Find a chain of fields $F subseteq E subseteq K$, where $E cong F[X]/(X^2 + X + 1)$ and $K cong E[X]/(X^3 -...












2












$begingroup$


Let $F subseteq E subseteq K$ 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 $min(z , E)(X) = X^3 - a$.



We suppose also that $E = F(x)$, with $[E : F] = 2$, such that $x$ is a root of the polynomial $X^2 + X + 1$ (so $x$ is a primitive cube root of unity). $K / E$ and $E / F$ have to be Galois extensions. Can we take $a$ such that $a in F$? I only look for an example.



I thought in $F = mathbb{Q}$, $E = F(e^{2 pi i/3})$ and $K = E(sqrt[3]{2})$ and $a = 2$, I am not sure if $K / E$ is normal though.






field-theory galois-theory examples-counterexamples







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$endgroup$








  • 2




    $begingroup$
    If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
    $endgroup$
    – Servaes
    Aug 14 '18 at 20:31


















2












$begingroup$


Let $F subseteq E subseteq K$ 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 $min(z , E)(X) = X^3 - a$.



We suppose also that $E = F(x)$, with $[E : F] = 2$, such that $x$ is a root of the polynomial $X^2 + X + 1$ (so $x$ is a primitive cube root of unity). $K / E$ and $E / F$ have to be Galois extensions. Can we take $a$ such that $a in F$? I only look for an example.



I thought in $F = mathbb{Q}$, $E = F(e^{2 pi i/3})$ and $K = E(sqrt[3]{2})$ and $a = 2$, I am not sure if $K / E$ is normal though.






field-theory galois-theory examples-counterexamples







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
    $endgroup$
    – Servaes
    Aug 14 '18 at 20:31
















2












2








2


0



$begingroup$


Let $F subseteq E subseteq K$ 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 $min(z , E)(X) = X^3 - a$.



We suppose also that $E = F(x)$, with $[E : F] = 2$, such that $x$ is a root of the polynomial $X^2 + X + 1$ (so $x$ is a primitive cube root of unity). $K / E$ and $E / F$ have to be Galois extensions. Can we take $a$ such that $a in F$? I only look for an example.



I thought in $F = mathbb{Q}$, $E = F(e^{2 pi i/3})$ and $K = E(sqrt[3]{2})$ and $a = 2$, I am not sure if $K / E$ is normal though.






field-theory galois-theory examples-counterexamples







share|cite|improve this question











$endgroup$




Let $F subseteq E subseteq K$ 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 $min(z , E)(X) = X^3 - a$.



We suppose also that $E = F(x)$, with $[E : F] = 2$, such that $x$ is a root of the polynomial $X^2 + X + 1$ (so $x$ is a primitive cube root of unity). $K / E$ and $E / F$ have to be Galois extensions. Can we take $a$ such that $a in F$? I only look for an example.



I thought in $F = mathbb{Q}$, $E = F(e^{2 pi i/3})$ and $K = E(sqrt[3]{2})$ and $a = 2$, I am not sure if $K / E$ is normal though.






field-theory galois-theory examples-counterexamples




field-theory galois-theory examples-counterexamples






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 13:34









Brahadeesh

6,36442363




6,36442363










asked Aug 14 '18 at 20:18









joseabp91joseabp91

1,248411




1,248411








  • 2




    $begingroup$
    If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
    $endgroup$
    – Servaes
    Aug 14 '18 at 20:31
















  • 2




    $begingroup$
    If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
    $endgroup$
    – Servaes
    Aug 14 '18 at 20:31










2




2




$begingroup$
If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
$endgroup$
– Servaes
Aug 14 '18 at 20:31






$begingroup$
If $z$ is a root of $X^3-a$, then so are $xz$ and $x^2z$, so $K/E$ is necessarily normal.
$endgroup$
– Servaes
Aug 14 '18 at 20:31












1 Answer
1






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oldest

votes


















1












$begingroup$

In general you cannot take $ain F$; your example is close to a counterexample:



Take $F = mathbb{Q}$ and $E = Fleft(e^{frac{2 pi i}{3}}right)$, and
$$K=Eleft(e^{frac{2 pi i}{9}}right).$$
Then we may take $z=e^{frac{2 pi i}{9}}$, but $min(z,E)=X^3-e^{frac{2 pi i}{3}}$.



In fact, for any choice of non-cube $ainmathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=mathbb{Q}(e^{frac{2 pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=mathbb{Q}(e^{frac{2 pi i}{9}})$ has Galois group $mathbb{Z}/6mathbb{Z}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean? Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 20:37










  • $begingroup$
    But this $a$ does not belong to $F$. Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 21:49










  • $begingroup$
    No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 0:27










  • $begingroup$
    Why cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 15 '18 at 11:28










  • $begingroup$
    Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 14:08













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1 Answer
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1 Answer
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active

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active

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1












$begingroup$

In general you cannot take $ain F$; your example is close to a counterexample:



Take $F = mathbb{Q}$ and $E = Fleft(e^{frac{2 pi i}{3}}right)$, and
$$K=Eleft(e^{frac{2 pi i}{9}}right).$$
Then we may take $z=e^{frac{2 pi i}{9}}$, but $min(z,E)=X^3-e^{frac{2 pi i}{3}}$.



In fact, for any choice of non-cube $ainmathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=mathbb{Q}(e^{frac{2 pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=mathbb{Q}(e^{frac{2 pi i}{9}})$ has Galois group $mathbb{Z}/6mathbb{Z}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean? Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 20:37










  • $begingroup$
    But this $a$ does not belong to $F$. Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 21:49










  • $begingroup$
    No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 0:27










  • $begingroup$
    Why cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 15 '18 at 11:28










  • $begingroup$
    Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 14:08


















1












$begingroup$

In general you cannot take $ain F$; your example is close to a counterexample:



Take $F = mathbb{Q}$ and $E = Fleft(e^{frac{2 pi i}{3}}right)$, and
$$K=Eleft(e^{frac{2 pi i}{9}}right).$$
Then we may take $z=e^{frac{2 pi i}{9}}$, but $min(z,E)=X^3-e^{frac{2 pi i}{3}}$.



In fact, for any choice of non-cube $ainmathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=mathbb{Q}(e^{frac{2 pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=mathbb{Q}(e^{frac{2 pi i}{9}})$ has Galois group $mathbb{Z}/6mathbb{Z}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you mean? Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 20:37










  • $begingroup$
    But this $a$ does not belong to $F$. Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 21:49










  • $begingroup$
    No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 0:27










  • $begingroup$
    Why cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 15 '18 at 11:28










  • $begingroup$
    Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 14:08
















1












1








1





$begingroup$

In general you cannot take $ain F$; your example is close to a counterexample:



Take $F = mathbb{Q}$ and $E = Fleft(e^{frac{2 pi i}{3}}right)$, and
$$K=Eleft(e^{frac{2 pi i}{9}}right).$$
Then we may take $z=e^{frac{2 pi i}{9}}$, but $min(z,E)=X^3-e^{frac{2 pi i}{3}}$.



In fact, for any choice of non-cube $ainmathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=mathbb{Q}(e^{frac{2 pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=mathbb{Q}(e^{frac{2 pi i}{9}})$ has Galois group $mathbb{Z}/6mathbb{Z}$.






share|cite|improve this answer











$endgroup$



In general you cannot take $ain F$; your example is close to a counterexample:



Take $F = mathbb{Q}$ and $E = Fleft(e^{frac{2 pi i}{3}}right)$, and
$$K=Eleft(e^{frac{2 pi i}{9}}right).$$
Then we may take $z=e^{frac{2 pi i}{9}}$, but $min(z,E)=X^3-e^{frac{2 pi i}{3}}$.



In fact, for any choice of non-cube $ainmathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=mathbb{Q}(e^{frac{2 pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=mathbb{Q}(e^{frac{2 pi i}{9}})$ has Galois group $mathbb{Z}/6mathbb{Z}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 14 '18 at 20:47

























answered Aug 14 '18 at 20:34









ServaesServaes

25.6k33996




25.6k33996












  • $begingroup$
    What do you mean? Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 20:37










  • $begingroup$
    But this $a$ does not belong to $F$. Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 21:49










  • $begingroup$
    No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 0:27










  • $begingroup$
    Why cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 15 '18 at 11:28










  • $begingroup$
    Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 14:08




















  • $begingroup$
    What do you mean? Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 20:37










  • $begingroup$
    But this $a$ does not belong to $F$. Cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 14 '18 at 21:49










  • $begingroup$
    No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 0:27










  • $begingroup$
    Why cannot I take $a in F$?
    $endgroup$
    – joseabp91
    Aug 15 '18 at 11:28










  • $begingroup$
    Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
    $endgroup$
    – Servaes
    Aug 15 '18 at 14:08


















$begingroup$
What do you mean? Cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 14 '18 at 20:37




$begingroup$
What do you mean? Cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 14 '18 at 20:37












$begingroup$
But this $a$ does not belong to $F$. Cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 14 '18 at 21:49




$begingroup$
But this $a$ does not belong to $F$. Cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 14 '18 at 21:49












$begingroup$
No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
$endgroup$
– Servaes
Aug 15 '18 at 0:27




$begingroup$
No; the latter argument shows that no element from $F$ can give rise to such an extension $K$.
$endgroup$
– Servaes
Aug 15 '18 at 0:27












$begingroup$
Why cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 15 '18 at 11:28




$begingroup$
Why cannot I take $a in F$?
$endgroup$
– joseabp91
Aug 15 '18 at 11:28












$begingroup$
Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
$endgroup$
– Servaes
Aug 15 '18 at 14:08






$begingroup$
Because for any $ain F$ we have $operatorname{Gal}(E[X]/(X^3-a)/F)cong S_3$, whereas $operatorname{Gal}(E[X]/(X^3-x)/F)congBbb{Z}/6Bbb{Z}$. Here $xin E$ is a root of $X^2+X+1$.
$endgroup$
– Servaes
Aug 15 '18 at 14:08




















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