Galois group of $prod_{i=1}^{p-1} (X^2-i)$
$begingroup$
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
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
$endgroup$
add a comment |
$begingroup$
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
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
$endgroup$
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
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
edited Jan 9 at 20:04
Jyrki Lahtonen
110k13171378
110k13171378
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
asked Dec 30 '16 at 17:40
RubertosRubertos
5,7372824
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $Z_2$.
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
$endgroup$
add a comment |
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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 $Z_2$.
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
$endgroup$
add a comment |
$begingroup$
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 $Z_2$.
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
$endgroup$
add a comment |
$begingroup$
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 $Z_2$.
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
$endgroup$
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 $Z_2$.
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
answered Dec 30 '16 at 17:58
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7021127
3,7021127
add a comment |
add a comment |
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$begingroup$
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