ and $a$ in the same formula. $\endgroup$
    – Lubin
    Jan 11, 2021 at 21:20
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    $\begingroup$ Do you know if the following stronger variant fails? If $d\geq 2$, then any element of $L^\times$ is a product of elements of the form $\alpha-a,a\in F$. $\endgroup$
    – Wojowu
    Jan 11, 2021 at 23:29
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    $\begingroup$ @Wojowu Here is a counterexample (found with the help of Pari/GP): consider $L=\mathbb{Q}(\zeta_8)$. Then $x=1+\zeta_8-\zeta_8^2$ has norm 9, while an element $a+b\zeta_8$ with $a,b \in \mathbb{Q}$ has norm $a^4+b^4$ and it is easy to see that the $3$-adic valuation of $a^4+b^4$ is divisible by 4. $\endgroup$ Jan 16, 2021 at 14:20

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