Galois group of $prod_{i=1}^{p-1} (X^2-i)$












2












$begingroup$


Let $p$ be an odd prime. How do I compute the Galois group of $F/Z_p$ where $F$ is the splitting field of $prod_{i=1}^{p-1} (X^2-i)$ over $Z_p$?



Since exactly $frac{p-1}{2}$ integers from $1leq ileq p-1$ are the roots of $X^2-i$ in $Z_p$, let's write $F=Z_p(sqrt{i_1},...,sqrt{i_k})$ where $k=frac{p-1}{2}$ and $sqrt{i_j} notin Z_p$.



This leads me to guess that the Galois group is $(Z_2)^k$, but is it really true? How do I prove this?






abstract-algebra field-theory galois-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
    $endgroup$
    – reuns
    Dec 30 '16 at 18:31












  • $begingroup$
    Edited the title to match the question body as well as the answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 20:05
















2












$begingroup$


Let $p$ be an odd prime. How do I compute the Galois group of $F/Z_p$ where $F$ is the splitting field of $prod_{i=1}^{p-1} (X^2-i)$ over $Z_p$?



Since exactly $frac{p-1}{2}$ integers from $1leq ileq p-1$ are the roots of $X^2-i$ in $Z_p$, let's write $F=Z_p(sqrt{i_1},...,sqrt{i_k})$ where $k=frac{p-1}{2}$ and $sqrt{i_j} notin Z_p$.



This leads me to guess that the Galois group is $(Z_2)^k$, but is it really true? How do I prove this?






abstract-algebra field-theory galois-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
    $endgroup$
    – reuns
    Dec 30 '16 at 18:31












  • $begingroup$
    Edited the title to match the question body as well as the answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 20:05














2












2








2





$begingroup$


Let $p$ be an odd prime. How do I compute the Galois group of $F/Z_p$ where $F$ is the splitting field of $prod_{i=1}^{p-1} (X^2-i)$ over $Z_p$?



Since exactly $frac{p-1}{2}$ integers from $1leq ileq p-1$ are the roots of $X^2-i$ in $Z_p$, let's write $F=Z_p(sqrt{i_1},...,sqrt{i_k})$ where $k=frac{p-1}{2}$ and $sqrt{i_j} notin Z_p$.



This leads me to guess that the Galois group is $(Z_2)^k$, but is it really true? How do I prove this?






abstract-algebra field-theory galois-theory







share|cite|improve this question











$endgroup$




Let $p$ be an odd prime. How do I compute the Galois group of $F/Z_p$ where $F$ is the splitting field of $prod_{i=1}^{p-1} (X^2-i)$ over $Z_p$?



Since exactly $frac{p-1}{2}$ integers from $1leq ileq p-1$ are the roots of $X^2-i$ in $Z_p$, let's write $F=Z_p(sqrt{i_1},...,sqrt{i_k})$ where $k=frac{p-1}{2}$ and $sqrt{i_j} notin Z_p$.



This leads me to guess that the Galois group is $(Z_2)^k$, but is it really true? How do I prove this?






abstract-algebra field-theory galois-theory




abstract-algebra field-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 20:04









Jyrki Lahtonen

110k13171378




110k13171378










asked Dec 30 '16 at 17:40









RubertosRubertos

5,7372824




5,7372824








  • 2




    $begingroup$
    The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
    $endgroup$
    – reuns
    Dec 30 '16 at 18:31












  • $begingroup$
    Edited the title to match the question body as well as the answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 20:05














  • 2




    $begingroup$
    The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
    $endgroup$
    – reuns
    Dec 30 '16 at 18:31












  • $begingroup$
    Edited the title to match the question body as well as the answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 20:05








2




2




$begingroup$
The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
$endgroup$
– reuns
Dec 30 '16 at 18:31






$begingroup$
The notation $mathrm{F}_p$ is preferred ($Z_p$ means the $p$-adic integers)
$endgroup$
– reuns
Dec 30 '16 at 18:31














$begingroup$
Edited the title to match the question body as well as the answer.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 20:05




$begingroup$
Edited the title to match the question body as well as the answer.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 20:05










1 Answer
1






active

oldest

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2












$begingroup$

You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $ain Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words,
$$Z_p(sqrt{i_1},ldots,sqrt{i_k})=Z_p(sqrt{i_l})$$
for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.






share|cite|improve this answer









$endgroup$













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    $begingroup$

    You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $ain Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words,
    $$Z_p(sqrt{i_1},ldots,sqrt{i_k})=Z_p(sqrt{i_l})$$
    for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $ain Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words,
      $$Z_p(sqrt{i_1},ldots,sqrt{i_k})=Z_p(sqrt{i_l})$$
      for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $ain Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words,
        $$Z_p(sqrt{i_1},ldots,sqrt{i_k})=Z_p(sqrt{i_l})$$
        for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.






        share|cite|improve this answer









        $endgroup$



        You should note that for any non-squares $x$ and $y$ of $Z_p$, there is some $ain Z_p$ such that $x=a^2y$. In this way, once you adjoin to the field the square root of one of such elements, you have adjoin the square root of all the other roots. In other words,
        $$Z_p(sqrt{i_1},ldots,sqrt{i_k})=Z_p(sqrt{i_l})$$
        for any $l$. Hence $F/Z_p$ is cuadratic and the Galois group is just $Z_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '16 at 17:58









        Josué Tonelli-CuetoJosué Tonelli-Cueto

        3,7021127




        3,7021127






























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