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What is the significance of considering characteristic $p$ dividing $n(n-1)$, in finding the Galois group of a equation of degree $n$
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I read in this paper
by K. Uchida, that, in finding Galois group of an equation $X^n-aX+b=0$, in theorem 1, author considers characteristic $p$ is not a divisor of $n(n-1)$.
I did not understand, this condition, p not dividing $n(n-1)$. Can any one explain this, Thank you very much.
group-theory galois-theory
asked Jul 16 at 4:05
user86925
615417
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up vote
0
down vote
favorite
I read in this paper
by K. Uchida, that, in finding Galois group of an equation $X^n-aX+b=0$, in theorem 1, author considers characteristic $p$ is not a divisor of $n(n-1)$.
I did not understand, this condition, p not dividing $n(n-1)$. Can any one explain this, Thank you very much.
group-theory galois-theory
asked Jul 16 at 4:05
user86925
615417
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I read in this paper
by K. Uchida, that, in finding Galois group of an equation $X^n-aX+b=0$, in theorem 1, author considers characteristic $p$ is not a divisor of $n(n-1)$.
I did not understand, this condition, p not dividing $n(n-1)$. Can any one explain this, Thank you very much.
group-theory galois-theory
asked Jul 16 at 4:05
user86925
615417
I read in this paper
by K. Uchida, that, in finding Galois group of an equation $X^n-aX+b=0$, in theorem 1, author considers characteristic $p$ is not a divisor of $n(n-1)$.
I did not understand, this condition, p not dividing $n(n-1)$. Can any one explain this, Thank you very much.
group-theory galois-theory
asked Jul 16 at 4:05
user86925
615417
edited Jul 16 at 4:18
asked Jul 16 at 4:05
user86925
615417
asked Jul 16 at 4:05
user86925
615417
asked Jul 16 at 4:05
user86925
615417
615417
add a comment |Â
add a comment |Â
1 Answer
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Since the characteristic $p$ is prime, it means we have both $pnotmid n$ and $pnotmid (n-1)$, by Euclid's lemma.
Note, the author explains that other than in this situation his results have been limited... (to the cases $n=p^m$ and $n=p^m+1$).
Note also that a large class of rings (namely those with no nontrivial zero divisors) have characteristic $0$ or prime...
To find additional details, read the paper...
answered Jul 16 at 5:42
Chris Custer
5,4582622
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since the characteristic $p$ is prime, it means we have both $pnotmid n$ and $pnotmid (n-1)$, by Euclid's lemma.
Note, the author explains that other than in this situation his results have been limited... (to the cases $n=p^m$ and $n=p^m+1$).
Note also that a large class of rings (namely those with no nontrivial zero divisors) have characteristic $0$ or prime...
To find additional details, read the paper...
answered Jul 16 at 5:42
Chris Custer
5,4582622
add a comment |Â
up vote
1
down vote
accepted
Since the characteristic $p$ is prime, it means we have both $pnotmid n$ and $pnotmid (n-1)$, by Euclid's lemma.
Note, the author explains that other than in this situation his results have been limited... (to the cases $n=p^m$ and $n=p^m+1$).
Note also that a large class of rings (namely those with no nontrivial zero divisors) have characteristic $0$ or prime...
To find additional details, read the paper...
answered Jul 16 at 5:42
Chris Custer
5,4582622
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since the characteristic $p$ is prime, it means we have both $pnotmid n$ and $pnotmid (n-1)$, by Euclid's lemma.
Note, the author explains that other than in this situation his results have been limited... (to the cases $n=p^m$ and $n=p^m+1$).
Note also that a large class of rings (namely those with no nontrivial zero divisors) have characteristic $0$ or prime...
To find additional details, read the paper...
answered Jul 16 at 5:42
Chris Custer
5,4582622
Since the characteristic $p$ is prime, it means we have both $pnotmid n$ and $pnotmid (n-1)$, by Euclid's lemma.
Note, the author explains that other than in this situation his results have been limited... (to the cases $n=p^m$ and $n=p^m+1$).
Note also that a large class of rings (namely those with no nontrivial zero divisors) have characteristic $0$ or prime...
To find additional details, read the paper...
answered Jul 16 at 5:42
Chris Custer
5,4582622
edited Jul 16 at 5:53
answered Jul 16 at 5:42
Chris Custer
5,4582622
answered Jul 16 at 5:42
Chris Custer
5,4582622
answered Jul 16 at 5:42
Chris Custer
5,4582622
5,4582622
add a comment |Â
add a comment |Â
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